3.11 Engineering physics (A-level only) — revision question pack

12 specification points · notes, questions, answers and worked methods

Checked against AQA 7408 section 3.11. Review basis: the qualification registry sourced from the AQA A-level Physics (7408) specification; registry verification recorded 11 July 2026.

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Answer all questions in the spaces provided.

3.11.1.1 · Concept of moment of inertia

Explanation

Moment of inertia $I$ measures an object's resistance to angular acceleration about a specified axis, so it is the rotational analogue of mass. For a point mass, $I=mr^2$ ; for several point masses, $I=\sum mr^2$ , where $r$ is the perpendicular distance from the axis.
Its unit is $\text{kg m}^2$ .
Both the total mass and its distribution matter: moving mass farther from the axis increases $I$ strongly because distance is squared.
For an extended object, the supplied expression represents the sum over all its small mass elements.
Examiners expect the stated axis to be used and every separate contribution to be included.

Point masses contribute according to their squared perpendicular distance from the rotation axis.

Worked example

Two point masses, $0.40\,\text{kg}$ and $0.25\,\text{kg}$ , are fixed $0.18\,\text{m}$ and $0.32\,\text{m}$ from an axis. Calculate their total moment of inertia.

1.Write one contribution for each mass: $I=\sum mr^2$ .
2.Substitute: $I=(0.40)(0.18)^2+(0.25)(0.32)^2$ .
3.Evaluate: $I=0.01296+0.02560=0.03856\,\text{kg m}^2$ .

Answer: $I=3.9\times10^{-2}\,\text{kg m}^2$ to two significant figures.

Common mistakes

Don't use the diameter or distance between masses instead of each perpendicular distance from the stated axis.
Don't add $mr$ contributions and forget that the distance is squared in $I=\sum mr^2$ .
Don't treat moment of inertia as depending only on total mass and ignore how that mass is distributed.

Exam tip

For a multi-part rotor, calculate and label each component's $I$ before adding the contributions.

Tier 1 · Easy

Two small balancing masses rotate about the same shaft. A $0.30\,\text{kg}$ mass is $0.22\,\text{m}$ from the shaft and a $0.45\,\text{kg}$ mass is $0.12\,\text{m}$ from it. Calculate their combined moment of inertia.
[2 marks]
Total for this question: 2
A $0.50\,\text{kg}$ point mass contributes $0.0180\,\text{kg m}^2$ to the moment of inertia of a rotor. Calculate its perpendicular distance from the rotation axis.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A training flywheel has moment of inertia $0.080\,\text{kg m}^2$ before four identical $0.25\,\text{kg}$ sliders are attached. Determine the total moment of inertia when every slider is fixed $0.30\,\text{m}$ from the axis.
[3 marks]
Total for this question: 3
Two $0.60\,\text{kg}$ trim masses are moved from $0.15\,\text{m}$ to $0.28\,\text{m}$ from a turbine shaft. Determine the increase in the turbine's moment of inertia.
[3 marks]
Total for this question: 3
A rotor has a central moment of inertia $0.024\,\text{kg m}^2$ . Four $0.18\,\text{kg}$ sensors are each $0.16\,\text{m}$ from the axis and two $0.32\,\text{kg}$ counterweights are each $0.30\,\text{m}$ from it. Determine the percentage of the rotor's total moment of inertia contributed by the counterweights.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A prototype rotor consists of a uniform disc of mass $6.0\,\text{kg}$ and radius $0.25\,\text{m}$ , a shaft with moment of inertia $0.028\,\text{kg m}^2$ , and three identical point masses fixed $0.35\,\text{m}$ from the axis. The total moment of inertia is $0.456\,\text{kg m}^2$ . The disc expression is $I=\frac12MR^2$ . Determine the mass of each point mass.
[5 marks]
Total for this question: 5
A rotor has a central assembly of moment of inertia $0.050\,\text{kg m}^2$ and four $0.30\,\text{kg}$ point masses, each $0.40\,\text{m}$ from the centre along a spoke. The spokes have negligible mass. The angle between each spoke and the rotation axis increases from $30^\circ$ to $60^\circ$ . Determine the percentage increase in the rotor's total moment of inertia.
[5 marks]
Total for this question: 5
A balancing rig has two identical inner point masses at $0.20\,\text{m}$ from its axis and two identical outer point masses at $0.45\,\text{m}$ . The four point masses have total mass $2.40\,\text{kg}$ . A central assembly contributes $0.0415\,\text{kg m}^2$ , and the complete rig has moment of inertia $0.300\,\text{kg m}^2$ . Determine the mass of each inner point mass and each outer point mass.
[5 marks]
Total for this question: 5
Two rotor designs use the same total mass of balance lugs. Design A has a uniform $5.40\,\text{kg}$ disc of radius $0.260\,\text{m}$ , four $0.180\,\text{kg}$ lugs at radius $0.340\,\text{m}$ and a hub of moment of inertia $0.0120\,\text{kg m}^2$ . Design B has a thin $5.40\,\text{kg}$ ring of radius $0.190\,\text{m}$ , six $0.120\,\text{kg}$ lugs at radius $0.300\,\text{m}$ and the same hub. Use $I_{\text{disc}}=\frac12MR^2$ and $I_{\text{ring}}=MR^2$ . Determine which design has the smaller moment of inertia and the percentage by which it is smaller than the other design.
[5 marks]
Total for this question: 5
A rotor has four identical uniform radial arms, each of mass $0.35\,\text{kg}$ and length $0.42\,\text{m}$ . Each arm has moment of inertia $I=\frac13ML^2$ about the shaft. A $0.11\,\text{kg}$ sensor is initially fixed at the end of every arm, and the hub contributes $0.020\,\text{kg m}^2$ . Two sensors are then moved to $0.18\,\text{m}$ from the shaft while the other two remain at the arm ends. Determine the percentage decrease in the rotor's moment of inertia.
[5 marks]
Total for this question: 5

3.11.1.2 · Rotational kinetic energy

Explanation

A rigid rotating object stores kinetic energy $E_k=\frac12I\omega^2$ , directly analogous to translational energy $\frac12mv^2$ . Increasing moment of inertia increases the stored energy in direct proportion, whereas increasing angular speed has a squared effect.
A flywheel therefore stores substantial energy when mass is placed far from its axis and it rotates rapidly.
In machinery, a flywheel absorbs energy when driving torque exceeds demand and releases energy when demand exceeds the drive, smoothing changes in torque and speed.
Flywheels can also store energy in vehicles and production machines.
Calculations require $I$ in $\text{kg m}^2$ and $\omega$ in $\text{rad s}^{-1}$ .

Worked example

A flywheel of moment of inertia $2.8\,\text{kg m}^2$ speeds up from $12\,\text{rad s}^{-1}$ to $30\,\text{rad s}^{-1}$ . Calculate the increase in rotational kinetic energy.

1.Use $\Delta E_k=\frac12I(\omega_2^2-\omega_1^2)$ .
2.Substitute: $\Delta E_k=\frac12(2.8)(30^2-12^2)$ .
3.Evaluate: $\Delta E_k=1.4(756)=1058.4\,\text{J}$ .

Answer: The rotational kinetic energy increases by $1.1\times10^3\,\text{J}$ .

Common mistakes

Don't use $E_k=I\omega^2$ and omit the factor of one half.
Don't substitute revolutions per second directly for $\omega$ instead of converting with $\omega=2\pi f$ .
Don't calculate the change using $(\omega_2-\omega_1)^2$ rather than $\omega_2^2-\omega_1^2$ .

Exam tip

For an energy change, write the final and initial $\frac12I\omega^2$ terms explicitly before subtracting.

Tier 1 · Easy

A small flywheel has moment of inertia $0.85\,\text{kg m}^2$ and angular speed $14\,\text{rad s}^{-1}$ . Calculate its rotational kinetic energy.
[2 marks]
Total for this question: 2
A polishing wheel of moment of inertia $0.64\,\text{kg m}^2$ slows from $25\,\text{rad s}^{-1}$ to $15\,\text{rad s}^{-1}$ . Calculate the rotational kinetic energy released.
[2 marks]
Total for this question: 2

Tier 2 · Standard

An energy-recovery flywheel of moment of inertia $40\,\text{kg m}^2$ stores $32\,\text{kJ}$ as rotational kinetic energy. Determine its angular speed.
[3 marks]
Total for this question: 3
A turbine rotor of moment of inertia $1.6\,\text{kg m}^2$ rotates at $600\,\text{rev min}^{-1}$ . A second rotor at $40\,\text{rad s}^{-1}$ stores the same rotational kinetic energy. Determine the second rotor's moment of inertia.
[3 marks]
Total for this question: 3
A flywheel of moment of inertia $3.5\,\text{kg m}^2$ initially rotates at $900\,\text{rev min}^{-1}$ . A motor transfers $18\,\text{kJ}$ to the flywheel, of which $80\%$ becomes rotational kinetic energy. Determine the final rotation rate in $\text{rev min}^{-1}$ to three significant figures.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A machine flywheel with moment of inertia $18\,\text{kg m}^2$ slows from $75\,\text{rad s}^{-1}$ to $45\,\text{rad s}^{-1}$ in $5.5\,\text{s}$ . During this interval, $72\%$ of the decrease in rotational kinetic energy is transferred usefully. Calculate the mean useful power.
[5 marks]
Total for this question: 5
A flywheel of moment of inertia $5.0\,\text{kg m}^2$ must deliver $1.80\,\text{kJ}$ of useful energy without slowing below $18\,\text{rad s}^{-1}$ . The useful output is $80\%$ of the decrease in the flywheel's rotational kinetic energy. Determine the minimum initial angular speed and the mean useful power if the delivery takes $2.5\,\text{s}$ .
[5 marks]
Total for this question: 5
Two flywheels have the same mass. Flywheel A may be treated as a thin ring with $I=MR^2$ and has maximum safe rim speed $240\,\text{m s}^{-1}$ . Flywheel B may be treated as a solid disc with $I=\frac12MR^2$ and has maximum safe rim speed $300\,\text{m s}^{-1}$ . Determine which flywheel stores more rotational kinetic energy at its maximum safe speed and the percentage by which it stores more.
[5 marks]
Total for this question: 5
A flywheel is fixed to a machine shaft to smooth its angular speed. The motor supplies a constant torque of $47.0\,\text{N m}$ . For $0.750\,\text{s}$ the process load torque is $71.0\,\text{N m}$ , after which it is $35.0\,\text{N m}$ for $1.50\,\text{s}$ . At the start of the high-load interval the angular speed is $56.0\,\text{rad s}^{-1}$ . Determine the minimum flywheel moment of inertia needed to prevent the speed falling below $52.0\,\text{rad s}^{-1}$ , then determine the angular speed at the end of the lower-load interval, and determine the rotational kinetic energy released by the flywheel during the high-load interval. Treat each torque as constant. Assume the moment of inertia of the shaft and machine is negligible compared with that of the flywheel.
[6 marks]
Total for this question: 6
Two independently rotating flywheels held in fixed bearings store a total of $2.10\,\text{kJ}$ . Flywheel A has moment of inertia $0.92\,\text{kg m}^2$ and angular speed $\omega$ . Flywheel B has moment of inertia $0.30\,\text{kg m}^2$ and angular speed $2.40\omega$ . B is brought to rest, and $65.0\%$ of the rotational kinetic energy it releases is transferred to A. Determine the initial value of $\omega$ and A's final angular speed.
[5 marks]
Total for this question: 5

3.11.1.3 · Rotational motion

Explanation

Angular displacement $\theta$ is measured in radians. Angular velocity is $\omega=\Delta\theta/\Delta t$ for a chosen positive direction; angular speed is its magnitude.
Angular acceleration is $\alpha=\Delta\omega/\Delta t$ . For uniform angular acceleration, the translational equations have direct rotational analogues: $\omega_2=\omega_1+\alpha t$ , $\theta=\frac12(\omega_1+\omega_2)t$ , $\theta=\omega_1t+\frac12\alpha t^2$ , and $\omega_2^2=\omega_1^2+2\alpha\theta$ .
On an angular-velocity–time graph, gradient gives angular acceleration and signed area gives angular displacement. A changing gradient represents non-uniform angular acceleration.
A straight graph section therefore represents constant angular acceleration over that interval.
Examiners may require both graphical interpretation and equation selection.

Gradient gives angular acceleration and area under an angular-velocity–time graph gives angular displacement.

Worked example

A rotor accelerates uniformly from $8.0\,\text{rad s}^{-1}$ to $26\,\text{rad s}^{-1}$ in $6.0\,\text{s}$ . Calculate its angular acceleration and angular displacement.

1. $\alpha=(\omega_2-\omega_1)/t=(26-8.0)/6.0=3.0\,\text{rad s}^{-2}$ .
2.For uniform acceleration, $\theta=\frac12(\omega_1+\omega_2)t$ .
3. $\theta=\frac12(8.0+26)(6.0)=102\,\text{rad}$ .

Answer: $\alpha=3.0\,\text{rad s}^{-2}$ and $\theta=1.0\times10^2\,\text{rad}$ .

Common mistakes

Don't use degrees in the rotational equations instead of converting angular displacement to radians.
Don't read the height of an $\omega$ – $t$ graph as angular acceleration instead of using its gradient.
Don't treat rotational frequency in hertz as angular velocity without multiplying by $2\pi$ .

Exam tip

On an $\omega$ – $t$ graph, label gradient and area with their physical quantities before calculating.

Tier 1 · Easy

The angular speed of a mixer rotor rises uniformly from $5.0\,\text{rad s}^{-1}$ to $17\,\text{rad s}^{-1}$ in $4.0\,\text{s}$ . Calculate its angular acceleration.
[2 marks]
Total for this question: 2
A conveyor drum rotates at a constant angular speed of $23\,\text{rad s}^{-1}$ for $4.5\,\text{s}$ . Calculate its angular displacement.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A rotor starts from rest and has constant angular acceleration $4.2\,\text{rad s}^{-2}$ for $6.0\,\text{s}$ . Determine the number of revolutions completed.
[3 marks]
Total for this question: 3
A centrifuge rotating at $36\,\text{rad s}^{-1}$ is brought uniformly to rest with angular acceleration $-4.5\,\text{rad s}^{-2}$ . Determine the angular displacement during braking.
[3 marks]
Total for this question: 3
A rotor's angular velocity changes uniformly from $+24\,\text{rad s}^{-1}$ to $-8.0\,\text{rad s}^{-1}$ in $5.0\,\text{s}$ . It then remains at $-8.0\,\text{rad s}^{-1}$ for $3.0\,\text{s}$ . Determine the rotor's net angular displacement over the $8.0\,\text{s}$ interval.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A flywheel's angular speed rises uniformly from $12\,\text{rad s}^{-1}$ to $48\,\text{rad s}^{-1}$ in $6.0\,\text{s}$ , remains at $48\,\text{rad s}^{-1}$ for $5.0\,\text{s}$ , then falls uniformly to $18\,\text{rad s}^{-1}$ in $10\,\text{s}$ . Determine its total angular displacement and the corresponding number of revolutions.
[5 marks]
Total for this question: 5
A test wheel has initial angular speed $3.0\,\text{rad s}^{-1}$ and completes $18$ revolutions in $8.0\,\text{s}$ with uniform angular acceleration. Determine its angular acceleration and final angular speed. The drive is then removed and the wheel slows uniformly to rest with an angular acceleration of the same magnitude. Calculate the stopping time.
[5 marks]
Total for this question: 5
A turntable starts from rest and accelerates uniformly at $3.2\,\text{rad s}^{-2}$ for a time $t$ . The drive is then removed and the turntable decelerates uniformly at $1.6\,\text{rad s}^{-2}$ until it stops. The total angular displacement is $45$ revolutions. Determine $t$ and the total time for the complete motion.
[5 marks]
Total for this question: 5
A test spindle begins at $7.50\,\text{rad s}^{-1}$ and reaches $25.5\,\text{rad s}^{-1}$ after accelerating uniformly for $4.80\,\text{s}$ . A constant-speed interval follows. The spindle then decelerates uniformly at $2.20\,\text{rad s}^{-2}$ for $7.50\,\text{s}$ . Its total angular displacement across the three stages is $38.0$ revolutions. Determine the duration of the constant-speed interval and the final angular speed.
[5 marks]
Total for this question: 5
A wheel rotates with constant angular acceleration. Between $t=3.0\,\text{s}$ and $t=4.0\,\text{s}$ it turns through $18.0\,\text{rad}$ , and between $t=8.0\,\text{s}$ and $t=9.0\,\text{s}$ it turns through $38.0\,\text{rad}$ . Determine its angular velocity when timing begins, its angular acceleration, and the time from the start of timing at which it first completes $25.0$ revolutions.
[6 marks]
Total for this question: 6

3.11.1.4 · Torque and angular acceleration

Explanation

Torque is the turning effect of a force. For a force perpendicular to the radius, $T=Fr$ , where $r$ is the perpendicular distance from the axis to the force's line of action; torque has unit $\text{N m}$ .
The rotational form of Newton's second law is $T=I\alpha$ , so the resultant torque produces angular acceleration in proportion to $1/I$ .
Driving and resisting torques act in opposite rotational senses and must be combined with signs before applying this equation.
If the force is not perpendicular, the perpendicular force component or perpendicular moment arm is required.
Examiners expect a resultant torque, not merely the applied driving torque.

A tangential force at radius

r

produces torque

T=Fr

about the axis.

Worked example

A motor supplies $36\,\text{N m}$ to a rotor while friction provides $5.0\,\text{N m}$ in the opposite direction. The rotor has $I=4.0\,\text{kg m}^2$ . Calculate its angular acceleration.

1.Find the resultant torque: $T=36-5.0=31\,\text{N m}$ .
2.Use $T=I\alpha$ and rearrange to $\alpha=T/I$ .
3. $\alpha=31/4.0=7.75\,\text{rad s}^{-2}$ .

Answer: $\alpha=7.8\,\text{rad s}^{-2}$ to two significant figures.

Common mistakes

Don't use the distance to the point of application when it is not the perpendicular distance to the force's line of action.
Don't substitute the driving torque into $T=I\alpha$ without subtracting the resisting torque.
Don't substitute force in newtons directly into $T=I\alpha$ without first calculating torque.

Exam tip

Draw a rotational sense beside every torque and form the signed resultant before using $T=I\alpha$ .

Tier 1 · Easy

A tangential force of $85\,\text{N}$ acts at the rim of a wheel of radius $0.16\,\text{m}$ . Calculate the torque about the axle.
[2 marks]
Total for this question: 2
A torque of $12\,\text{N m}$ is produced by a tangential force acting $0.24\,\text{m}$ from an axle. Calculate the force.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A constant resultant torque of $24\,\text{N m}$ acts on a rotor with moment of inertia $3.2\,\text{kg m}^2$ . Its initial angular speed is $6.0\,\text{rad s}^{-1}$ . Determine its angular speed after $3.0\,\text{s}$ .
[3 marks]
Total for this question: 3
A braking force of $30\,\text{N}$ acts tangentially at radius $0.20\,\text{m}$ . The angular speed of a rotor falls uniformly from $15\,\text{rad s}^{-1}$ to $9.0\,\text{rad s}^{-1}$ in $3.0\,\text{s}$ . Assume the braking force provides the only resistive torque. Determine the rotor's moment of inertia.
[3 marks]
Total for this question: 3
A force of $120\,\text{N}$ acts at a point $0.35\,\text{m}$ from a shaft axis. The angle between the force and the radius is $55^\circ$ . A resisting torque of $8.0\,\text{N m}$ also acts. The shaft assembly has moment of inertia $4.5\,\text{kg m}^2$ . Determine its angular acceleration.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A drive supplies a constant torque of $54\,\text{N m}$ to a flywheel of moment of inertia $6.2\,\text{kg m}^2$ . Bearing friction provides a constant opposing torque of $7.5\,\text{N m}$ . The initial angular speed is $10\,\text{rad s}^{-1}$ . Calculate the angular speed after $4.5\,\text{s}$ .
[5 marks]
Total for this question: 5
A hoist motor supplies $70\,\text{N m}$ to a drum of radius $0.25\,\text{m}$ and moment of inertia $8.0\,\text{kg m}^2$ . Bearing friction supplies an opposing torque of $6.0\,\text{N m}$ . The drum has angular acceleration $4.0\,\text{rad s}^{-2}$ as the load rises. Determine the cable tension and the mass of the load. Use $g=9.81\,\text{m s}^{-2}$ .
[5 marks]
Total for this question: 5
A motor's driving torque decreases linearly from $60\,\text{N m}$ at zero angular speed to zero at $120\,\text{rad s}^{-1}$ . It drives a rotor of moment of inertia $2.4\,\text{kg m}^2$ against a tangential load force of $40\,\text{N}$ acting $0.30\,\text{m}$ from the axis. Determine the angular speed when the rotor's angular acceleration is $8.0\,\text{rad s}^{-2}$ .
[5 marks]
Total for this question: 5
A $96\,\text{N}$ driving force acts tangentially at radius $0.28\,\text{m}$ on a rotor. A $70\,\text{N}$ force acts in the opposite rotational sense at radius $0.20\,\text{m}$ , making an angle of $35^\circ$ with the radius. Bearing friction supplies a further opposing torque of $3.6\,\text{N m}$ . The angular speed rises uniformly from $4.0\,\text{rad s}^{-1}$ to $19\,\text{rad s}^{-1}$ in $6.5\,\text{s}$ . Determine the resultant torque and the rotor's moment of inertia.
[5 marks]
Total for this question: 5
A motor rotor of moment of inertia $2.6\,\text{kg m}^2$ is locked by a clutch to a coaxial driven rotor of moment of inertia $1.4\,\text{kg m}^2$ . The motor supplies $48\,\text{N m}$ . Resistive torques of $3.5\,\text{N m}$ and $2.0\,\text{N m}$ act on the motor and driven rotors respectively. Determine their common angular acceleration and the torque transmitted by the clutch to the driven rotor.
[5 marks]
Total for this question: 5

3.11.1.5 · Angular momentum

Explanation

For a rigid body rotating about a fixed axis, angular momentum is $L=I\omega$ , measured in $\text{kg m}^2\text{ s}^{-1}$ . Total angular momentum is conserved when the resultant external torque is zero.
Consequently, reducing $I$ increases $\omega$ so that the signed total $I\omega$ remains constant, as in a spinning athlete drawing in the arms.
A constant torque acting for time $\Delta t$ provides angular impulse $T\Delta t=\Delta(I\omega)$ .
Direction matters: opposite rotations have opposite angular momenta.
When rotating objects couple and move together, angular momentum can be conserved even though rotational kinetic energy decreases, because the coupling is inelastic.

Worked example

A skater has $I=3.6\,\text{kg m}^2$ at $5.0\,\text{rad s}^{-1}$ and reduces the moment of inertia to $2.0\,\text{kg m}^2$ . Calculate the new angular speed when external torque is negligible.

1.State conservation: $I_1\omega_1=I_2\omega_2$ .
2.Substitute: $(3.6)(5.0)=(2.0)\omega_2$ .
3.Rearrange: $\omega_2=18/2.0=9.0\,\text{rad s}^{-1}$ .

Answer: The new angular speed is $9.0\,\text{rad s}^{-1}$ .

Common mistakes

Don't conserve angular momentum despite a non-zero external torque acting during the stated interval.
Don't conserve rotational kinetic energy when two rotating parts lock together in an inelastic coupling.
Don't drop the sign of angular velocity when two bodies initially rotate in opposite directions.

Exam tip

A conservation answer must state that the resultant external torque is zero or negligible.

Tier 1 · Easy

A turntable has moment of inertia $2.4\,\text{kg m}^2$ and rotates at $18\,\text{rad s}^{-1}$ . Calculate its angular momentum.
[2 marks]
Total for this question: 2
A rotating assembly has angular momentum $54\,\text{kg m}^2\text{ s}^{-1}$ at angular speed $12\,\text{rad s}^{-1}$ . Calculate its moment of inertia.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A flywheel with moment of inertia $0.80\,\text{kg m}^2$ rotates at $30\,\text{rad s}^{-1}$ . It is coupled to a stationary coaxial wheel of moment of inertia $1.20\,\text{kg m}^2$ , and the pair then rotate together. Determine their common angular speed.
[3 marks]
Total for this question: 3
A rotor of moment of inertia $2.5\,\text{kg m}^2$ is slowed from $22\,\text{rad s}^{-1}$ to $10\,\text{rad s}^{-1}$ by a constant braking torque of $6.0\,\text{N m}$ . Determine the braking time.
[3 marks]
Total for this question: 3
A rotor of moment of inertia $3.2\,\text{kg m}^2$ initially has angular velocity $-4.0\,\text{rad s}^{-1}$ . A positive torque pulse rises uniformly from zero to $12\,\text{N m}$ and then falls uniformly to zero over a total time of $2.5\,\text{s}$ . The angular impulse is the area under the torque-time graph. Determine the rotor's final angular velocity.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A rotating platform and athlete initially have moment of inertia $4.8\,\text{kg m}^2$ and angular speed $9.0\,\text{rad s}^{-1}$ . An external constant opposing torque of $3.2\,\text{N m}$ acts for $1.5\,\text{s}$ . The athlete then changes position so that the combined moment of inertia is $3.0\,\text{kg m}^2$ . Determine the final angular speed, assuming the position change itself involves negligible external torque.
[5 marks]
Total for this question: 5
A disc of moment of inertia $1.8\,\text{kg m}^2$ rotates at $+20\,\text{rad s}^{-1}$ . It couples to a coaxial disc of moment of inertia $1.2\,\text{kg m}^2$ rotating at $-10\,\text{rad s}^{-1}$ . The discs then rotate together with negligible external torque. Determine the percentage of the initial rotational kinetic energy that is lost.
[5 marks]
Total for this question: 5
A platform has moment of inertia $4.0\,\text{kg m}^2$ . A $0.50\,\text{kg}$ point mass is held $0.80\,\text{m}$ from the axis while the platform rotates at $5.0\,\text{rad s}^{-1}$ . The mass is then thrown tangentially, in the direction opposite to the platform's rotation, at $6.0\,\text{m s}^{-1}$ relative to the platform. For the thrown mass, its angular momentum about the axis is $mrv$ . Immediately after the throw, its tangential velocity component in the laboratory frame is $0.80\omega-6.0$ , where $\omega$ is the platform's new angular speed. Determine $\omega$ , assuming negligible external torque.
[5 marks]
Total for this question: 5
A spacecraft body has moment of inertia $18.0\,\text{kg m}^2$ about its control axis. A coaxial reaction wheel of moment of inertia $0.420\,\text{kg m}^2$ is initially at rest relative to the body. An internal motor then gives the wheel an angular velocity of $+240\,\text{rad s}^{-1}$ relative to the body. Determine the angular velocities of the spacecraft body and the wheel relative to an external inertial frame. Assume the resultant external torque is negligible.
[5 marks]
Total for this question: 5
A turntable of moment of inertia $0.720\,\text{kg m}^2$ rotates at $+8.50\,\text{rad s}^{-1}$ . A $0.0600\,\text{kg}$ pellet travels tangentially in the opposite direction and embeds at radius $0.340\,\text{m}$ . Immediately afterwards the turntable and pellet rotate together at $+5.00\,\text{rad s}^{-1}$ . Determine the pellet's speed before impact and the percentage of the initial kinetic energy converted to other forms. Assume external torque is negligible during the impact.
[6 marks]
Total for this question: 6

3.11.1.6 · Work and power

Explanation

A constant torque acting through angular displacement transfers work $W=T\theta$ , with $\theta$ in radians. The instantaneous rate of energy transfer is rotational power $P=T\omega$ .
These equations parallel $W=Fs$ and $P=Fv$ .
In real rotating machinery, frictional torque removes mechanical energy and must appear in the torque, work or power balance.
At constant angular speed, resultant torque is zero, but the driving torque can still balance load and friction torques and therefore transfer power.
When torque or angular speed varies, $P=T\omega$ uses simultaneous values; total work may instead be found from the energy change or an appropriate torque-angle area.

Worked example

A shaft turns at $900\,\text{rev min}^{-1}$ . It supplies a load torque of $48\,\text{N m}$ and overcomes friction torque $6.0\,\text{N m}$ . Calculate the motor power.

1.The motor torque is $T=48+6.0=54\,\text{N m}$ .
2. $900\,\text{rev min}^{-1}=15\,\text{rev s}^{-1}$ , so $\omega=2\pi(15)=94.2\,\text{rad s}^{-1}$ .
3. $P=T\omega=(54)(94.2)=5.09\times10^3\,\text{W}$ .

Answer: The motor power is $5.1\,\text{kW}$ to two significant figures.

Common mistakes

Don't use angular displacement in revolutions in $W=T\theta$ instead of converting it to radians.
Don't omit frictional torque when finding the driving power required by real machinery.
Don't conclude that power transfer is zero at steady speed because the resultant torque is zero.

Exam tip

Distinguish driving, load and friction torques before selecting the torque used in $P=T\omega$ .

Tier 1 · Easy

A constant torque of $18\,\text{N m}$ turns a shaft through $3.5\,\text{rad}$ . Calculate the work done by the torque.
[2 marks]
Total for this question: 2
A shaft receives $550\,\text{J}$ of work from a constant torque of $24\,\text{N m}$ . Calculate the angular displacement.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A motor provides a torque of $42\,\text{N m}$ while its shaft rotates at $1200\,\text{rev min}^{-1}$ . Determine the power transferred by the shaft.
[3 marks]
Total for this question: 3
A capstan turns through $120\,\text{rad}$ at constant angular speed while a load torque of $35\,\text{N m}$ and a friction torque of $5.0\,\text{N m}$ oppose its rotation. Determine the work supplied by the motor.
[3 marks]
Total for this question: 3
The driving torque on a shaft increases uniformly from $10.0\,\text{N m}$ to $30.0\,\text{N m}$ while the shaft turns through $40\,\text{rad}$ . The torque then remains at $30\,\text{N m}$ for a further $25\,\text{rad}$ . Work is the area under the torque-angle graph. Determine the total work done by the driving torque.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A motor maintains a constant angular speed of $32\,\text{rad s}^{-1}$ for $25\,\text{s}$ while supplying torque $84\,\text{N m}$ . Bearing friction opposes the motion with torque $9.0\,\text{N m}$ ; the remaining torque acts on the load. Calculate the useful power and the useful energy transferred to the load.
[5 marks]
Total for this question: 5
A mixer shaft turns through $80\,\text{rad}$ in $5.0\,\text{s}$ . The mean motor torque over this angular displacement is $70\,\text{N m}$ , while friction provides a constant opposing torque of $8.0\,\text{N m}$ . The shaft has the same rotational kinetic energy at the start and end. Determine the useful work transferred to the mixture and the mean useful power.
[5 marks]
Total for this question: 5
A rotor of moment of inertia $2.5\,\text{kg m}^2$ initially rotates at $8.0\,\text{rad s}^{-1}$ . Over the next $60\,\text{rad}$ , the driving torque decreases uniformly from $50\,\text{N m}$ to $20\,\text{N m}$ while a constant friction torque of $6.0\,\text{N m}$ opposes the motion. Work is the area under the torque-angle graph. Determine the rotor's final angular speed.
[5 marks]
Total for this question: 5
A motor's torque decreases linearly from $80\,\text{N m}$ at zero angular speed to $20\,\text{N m}$ at $120\,\text{rad s}^{-1}$ . The opposing load torque is given by $T_L=5.0+0.250\omega$ , where torque is in $\text{N m}$ and angular speed is in $\text{rad s}^{-1}$ . At steady speed, determine the angular speed, the useful power transferred to the load, and the useful energy transferred in $45\,\text{s}$ .
[5 marks]
Total for this question: 5
A motor applies a constant torque of $72\,\text{N m}$ to a shaft assembly of moment of inertia $4.2\,\text{kg m}^2$ . Its angular speed increases uniformly from $12\,\text{rad s}^{-1}$ to $28\,\text{rad s}^{-1}$ in $8.0\,\text{s}$ . A constant friction torque of $6.5\,\text{N m}$ acts, and the remaining transferred energy goes to a useful load. Determine the useful work and the mean useful power during this interval.
[6 marks]
Total for this question: 6

3.11.2.1 · First law of thermodynamics

Explanation

AQA writes the First Law as $Q=\Delta U+W$ . Here $Q$ is energy transferred to the system by heating, $\Delta U$ is the increase in internal energy, and $W$ is work done by the system.
Thus heating gives positive $Q$ , cooling negative $Q$ , expansion normally positive $W$ , and compression negative $W$ .
Rearranging to $\Delta U=Q-W$ helps preserve this sign convention.
Internal energy is a state function, so its change depends only on the initial and final states.
Over a complete cycle $\Delta U=0$ , although non-zero net heat transfer can equal the net work done by the gas.

Worked example

During compression, $300\,\text{J}$ of work is done on a gas while $80\,\text{J}$ leaves by heating. Calculate $\Delta U$ using $Q=\Delta U+W$ .

1.Energy leaving by heating gives $Q=-80\,\text{J}$ .
2.Work done on the gas means work done by the gas is $W=-300\,\text{J}$ .
3. $\Delta U=Q-W=-80-(-300)=+220\,\text{J}$ .

Answer: The internal energy increases by $220\,\text{J}$ .

Common mistakes

Don't make work done on the gas positive even though $W$ is defined as work done by the gas.
Don't treat energy leaving by heating as positive $Q$ .
Don't assume $\Delta U=0$ for any process rather than only a return to the same thermodynamic state.

Exam tip

Write the signs of $Q$ and $W$ in words before substituting into $\Delta U=Q-W$ .

Tier 1 · Easy

A gas receives $520\,\text{J}$ by heating and does $180\,\text{J}$ of work. Calculate its change in internal energy using AQA's convention $Q=\Delta U+W$ .
[2 marks]
Total for this question: 2
The internal energy of a gas decreases by $120\,\text{J}$ while the gas does $80\,\text{J}$ of work. Calculate the energy transferred to the gas by heating using $Q=\Delta U+W$ .
[2 marks]
Total for this question: 2

Tier 2 · Standard

A gas is compressed so that $240\,\text{J}$ of work is done on it. During the compression, $90\,\text{J}$ of energy leaves the gas by heating. Using $Q=\Delta U+W$ and taking work output from the gas as positive, determine $\Delta U$ .
[3 marks]
Total for this question: 3
During one operation, a gas first does $180\,\text{J}$ of work and later has $80\,\text{J}$ of work done on it. Its total internal energy decreases by $50\,\text{J}$ . Using $Q=\Delta U+W$ , determine the net heat transfer and its direction.
[3 marks]
Total for this question: 3
A gas undergoes two successive processes. In the first, it receives $600\,\text{J}$ by heating and does $250\,\text{J}$ of work. In the second, $180\,\text{J}$ leaves by heating while $90\,\text{J}$ of work is done on the gas. Using $Q=\Delta U+W$ , determine the gas's total change in internal energy.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A gas completes a two-process cycle. In process A it receives $3.2\,\text{kJ}$ by heating and its internal energy increases by $1.1\,\text{kJ}$ . In process B the gas returns to its initial state while $1.4\,\text{kJ}$ leaves it by heating. Determine the net work done by the gas during the cycle. State the signs used in $Q=\Delta U+W$ .
[5 marks]
Total for this question: 5
A gas undergoes three processes. In A, it receives $1.20\,\text{kJ}$ by heating and does $0.45\,\text{kJ}$ of work. In B, $0.30\,\text{kJ}$ leaves by heating and $0.60\,\text{kJ}$ of work is done on the gas. During C, the gas does $0.25\,\text{kJ}$ of work. After C, its internal energy is $0.20\,\text{kJ}$ below its initial value. Determine the heat transfer in C and the net work done by the gas over all three processes.
[5 marks]
Total for this question: 5
A gas can move from state A to state C by either of two routes. On the direct route, it receives $2.40\,\text{kJ}$ by heating and does $0.90\,\text{kJ}$ of work. On the alternative route from A to B, $0.30\,\text{kJ}$ leaves by heating and $0.80\,\text{kJ}$ of work is done on the gas. From B to C, the gas does $0.60\,\text{kJ}$ of work. Determine the heat transfer from B to C, using $Q=\Delta U+W$ .
[5 marks]
Total for this question: 5
A fixed mass of gas passes through states A, B and C before returning to A. Its internal energies at A, B and C are $2.20\,\text{kJ}$ , $3.05\,\text{kJ}$ and $2.55\,\text{kJ}$ respectively. The heat transfers to the gas are $+1.40\,\text{kJ}$ for A to B, $-0.250\,\text{kJ}$ for B to C and $-0.600\,\text{kJ}$ for C to A. Using $Q=\Delta U+W$ , determine the work done by the gas in each process and the net work per cycle.
[6 marks]
Total for this question: 6
A gas completes $18.0$ identical cycles each second. During the expansion part of each cycle it receives $480\,\text{J}$ by heating and does $310\,\text{J}$ of work. During compression, $260\,\text{J}$ of work is done on the gas, which then returns to its initial state. Using $Q=\Delta U+W$ , where $W$ is the work done by the gas, determine the heat transfer during compression, the net mechanical power output and the rate at which energy is rejected by heating.
[5 marks]
Total for this question: 5

3.11.2.2 · Non-flow processes

Explanation

The required non-flow changes are isothermal, adiabatic, constant pressure and constant volume. An ideal gas obeys $pV=nRT$ .
During an isothermal change, temperature is constant and $pV=\text{constant}$ ; for an ideal gas, $\Delta U=0$ , so $Q=W$ . During an adiabatic change, $Q=0$ and $pV^\gamma=\text{constant}$ , so $\Delta U=-W$ .
At constant pressure, work done by the gas is $W=p\Delta V$ . At constant volume, $\Delta V=0$ and $W=0$ .
Each process must also satisfy $Q=\Delta U+W$ .
Rapid well-insulated changes are commonly close to adiabatic because little energy transfers by heating.

Worked example

An ideal gas expands isothermally from $1.50\times10^{-3}\,\text{m}^3$ at $320\,\text{kPa}$ to $4.0\times10^{-3}\,\text{m}^3$ . Calculate the final pressure and state $\Delta U$ .

1.For an isothermal change, $p_1V_1=p_2V_2$ .
2. $p_2=(320)(1.5/4.0)=120\,\text{kPa}$ .
3.The ideal-gas temperature is unchanged, so its internal energy is unchanged.

Answer: $p_2=120\,\text{kPa}$ and $\Delta U=0$ .

Common mistakes

Don't equate adiabatic with constant temperature rather than with zero heat transfer.
Don't use $pV=\text{constant}$ for an adiabatic change instead of $pV^\gamma=\text{constant}$ .
Don't calculate non-zero work for a constant-volume process even though $\Delta V=0$ .

Exam tip

Name the defining constraint first, then apply both its process equation and the First Law.

Tier 1 · Easy

A rigid sealed vessel receives $450\,\text{J}$ by heating. State the work done by the gas and calculate its increase in internal energy.
[2 marks]
Total for this question: 2
An ideal gas expands isothermally and does $210\,\text{J}$ of work. State its change in internal energy and calculate the heat transferred to the gas.
[2 marks]
Total for this question: 2

Tier 2 · Standard

An ideal gas expands isothermally from pressure $240\,\text{kPa}$ and volume $2.0\times10^{-3}\,\text{m}^3$ to volume $5.0\times10^{-3}\,\text{m}^3$ . Determine its final pressure and state its change in internal energy.
[3 marks]
Total for this question: 3
A gas receives $420\,\text{J}$ by heating while expanding at constant pressure $200\,\text{kPa}$ from $1.50\times10^{-3}\,\text{m}^3$ to $2.2\times10^{-3}\,\text{m}^3$ . Determine its change in internal energy.
[3 marks]
Total for this question: 3
An ideal gas at $260\,\text{kPa}$ and volume $1.50\times10^{-3}\,\text{m}^3$ expands isothermally to $4.50\times10^{-3}\,\text{m}^3$ . It is then heated at constant volume from $290\,\text{K}$ to $348\,\text{K}$ . Determine the final pressure.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A gas is compressed adiabatically from $p_1=100\,\text{kPa}$ and $V_1=4.8\times10^{-3}\,\text{m}^3$ to $V_2=1.8\times10^{-3}\,\text{m}^3$ . Take $\gamma=1.40$ . During the compression, $720\,\text{J}$ of work is done on the gas. Calculate the final pressure and the change in internal energy.
[5 marks]
Total for this question: 5
A gas expands adiabatically from pressure $500\,\text{kPa}$ and volume $1.20\times10^{-3}\,\text{m}^3$ to pressure $139\,\text{kPa}$ and volume $3.00\times10^{-3}\,\text{m}^3$ . During the expansion, the gas does $4.6\times10^2\,\text{J}$ of work. It is then held at the final volume and receives $1.8\times10^2\,\text{J}$ by heating. Determine $\gamma$ , the change in internal energy during the adiabatic expansion, and the total change in internal energy after both processes.
[5 marks]
Total for this question: 5
A gas initially at $300\,\text{kPa}$ and $1.20\times10^{-3}\,\text{m}^3$ expands isothermally to an unknown volume $V_B$ . It is then compressed adiabatically from $V_B$ to $1.50\times10^{-3}\,\text{m}^3$ , reaching $500\,\text{kPa}$ . For the gas, $\gamma=1.40$ . Determine $V_B$ .
[5 marks]
Total for this question: 5
A gas at $180\,\text{kPa}$ and volume $2.40\times10^{-3}\,\text{m}^3$ is compressed adiabatically to $1.20\times10^{-3}\,\text{m}^3$ . For the gas, $\gamma=1.33$ and the work done by the gas in an adiabatic change may be found from $W=(p_1V_1-p_2V_2)/(\gamma-1)$ . The gas then expands at the resulting constant pressure to $1.80\times10^{-3}\,\text{m}^3$ while receiving $520\,\text{J}$ by heating. Determine the pressure after compression, the net work done by the gas and the total change in internal energy over both processes.
[6 marks]
Total for this question: 6
The manufacturer quotes $\gamma=1.55$ . A gas expands adiabatically from $640\,\text{kPa}$ and $0.900\times10^{-3}\,\text{m}^3$ to $150\,\text{kPa}$ and $2.30\times10^{-3}\,\text{m}^3$ . A sensor records $390\,\text{J}$ of work output. Use $pV^\gamma=\text{constant}$ and $W=(p_1V_1-p_2V_2)/(\gamma-1)$ to calculate $\gamma$ from the pressure-volume data and the theoretical work output. Evaluate whether the sensor value is within $5.0\%$ of the theoretical value.
[5 marks]
Total for this question: 5

3.11.2.3 · The p-V diagram

Explanation

A thermodynamic process is shown as a path on a pressure–volume diagram. Motion to larger volume is expansion; motion to smaller volume is compression.
The work done by the gas is the signed area under the path. For constant pressure, $W=p\Delta V$ ; an expansion gives positive work and a compression negative work.
For a cycle, the net work per cycle is the area enclosed by the loop, not the total area beneath one branch.
A clockwise loop represents positive net work by the gas, while an anticlockwise loop requires net work input.
Pressure must be in pascals and volume in $\text{m}^3$ for the area to be in joules.

The enclosed area of a clockwise

p

–

V

cycle is the positive net work done by the gas.

Worked example

A rectangular clockwise cycle spans $V=2.0$ to $5.0\times10^{-3}\,\text{m}^3$ and $p=120$ to $360\,\text{kPa}$ . Calculate the net work per cycle.

1.The net work is the enclosed rectangular area: $W=\Delta p\,\Delta V$ .
2. $\Delta p=(360-120)\times10^3=2.40\times10^5\,\text{Pa}$ .
3. $W=(2.40\times10^5)(3.0\times10^{-3})=720\,\text{J}$ .

Answer: The gas does $+720\,\text{J}$ of net work per cycle.

Common mistakes

Don't use the full area from the pressure axis to the upper branch instead of the area enclosed by the cycle.
Don't leave pressure in kilopascals when calculating an area expected in joules.
Don't assign positive work to an anticlockwise cycle rather than to a clockwise cycle.

Exam tip

For an estimate from a curved path, state that work is the area under the graph and show how that area was approximated.

Tier 1 · Easy

A gas expands at constant pressure $180\,\text{kPa}$ from volume $1.2\times10^{-3}\,\text{m}^3$ to $3.7\times10^{-3}\,\text{m}^3$ . Calculate the work done by the gas.
[2 marks]
Total for this question: 2
A gas is compressed at constant pressure $250\,\text{kPa}$ from volume $4.0\times10^{-3}\,\text{m}^3$ to $1.6\times10^{-3}\,\text{m}^3$ . Calculate the work done by the gas.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A gas follows a clockwise rectangular cycle on a $p$ - $V$ diagram between pressures $150\,\text{kPa}$ and $400\,\text{kPa}$ and between volumes $1.0\times10^{-3}\,\text{m}^3$ and $3.5\times10^{-3}\,\text{m}^3$ . Determine the net work done per cycle.
[3 marks]
Total for this question: 3
On a $p$ - $V$ diagram, a gas follows the triangular cycle $A\rightarrow B\rightarrow C\rightarrow A$ , where $A=(1.0\times10^{-3}\,\text{m}^3,100\,\text{kPa})$ , $B=(4.0\times10^{-3}\,\text{m}^3,100\,\text{kPa})$ and $C=(1.0\times10^{-3}\,\text{m}^3,300\,\text{kPa})$ . Determine the net work done by the gas per cycle.
[3 marks]
Total for this question: 3
A gas expands at constant pressure $300\,\text{kPa}$ from $1.0\times10^{-3}\,\text{m}^3$ to $3.0\times10^{-3}\,\text{m}^3$ . It then expands to $5.0\times10^{-3}\,\text{m}^3$ while the pressure falls linearly to $100\,\text{kPa}$ . Determine the total work done by the gas.
[4 marks]
Total for this question: 4

Tier 3 · Hard

On a $p$ - $V$ diagram, a gas expands along a straight line from $(0.8\times10^{-3}\,\text{m}^3,600\,\text{kPa})$ to $(3.2\times10^{-3}\,\text{m}^3,200\,\text{kPa})$ . It then compresses at constant pressure $200\,\text{kPa}$ to its initial volume before returning at constant volume to its initial state. Calculate the net work done by the gas in the cycle.
[5 marks]
Total for this question: 5
An anticlockwise loop on a $p$ - $V$ diagram encloses $84$ grid squares. One grid square represents a pressure interval of $25.0\,\text{kPa}$ and a volume interval of $0.20\times10^{-3}\,\text{m}^3$ . The gas completes $18.0$ cycles each second. Determine the signed work done by the gas per cycle and its signed mean mechanical power.
[5 marks]
Total for this question: 5
A gas expands along a curved path on a $p$ - $V$ diagram. At volumes $(1.0,\,2.0,\,3.5,\,5.0)\times10^{-3}\,\text{m}^3$ , its pressures are respectively $500$ , $360$ , $240$ and $180\,\text{kPa}$ . The gas is then cooled at constant volume to $120\,\text{kPa}$ before returning to its initial volume at that constant pressure, and a final constant-volume process restores the initial state. Using trapezia between the given points, determine the net work done by the gas in the cycle.
[5 marks]
Total for this question: 5
A gas follows the clockwise cycle $A\rightarrow B\rightarrow C\rightarrow D\rightarrow A$ on a $p$ - $V$ diagram. The coordinates are $A=(1.20\times10^{-3}\,\text{m}^3,140\,\text{kPa})$ , $B=(1.20\times10^{-3}\,\text{m}^3,p_B)$ , $C=(4.80\times10^{-3}\,\text{m}^3,260\,\text{kPa})$ and $D=(4.80\times10^{-3}\,\text{m}^3,140\,\text{kPa})$ . Paths B to C and D to A are straight. The net work done by the gas is $720\,\text{J}$ per cycle. Determine $p_B$ .
[5 marks]
Total for this question: 5
A gas follows the closed path $A\rightarrow B\rightarrow C\rightarrow D\rightarrow A$ using straight lines on a $p$ - $V$ diagram. In units of $(10^{-3}\,\text{m}^3,\text{kPa})$ , the points are $A=(1.0,200)$ , $B=(3.0,500)$ , $C=(5.0,200)$ and $D=(3.0,80)$ . Determine the signed work done by the gas on each path, the net work per cycle and whether the cycle is clockwise or anticlockwise.
[6 marks]
Total for this question: 6

3.11.2.4 · Engine cycles

Explanation

A four-stroke petrol engine completes induction, compression, power and exhaust strokes; it compresses a fuel–air mixture before spark ignition. A diesel engine compresses air before fuel injection and ignition.
Their indicator diagrams show measured pressure–volume cycles, which can be compared with theoretical cycles without requiring constructional detail. Indicated power equals loop area multiplied by cycles per second per cylinder and cylinder count.
Input power is calorific value times fuel mass-flow rate. Brake power is $T\omega$ , and friction power is indicated power minus brake power.
Overall, thermal and mechanical efficiencies are respectively brake/input, indicated/input and brake/indicated power.
Other supplied cycles require interpretation using the same ideas.

Worked example

A four-cylinder four-stroke engine runs at $1800\,\text{rev min}^{-1}$ . Each indicator loop has area $500\,\text{J}$ . Calculate the indicated power.

1. $1800\,\text{rev min}^{-1}=30\,\text{rev s}^{-1}$ .
2.A four-stroke cylinder completes one cycle per two revolutions, giving $15\,\text{cycles s}^{-1}$ .
3. $P_i=(500)(15)(4)=3.0\times10^4\,\text{W}$ .

Answer: The indicated power is $30\,\text{kW}$ .

Common mistakes

Don't use crankshaft revolutions per second as cycles per second for a four-stroke cylinder and double the indicated power.
Don't calculate thermal efficiency using brake power instead of indicated power.
Don't subtract indicated power from brake power and obtain a negative friction power.

Exam tip

Write the three efficiency ratios in words before substituting because their numerators are different.

Tier 1 · Easy

An engine produces a steady crankshaft torque of $140\,\text{N m}$ at $3000\,\text{rev min}^{-1}$ . Calculate its brake power.
[2 marks]
Total for this question: 2
A single cylinder in a four-stroke engine turns at $3600\,\text{rev min}^{-1}$ . Calculate the number of engine cycles completed by the cylinder each second.
[2 marks]
Total for this question: 2

Tier 2 · Standard

The area enclosed by the indicator diagram for one cylinder of a four-stroke engine is $620\,\text{J}$ per cycle. The four-cylinder engine runs at $2400\,\text{rev min}^{-1}$ . Determine the indicated power.
[3 marks]
Total for this question: 3
An engine has indicated power $62\,\text{kW}$ and brake torque $200\,\text{N m}$ at $2500\,\text{rev min}^{-1}$ . Determine its friction power.
[3 marks]
Total for this question: 3
An engine produces brake torque $150\,\text{N m}$ at $2400\,\text{rev min}^{-1}$ . Fuel of calorific value $44.0\,\text{MJ kg}^{-1}$ is supplied at $3.00\times10^{-3}\,\text{kg s}^{-1}$ . Determine the engine's overall efficiency.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A four-cylinder, four-stroke engine runs at $3000\,\text{rev min}^{-1}$ . Each cylinder's indicator-loop area is $380\,\text{J}$ . The brake torque is $105\,\text{N m}$ . Fuel of calorific value $44.0\,\text{MJ kg}^{-1}$ is supplied at $1.60\times10^{-3}\,\text{kg s}^{-1}$ . Determine the indicated power, brake power, friction power, and thermal efficiency.
[6 marks]
Total for this question: 6
A six-cylinder four-stroke engine runs at $1.80\times10^3\,\text{rev min}^{-1}$ . Each cylinder has indicator-loop area $4.50\times10^2\,\text{J}$ . The thermal efficiency is $30.0\%$ , the fuel calorific value is $45.0\,\text{MJ kg}^{-1}$ and the brake torque is $1.80\times10^2\,\text{N m}$ . Determine the fuel mass-flow rate and the mechanical efficiency.
[6 marks]
Total for this question: 6
A four-cylinder four-stroke engine runs at $2400\,\text{rev min}^{-1}$ and has indicator-loop area $390\,\text{J}$ per cylinder. Before servicing, its brake torque is $90\,\text{N m}$ and its fuel input power is $120\,\text{kW}$ . After servicing, the indicator-loop area and speed are unchanged, the brake torque is $105\,\text{N m}$ and the fuel input power is $108\,\text{kW}$ . An engineer claims that only the mechanical efficiency has improved. Deduce whether the claim is correct.
[6 marks]
Total for this question: 6
A three-cylinder four-stroke engine runs at $3600\,\text{rev min}^{-1}$ and produces brake torque $110\,\text{N m}$ . Its mechanical efficiency is $82.0\%$ and its thermal efficiency is $34.0\%$ . Fuel of calorific value $42.5\,\text{MJ kg}^{-1}$ is used. Determine the area of the indicator diagram for each cylinder and the fuel mass-flow rate.
[6 marks]
Total for this question: 6
At one operating point, an engine runs at $2775\,\text{rev min}^{-1}$ with measured brake torque $143\,\text{N m}$ . Its indicated power is $52.6\,\text{kW}$ and its thermal efficiency is $31.8\%$ . The fuel calorific value is $42.7\,\text{MJ kg}^{-1}$ . Determine the friction power, the fuel mass-flow rate and the mechanical efficiency at this operating point.
[6 marks]
Total for this question: 6

3.11.2.5 · Second Law and engines

Explanation

The First Law alone does not forbid complete conversion of heat into work, but the Second Law requires a heat engine to operate between a hot source and a colder sink.
Each cycle receives $Q_H$ , produces work $W$ and rejects $Q_C$ , so $Q_H=W+Q_C$ and $\eta=W/Q_H=(Q_H-Q_C)/Q_H$ .
The maximum theoretical efficiency is $\eta_{\max}=(T_H-T_C)/T_H$ , with temperatures in kelvin.
Practical efficiency is lower because of friction, unwanted heat transfer, incomplete combustion and irreversible finite-rate processes.
Combined heat and power schemes improve total energy use by using both $W$ and useful rejected heat, but this utilisation fraction is not the engine's heat-to-work efficiency.

A heat engine must reject energy

Q_C

to a cold sink while producing work.

Worked example

An engine receives $8.0\,\text{kJ}$ and rejects $5.2\,\text{kJ}$ per cycle. Calculate its efficiency.

1.Use energy conservation: $W=Q_H-Q_C=8.0-5.2=2.8\,\text{kJ}$ .
2.Use $\eta=W/Q_H$ .
3. $\eta=2.8/8.0=0.35$ .

Answer: The efficiency is $0.35$ , or $35\%$ .

Common mistakes

Don't claim a cyclic engine can turn all $Q_H$ into work without rejecting energy to a sink.
Don't use temperatures in degrees Celsius in $\eta_{\max}=(T_H-T_C)/T_H$ .
Don't call the combined work-plus-heating utilisation fraction a heat-engine efficiency.

Exam tip

When comparing actual and maximum efficiency, identify at least one irreversible loss mechanism in the practical engine.

Tier 1 · Easy

A heat engine receives $5.0\,\text{kJ}$ from its hot source and rejects $3.2\,\text{kJ}$ to its sink each cycle. Calculate its efficiency.
[2 marks]
Total for this question: 2
A cyclic heat engine receives $7.2\,\text{kJ}$ from a hot source and produces $2.1\,\text{kJ}$ of work. Calculate the energy rejected to the cold sink.
[2 marks]
Total for this question: 2

Tier 2 · Standard

An engine operates between a source at $720\,\text{K}$ and a sink at $310\,\text{K}$ . Determine its maximum theoretical efficiency and explain why a practical engine operating at these temperatures has a lower efficiency.
[3 marks]
Total for this question: 3
A practical engine operates between a source at $800\,\text{K}$ and a sink at $320\,\text{K}$ . Its actual efficiency is $75\%$ of the maximum theoretical value. Calculate the work produced when the engine receives $6.0\,\text{kJ}$ from the source.
[4 marks]
Total for this question: 4
A heat engine's hot source is at $650\,{}^\circ\text{C}$ . Determine the highest sink temperature, in $^\circ\text{C}$ , for which the engine's maximum theoretical efficiency is at least $62.0\%$ .
[4 marks]
Total for this question: 4

Tier 3 · Hard

A combined heat and power plant receives thermal energy at $120\,\text{MW}$ and produces $45\,\text{MW}$ of electrical work. It supplies $60\,\text{MW}$ of otherwise rejected heat to nearby buildings. The engine source and sink temperatures are $850\,\text{K}$ and $300\,\text{K}$ . Calculate the engine efficiency, the total useful-energy fraction, and the maximum theoretical efficiency. Explain the difference between the last two quantities.
[5 marks]
Total for this question: 5
A heat engine operates between a source at $900\,\text{K}$ and a sink at $300\,\text{K}$ . It receives $12.0\,\text{kJ}$ from the source each cycle and has an actual efficiency of $45.0\%$ . Determine the maximum theoretical work per cycle, the actual energy rejected to the sink, and the actual work as a percentage of the maximum theoretical work.
[5 marks]
Total for this question: 5
A proposed heat engine receives $18.0\,\text{kJ}$ from a source at $720\,\text{K}$ each cycle. An engineer claims that it can produce $9.60\,\text{kJ}$ of work whether its sink is at $300\,\text{K}$ or $360\,\text{K}$ . Deduce, for each sink temperature, whether the claim is possible. Determine the minimum energy that any engine receiving $18.0\,\text{kJ}$ from the $720\,\text{K}$ source must reject per cycle at each sink temperature.
[5 marks]
Total for this question: 5
A heat engine must produce $4.20\,\text{kJ}$ of work for every $10.5\,\text{kJ}$ received from its hot source. Its actual efficiency is $64.0\%$ of the maximum theoretical efficiency, and the sink is at $295\,\text{K}$ . Determine the minimum hot-source temperature and the energy rejected to the sink per cycle. Explain why a lower source temperature cannot meet the stated performance.
[5 marks]
Total for this question: 5
A heat engine receives energy from a source at $745\,\text{K}$ at a rate of $84.6\,\text{kW}$ . Measurements show that it rejects energy to its sink at $50.7\,\text{kW}$ . An operator reports that the sink is at $460\,\text{K}$ . Determine the measured work output and efficiency, then calculate the maximum theoretical work output for the reported temperatures and deduce whether the reported sink temperature is achievable.
[6 marks]
Total for this question: 6

3.11.2.6 · Reversed heat engines

Explanation

A reversed heat engine uses work input $W$ to transfer energy $Q_C$ from a cold region to a hot region, delivering $Q_H=Q_C+W$ . A refrigerator's useful transfer is $Q_C$ , so $\mathrm{COP}_{\text{ref}}=Q_C/W=Q_C/(Q_H-Q_C)$ .
A heat pump's useful transfer is $Q_H$ , so $\mathrm{COP}_{\text{hp}}=Q_H/W=Q_H/(Q_H-Q_C)$ . Hence $\mathrm{COP}_{\text{hp}}=\mathrm{COP}_{\text{ref}}+1$ .
Ideal limits are $T_C/(T_H-T_C)$ and $T_H/(T_H-T_C)$ respectively, using kelvin. COP can exceed one because it compares heat transferred with work input; it is not a conversion efficiency.
Only the basic principles and uses of refrigerators and heat pumps are required.
Detailed device cycles are not required.

Work input drives heat from a cold region to a hot region in a reversed heat engine.

Worked example

A refrigerator removes $7.5\,\text{kJ}$ from its cold space while using $2.5\,\text{kJ}$ of work. Calculate both refrigerator and heat-pump COP values for the same device.

1. $\mathrm{COP}_{\text{ref}}=Q_C/W=7.5/2.5=3.0$ .
2. $Q_H=Q_C+W=7.5+2.5=10.0\,\text{kJ}$ .
3. $\mathrm{COP}_{\text{hp}}=Q_H/W=10.0/2.5=4.0$ .

Answer: $\mathrm{COP}_{\text{ref}}=3.0$ and $\mathrm{COP}_{\text{hp}}=4.0$ .

Common mistakes

Don't use $Q_H$ as the useful transfer when calculating refrigerator COP.
Don't use temperatures in degrees Celsius in the ideal COP expressions.
Don't reject a COP above one as impossible by confusing COP with efficiency.

Exam tip

Identify the desired heat transfer first: $Q_C$ for refrigeration and $Q_H$ for space heating.

Tier 1 · Easy

A refrigerator removes $360\,\text{J}$ from its cold compartment for every $120\,\text{J}$ of electrical work supplied. Calculate its coefficient of performance.
[2 marks]
Total for this question: 2
A freezer removes $9.00\times10^2\,\text{J}$ from its interior while receiving $3.00\times10^2\,\text{J}$ of electrical work. Calculate the energy delivered to the room.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A heat pump delivers $14\,\text{kW}$ to a building while consuming $4.0\,\text{kW}$ of electrical power. Determine its heat-pump coefficient of performance and the rate at which it extracts energy from outside.
[3 marks]
Total for this question: 3
A refrigerator removes energy from a cold store at $8.40\,\text{kW}$ while using $2.10\,\text{kW}$ of electrical power. The cold store is at $258\,\text{K}$ and the surroundings are at $298\,\text{K}$ . Calculate the rate of heat transfer to the surroundings and express the actual coefficient of performance as a percentage of the maximum theoretical value.
[4 marks]
Total for this question: 4
An ideal refrigerator keeps a compartment at $-18\,{}^\circ\text{C}$ while the room is at $22\,{}^\circ\text{C}$ . It removes energy from the compartment at $1.20\,\text{kW}$ . Determine the minimum electrical power required.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A heat pump keeps a building at $293\,\text{K}$ when the outside temperature is $268\,\text{K}$ . Its actual coefficient of performance is $42\%$ of the maximum theoretical value. Calculate the electrical power required when the heat delivered to the building is $18\,\text{kW}$ .
[5 marks]
Total for this question: 5
A reversed heat-engine device operates between a cold space at $263\,\text{K}$ and a room at $296\,\text{K}$ . Its actual refrigerator coefficient of performance is $60\%$ of the maximum theoretical value. The device is used as a heat pump with an electrical input of $1.50\,\text{kW}$ . Determine its actual heat-pump coefficient of performance, the heating rate delivered to the room, and the electricity cost for $8.0\,\text{h}$ at $\pounds0.28$ per $\text{kW h}$ .
[5 marks]
Total for this question: 5
A refrigerator operates in a room at $300\,\text{K}$ with electrical input power $200\,\text{W}$ . At cold-space temperature $T_C$ , its actual refrigerator coefficient of performance is $50\%$ of the maximum theoretical value. Energy leaks into the cold space at a rate $\left(60\,\text{W K}^{-1}\right)(300\,\text{K}-T_C)$ . Determine the steady cold-space temperature.
[5 marks]
Total for this question: 5
A refrigerator keeps a cold store at $258\,\text{K}$ in a room at $303\,\text{K}$ . Its actual refrigerator coefficient of performance is $48.0\%$ of the maximum theoretical value. Its $350\,\text{W}$ motor runs for $70.0\%$ of the time. At steady temperature, assume the mean energy leak into the store is $k(T_H-T_C)$ . Determine $k$ and the mean rate at which the refrigerator transfers energy to the room.
[5 marks]
Total for this question: 5
A heat pump delivers $14.6\,\text{kW}$ to a building maintained at $297\,\text{K}$ while consuming $3.20\,\text{kW}$ of electrical power. Its actual heat-pump coefficient of performance is $57.0\%$ of the maximum theoretical value. Determine the actual and maximum theoretical coefficients of performance, the effective outside temperature and the rate at which energy is extracted from outside.
[6 marks]
Total for this question: 6

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

3.11.1.1 · Concept of moment of inertia

Tier 1 · Easy

Mark scheme for 3.11.1.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$2.1\times10^{-2}\,\text{kg m}^2$	Use $I=\sum mr^2$ . Hence $I=(0.30)(0.22)^2+(0.45)(0.12)^2=0.01452+0.00648=0.02100\,\text{kg m}^2$ , which is $2.1\times10^{-2}\,\text{kg m}^2$ to two significant figures.	2
02.1	$0.19\,\text{m}$	For a point mass, $I=mr^2$ , so $r=\sqrt{I/m}=\sqrt{0.0180/0.50}=0.1897\,\text{m}$ . This is $0.19\,\text{m}$ to two significant figures.	2

Tier 2 · Standard

Mark scheme for 3.11.1.1 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$0.17\,\text{kg m}^2$	Each slider contributes $mr^2=(0.25)(0.30)^2=0.0225\,\text{kg m}^2$ . Four contribute $4(0.0225)=0.0900\,\text{kg m}^2$ . Add the original flywheel value: $I=0.080+0.0900=0.170\,\text{kg m}^2$ , which is $0.17\,\text{kg m}^2$ to two significant figures.	3
02.1	$6.7\times10^{-2}\,\text{kg m}^2$	Only the trim-mass contributions change. Thus $\Delta I=2m(r_2^2-r_1^2)=2(0.60)[(0.28)^2-(0.15)^2]=0.06708\,\text{kg m}^2$ . This is $6.7\times10^{-2}\,\text{kg m}^2$ to two significant figures.	3
03.1	$58\%$	The sensors contribute $4mr^2=4(0.18)(0.16)^2=0.018432\,\text{kg m}^2$ . The counterweights contribute $2(0.32)(0.30)^2=0.05760\,\text{kg m}^2$ . The total is $0.024+0.018432+0.05760=0.100032\,\text{kg m}^2$ . Hence the counterweight percentage is $100(0.05760/0.100032)=57.58\%$ , or $58\%$ to two significant figures.	4

Tier 3 · Hard

Mark scheme for 3.11.1.1 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$0.65\,\text{kg}$	The disc contributes $I_{\text{disc}}=\frac12(6.0)(0.25)^2=0.1875\,\text{kg m}^2$ . The three masses therefore contribute $0.456-0.1875-0.028=0.2405\,\text{kg m}^2$ . With $0.2405=3m(0.35)^2$ , $m=0.2405/[3(0.35)^2]=0.654\,\text{kg}$ , which is $0.65\,\text{kg}$ to two significant figures.	5
02.1	$98\%$	The perpendicular distances are $r_1=(0.40)\sin30^\circ=0.200\,\text{m}$ and $r_2=(0.40)\sin60^\circ=0.346\,\text{m}$ . Initially, $I_1=0.050+4(0.30)(0.200)^2=0.0980\,\text{kg m}^2$ . Finally, $I_2=0.050+4(0.30)(0.346)^2=0.194\,\text{kg m}^2$ . The percentage increase is $100(I_2-I_1)/I_1=97.96\%$ , which is $98\%$ to two significant figures.	5
03.1	Each inner mass is $0.700\,\text{kg}$ and each outer mass is $0.500\,\text{kg}$	Let each inner mass be $m_i$ and each outer mass be $m_o$ . The mass condition gives $2m_i+2m_o=2.40$ , so $m_i+m_o=1.20$ . The point masses contribute $0.300-0.0415=0.2585\,\text{kg m}^2$ , giving $2m_i(0.20)^2+2m_o(0.45)^2=0.2585$ . Thus $0.0800m_i+0.405m_o=0.2585$ . Substituting $m_o=1.20-m_i$ gives $m_i=0.700\,\text{kg}$ and hence $m_o=0.500\,\text{kg}$ .	5
04.1	Design B; its moment of inertia is $2.16\%$ smaller than that of design A	For A, $I_A=\frac12(5.40)(0.260)^2+4(0.180)(0.340)^2+0.0120=0.18252+0.083232+0.0120=0.277752\,\text{kg m}^2$ . For B, $I_B=(5.40)(0.190)^2+6(0.120)(0.300)^2+0.0120=0.19494+0.06480+0.0120=0.271740\,\text{kg m}^2$ . Therefore B has the smaller value. Relative to A, the reduction is $100(0.277752-0.271740)/0.277752=2.1645\%$ , giving $2.16\%$ to three significant figures.	5
05.1	$17.6\%$	The four arms contribute $4[\frac13(0.35)(0.42)^2]=0.082320\,\text{kg m}^2$ . Initially the sensors contribute $4(0.11)(0.42)^2=0.077616\,\text{kg m}^2$ , so $I_i=0.020+0.082320+0.077616=0.179936\,\text{kg m}^2$ . Finally their contribution is $2(0.11)(0.42)^2+2(0.11)(0.18)^2=0.045936\,\text{kg m}^2$ , so $I_f=0.148256\,\text{kg m}^2$ . The percentage decrease is $100(0.179936-0.148256)/0.179936=17.606\%$ , or $17.6\%$ to three significant figures.	5

3.11.1.2 · Rotational kinetic energy

Tier 1 · Easy

Mark scheme for 3.11.1.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$83\,\text{J}$	Use $E_k=\frac12I\omega^2$ . Thus $E_k=\frac12(0.85)(14)^2=83.3\,\text{J}$ , giving $83\,\text{J}$ to two significant figures.	2
02.1	$1.3\times10^2\,\text{J}$	The released energy is $\frac12I(\omega_1^2-\omega_2^2)=\frac12(0.64)(25^2-15^2)=128\,\text{J}$ , which is $1.3\times10^2\,\text{J}$ to two significant figures.	2

Tier 2 · Standard

Mark scheme for 3.11.1.2 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$40\,\text{rad s}^{-1}$	Convert the energy: $E_k=3.2\times10^4\,\text{J}$ . Rearranging $E_k=\frac12I\omega^2$ gives $\omega=\sqrt{2E_k/I}=\sqrt{2(3.2\times10^4)/40}=\sqrt{1600}=40\,\text{rad s}^{-1}$ .	3
02.1	$3.9\,\text{kg m}^2$	The first angular speed is $\omega_1=2\pi(600/60)=62.83\,\text{rad s}^{-1}$ . Equal rotational kinetic energies give $\frac12I_1\omega_1^2=\frac12I_2\omega_2^2$ . Hence $I_2=I_1(\omega_1/\omega_2)^2=1.6(62.83/40)^2=3.948\,\text{kg m}^2$ , which is $3.9\,\text{kg m}^2$ to two significant figures.	3
03.1	$1.25\times10^3\,\text{rev min}^{-1}$	Initially, $\omega_i=2\pi(900/60)=94.25\,\text{rad s}^{-1}$ and $E_i=\frac12(3.5)(94.25)^2=1.554\times10^4\,\text{J}$ . The stored energy gain is $0.80(18000)=14400\,\text{J}$ , so $E_f=2.994\times10^4\,\text{J}$ . Hence $\omega_f=\sqrt{2E_f/I}=130.8\,\text{rad s}^{-1}$ . Converting back gives $60\omega_f/(2\pi)=1249\,\text{rev min}^{-1}$ , or $1.25\times10^3\,\text{rev min}^{-1}$ to three significant figures.	4

Tier 3 · Hard

Mark scheme for 3.11.1.2 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$4.2\,\text{kW}$	The energy decrease is $\Delta E_k=\frac12I(\omega_1^2-\omega_2^2)=\frac12(18)(75^2-45^2)=32400\,\text{J}$ . The useful energy is $0.72(32400)=23328\,\text{J}$ . Hence the mean useful power is $P=23328/5.5=4.24\times10^3\,\text{W}=4.2\,\text{kW}$ to two significant figures.	5
02.1	$35\,\text{rad s}^{-1}$ and $7.2\times10^2\,\text{W}$	The required decrease in rotational kinetic energy is $1800/0.80=2250\,\text{J}$ . Therefore $\frac12I(\omega_i^2-\omega_f^2)=2250$ , so $\omega_i=\sqrt{(18)^2+2(2250)/5.0}=34.99\,\text{rad s}^{-1}$ . The minimum initial speed is $35\,\text{rad s}^{-1}$ to two significant figures. The mean useful power is $1800/2.5=720\,\text{W}=7.2\times10^2\,\text{W}$ .	5
03.1	Flywheel A stores $28.0\%$ more rotational kinetic energy than flywheel B	At the rim, $v=R\omega$ . For A, $E_A=\frac12(MR^2)(v_A/R)^2=\frac12M(240)^2=28800M\,\text{J}$ . For B, $E_B=\frac12(\frac12MR^2)(v_B/R)^2=\frac14M(300)^2=22500M\,\text{J}$ . The common mass cancels in the comparison. Therefore A stores more, by $100(28800-22500)/22500=28.0\%$ .	5
04.1	$I=4.50\,\text{kg m}^2$ ; the final angular speed is $56.0\,\text{rad s}^{-1}$ ; the flywheel releases $972\,\text{J}$ during the high-load interval	During the high-load interval the resultant torque is $47.0-71.0=-24.0\,\text{N m}$ . At the limiting moment of inertia, $\alpha=(52.0-56.0)/0.750=-5.33333\,\text{rad s}^{-2}$ . From $T=I\alpha$ , $I=24.0/5.33333=4.50\,\text{kg m}^2$ . During the lower-load interval the resultant torque is $47.0-35.0=+12.0\,\text{N m}$ , so $\alpha=12.0/4.50=2.66667\,\text{rad s}^{-2}$ . The speed rise is $(2.66667)(1.50)=4.00\,\text{rad s}^{-1}$ , giving a final speed of $52.0+4.00=56.0\,\text{rad s}^{-1}$ . The energy released at high load is $\Delta E_k=\tfrac12I(\omega_1^2-\omega_2^2)=\tfrac12(4.50)(56.0^2-52.0^2)=972\,\text{J}$ , regained at lower load, smoothing the shaft speed.	6
05.1	$\omega=39.8\,\text{rad s}^{-1}$ and A's final angular speed is $59.4\,\text{rad s}^{-1}$ (accept $59.3\,\text{rad s}^{-1}$ )	Initially, $2100=\frac12(0.92)\omega^2+\frac12(0.30)(2.40\omega)^2=1.324\omega^2$ , giving $\omega=39.82591\,\text{rad s}^{-1}$ . The separate energies are $E_A=\frac12(0.92)(39.82591)^2=729.607\,\text{J}$ and $E_B=2100-729.607=1370.393\,\text{J}$ . A finally stores $729.607+0.650(1370.393)=1620.363\,\text{J}$ . Thus $\omega_{A,f}=\sqrt{2(1620.363)/0.92}=59.3509\,\text{rad s}^{-1}$ , giving $39.8\,\text{rad s}^{-1}$ and $59.4\,\text{rad s}^{-1}$ to three significant figures.	5

3.11.1.3 · Rotational motion

Tier 1 · Easy

Mark scheme for 3.11.1.3 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$3.0\,\text{rad s}^{-2}$	Use $\alpha=\Delta\omega/\Delta t=(17-5.0)/4.0=3.0\,\text{rad s}^{-2}$ .	2
02.1	$1.0\times10^2\,\text{rad}$	At constant angular speed, $\theta=\omega t=(23)(4.5)=103.5\,\text{rad}$ , which is $1.0\times10^2\,\text{rad}$ to two significant figures.	2

Tier 2 · Standard

Mark scheme for 3.11.1.3 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$12$ revolutions	The angular displacement is $\theta=\omega_1t+\frac12\alpha t^2=0+\frac12(4.2)(6.0)^2=75.6\,\text{rad}$ . Divide by $2\pi$ : $N=75.6/(2\pi)=12.0$ revolutions.	3
02.1	$1.4\times10^2\,\text{rad}$	Use $\omega_2^2=\omega_1^2+2\alpha\theta$ . Therefore $0=36^2+2(-4.5)\theta$ , so $\theta=36^2/9.0=144\,\text{rad}=1.4\times10^2\,\text{rad}$ to two significant figures.	3
03.1	$+16\,\text{rad}$	For the uniform change, the mean signed angular velocity is $[24+(-8.0)]/2=8.0\,\text{rad s}^{-1}$ . Its angular displacement is therefore $(8.0)(5.0)=+40\,\text{rad}$ . The constant negative angular velocity gives $\theta_2=(-8.0)(3.0)=-24\,\text{rad}$ . Hence the net angular displacement is $40-24=+16\,\text{rad}$ .	3

Tier 3 · Hard

Mark scheme for 3.11.1.3 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$7.5\times10^2\,\text{rad}$ , or $1.2\times10^2$ revolutions	Angular displacement is the area under the $\omega$ - $t$ graph. During acceleration, $\theta_1=\frac12(12+48)(6.0)=180\,\text{rad}$ . At constant speed, $\theta_2=(48)(5.0)=240\,\text{rad}$ . During slowing, $\theta_3=\frac12(48+18)(10)=330\,\text{rad}$ . Therefore $\theta=180+240+330=750\,\text{rad}$ and $N=750/(2\pi)=119.4$ . To two significant figures these are $7.5\times10^2\,\text{rad}$ and $1.2\times10^2$ revolutions.	5
02.1	$2.8\,\text{rad s}^{-2}$ , $25\,\text{rad s}^{-1}$ and $9.1\,\text{s}$	The angular displacement is $\theta=18(2\pi)=36\pi\,\text{rad}$ . From $\theta=\omega_1t+\frac12\alpha t^2$ , $36\pi=(3.0)(8.0)+\frac12\alpha(8.0)^2$ , giving $\alpha=2.784\,\text{rad s}^{-2}$ . Then $\omega_2=\omega_1+\alpha t=3.0+(2.784)(8.0)=25.27\,\text{rad s}^{-1}$ . During braking, $0=25.27-(2.784)t$ , so $t=9.077\,\text{s}$ . To two significant figures, the results are $2.8\,\text{rad s}^{-2}$ , $25\,\text{rad s}^{-1}$ and $9.1\,\text{s}$ .	5
03.1	$t=7.7\,\text{s}$ and the total time is $23\,\text{s}$	At the end of the powered interval, $\omega=3.2t$ and $\theta_1=\frac12(3.2)t^2=1.6t^2$ . During deceleration, $0=\omega^2-2(1.6)\theta_2$ , so $\theta_2=(3.2t)^2/3.2=3.2t^2$ . Therefore $45(2\pi)=4.8t^2$ , giving $t=7.675\,\text{s}=7.7\,\text{s}$ . The deceleration time is $\omega/1.6=(3.2t)/1.6=2t$ , so the total time is $3t=23.0\,\text{s}$ , or $23\,\text{s}$ to two significant figures.	5
04.1	$1.18\,\text{s}$ and $9.00\,\text{rad s}^{-1}$	The final angular speed is $25.5-(2.20)(7.50)=9.00\,\text{rad s}^{-1}$ . The first stage contributes $\frac12(7.50+25.5)(4.80)=79.200\,\text{rad}$ and the last contributes $\frac12(25.5+9.00)(7.50)=129.375\,\text{rad}$ . The total displacement is $38.0(2\pi)=238.761\,\text{rad}$ . If the constant-speed duration is $t$ , then $79.200+25.5t+129.375=238.761$ , so $t=1.18377\,\text{s}$ . Therefore the required results are $1.18\,\text{s}$ and $9.00\,\text{rad s}^{-1}$ .	5
05.1	$4.00\,\text{rad s}^{-1}$ , $4.00\,\text{rad s}^{-2}$ and $7.92\,\text{s}$	For constant angular acceleration, the displacement during the $n$ th second is $\theta_n=\omega_0+\frac12\alpha[n^2-(n-1)^2]=\omega_0+\alpha(n-\frac12)$ . Thus $\omega_0+3.5\alpha=18.0$ and $\omega_0+8.5\alpha=38.0$ . Subtraction gives $5.0\alpha=20.0$ , so $\alpha=4.00\,\text{rad s}^{-2}$ and $\omega_0=4.00\,\text{rad s}^{-1}$ . For $25.0$ revolutions, $50\pi=4.00t+\frac12(4.00)t^2$ . Solving $2t^2+4t-50\pi=0$ gives the positive root $t=7.91851\,\text{s}$ , or $7.92\,\text{s}$ .	6

3.11.1.4 · Torque and angular acceleration

Tier 1 · Easy

Mark scheme for 3.11.1.4 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$14\,\text{N m}$	The force is perpendicular to the radius, so $T=Fr=(85)(0.16)=13.6\,\text{N m}$ , which is $14\,\text{N m}$ to two significant figures.	2
02.1	$50\,\text{N}$	For a tangential force, $T=Fr$ . Hence $F=T/r=12/0.24=50\,\text{N}$ .	2

Tier 2 · Standard

Mark scheme for 3.11.1.4 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$29\,\text{rad s}^{-1}$	From $T=I\alpha$ , $\alpha=T/I=24/3.2=7.5\,\text{rad s}^{-2}$ . Then $\omega=\omega_1+\alpha t=6.0+(7.5)(3.0)=28.5\,\text{rad s}^{-1}$ , giving $29\,\text{rad s}^{-1}$ to two significant figures.	3
02.1	$3.0\,\text{kg m}^2$	The braking torque is $T=Fr=(30)(0.20)=6.0\,\text{N m}$ . The magnitude of the angular acceleration is $\|\alpha\|=(15-9.0)/3.0=2.0\,\text{rad s}^{-2}$ . From $T=I\alpha$ , $I=6.0/2.0=3.0\,\text{kg m}^2$ .	3
03.1	$5.9\,\text{rad s}^{-2}$	The driving torque is $T_d=Fr\sin55^\circ=(120)(0.35)\sin55^\circ=34.40\,\text{N m}$ . The resultant torque is $34.40-8.0=26.40\,\text{N m}$ . From $T=I\alpha$ , $\alpha=26.40/4.5=5.867\,\text{rad s}^{-2}$ , or $5.9\,\text{rad s}^{-2}$ to two significant figures.	4

Tier 3 · Hard

Mark scheme for 3.11.1.4 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$44\,\text{rad s}^{-1}$	The resultant torque is $T=54-7.5=46.5\,\text{N m}$ . Hence $\alpha=T/I=46.5/6.2=7.50\,\text{rad s}^{-2}$ . Using $\omega=\omega_1+\alpha t$ gives $\omega=10+(7.50)(4.5)=43.75\,\text{rad s}^{-1}$ , which is $44\,\text{rad s}^{-1}$ to two significant figures.	5
02.1	The cable tension is $1.3\times10^2\,\text{N}$ ; the load mass is $12\,\text{kg}$	The required resultant torque is $I\alpha=(8.0)(4.0)=32\,\text{N m}$ . If the cable tension is $F$ , then $70-6.0-F(0.25)=32$ , so $F=128\,\text{N}$ . The load's linear acceleration is $a=r\alpha=(0.25)(4.0)=1.0\,\text{m s}^{-2}$ . Applying $F-mg=ma$ gives $m=128/(9.81+1.0)=11.84\,\text{kg}$ , which is $12\,\text{kg}$ to two significant figures.	5
03.1	$57.6\,\text{rad s}^{-1}$	The load torque is $Fr=(40)(0.30)=12\,\text{N m}$ . The required resultant torque is $I\alpha=(2.4)(8.0)=19.2\,\text{N m}$ . The motor must therefore supply $19.2+12=31.2\,\text{N m}$ . Linearity of the torque-speed characteristic gives $T_m=60(1-\omega/120)$ . Hence $31.2=60(1-\omega/120)$ , so $\omega=57.6\,\text{rad s}^{-1}$ .	5
04.1	$15.2\,\text{N m}$ and $6.61\,\text{kg m}^2$ (accept $6.59$ - $6.61\,\text{kg m}^2$ )	The driving torque is $(96)(0.28)=26.88\,\text{N m}$ . The angled opposing force contributes $(70)(0.20)\sin35^\circ=8.03007\,\text{N m}$ . Therefore $T_{\text{resultant}}=26.88-8.03007-3.6=15.24993\,\text{N m}$ . The angular acceleration is $(19-4.0)/6.5=2.307692\,\text{rad s}^{-2}$ . From $T=I\alpha$ , $I=15.24993/2.307692=6.60830\,\text{kg m}^2$ . The results are $15.2\,\text{N m}$ and $6.61\,\text{kg m}^2$ to three significant figures.	5
05.1	$10.6\,\text{rad s}^{-2}$ and $16.9\,\text{N m}$ (accept $16.8\,\text{N m}$ )	Treating both rotors as one system, the resultant external torque is $48-3.5-2.0=42.5\,\text{N m}$ and the total moment of inertia is $2.6+1.4=4.0\,\text{kg m}^2$ . Hence $\alpha=42.5/4.0=10.625\,\text{rad s}^{-2}$ . For the driven rotor alone, $T_{\text{clutch}}-2.0=(1.4)(10.625)$ , so $T_{\text{clutch}}=16.875\,\text{N m}$ . Thus the common acceleration is $10.6\,\text{rad s}^{-2}$ and the transmitted torque is $16.9\,\text{N m}$ to three significant figures.	5

3.11.1.5 · Angular momentum

Tier 1 · Easy

Mark scheme for 3.11.1.5 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$43\,\text{kg m}^2\text{ s}^{-1}$	Use $L=I\omega=(2.4)(18)=43.2\,\text{kg m}^2\text{ s}^{-1}$ , giving $43\,\text{kg m}^2\text{ s}^{-1}$ to two significant figures.	2
02.1	$4.5\,\text{kg m}^2$	Using $L=I\omega$ , $I=L/\omega=54/12=4.5\,\text{kg m}^2$ .	2

Tier 2 · Standard

Mark scheme for 3.11.1.5 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$12\,\text{rad s}^{-1}$	With negligible external torque, angular momentum is conserved: $(0.80)(30)+(1.20)(0)=(0.80+1.20)\omega$ . Therefore $24=2.00\omega$ , so $\omega=12\,\text{rad s}^{-1}$ .	3
02.1	$5.0\,\text{s}$	The magnitude of the angular-momentum change is $\Delta L=I(22-10)=(2.5)(12)=30\,\text{kg m}^2\text{ s}^{-1}$ . Angular impulse gives $Tt=\Delta L$ , so $t=30/6.0=5.0\,\text{s}$ .	3
03.1	$+0.69\,\text{rad s}^{-1}$	The angular impulse is the triangular area under the torque-time graph: $\Delta L=\frac12(2.5)(12)=15\,\text{N m s}$ . The initial angular momentum is $L_i=I\omega_i=(3.2)(-4.0)=-12.8\,\text{kg m}^2\text{ s}^{-1}$ . Hence $L_f=-12.8+15=+2.2\,\text{kg m}^2\text{ s}^{-1}$ . Therefore $\omega_f=L_f/I=2.2/3.2=+0.6875\,\text{rad s}^{-1}$ , or $+0.69\,\text{rad s}^{-1}$ to two significant figures.	4

Tier 3 · Hard

Mark scheme for 3.11.1.5 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$13\,\text{rad s}^{-1}$	Initially $L_i=I\omega=(4.8)(9.0)=43.2\,\text{kg m}^2\text{ s}^{-1}$ . The opposing angular impulse is $T\Delta t=-(3.2)(1.5)=-4.8\,\text{kg m}^2\text{ s}^{-1}$ , so the angular momentum becomes $L=43.2-4.8=38.4\,\text{kg m}^2\text{ s}^{-1}$ . During the position change this is conserved, hence $\omega=L/I=38.4/3.0=12.8\,\text{rad s}^{-1}$ , or $13\,\text{rad s}^{-1}$ to two significant figures.	5
02.1	$77\%$	Conservation of angular momentum gives $(1.8)(20)+(1.2)(-10)=(1.8+1.2)\omega$ , so $\omega=8.0\,\text{rad s}^{-1}$ . Initially, $E_i=\frac12(1.8)(20)^2+\frac12(1.2)(10)^2=420\,\text{J}$ . Finally, $E_f=\frac12(3.0)(8.0)^2=96\,\text{J}$ . The energy lost is $420-96=324\,\text{J}$ , so the percentage loss is $100(324/420)=77.14\%$ , or $77\%$ to two significant figures.	5
03.1	$5.6\,\text{rad s}^{-1}$	Initially, the total moment of inertia is $4.0+(0.50)(0.80)^2=4.32\,\text{kg m}^2$ , so $L_i=(4.32)(5.0)=21.6\,\text{kg m}^2\text{ s}^{-1}$ . After release, the platform has angular momentum $4.0\omega$ . The mass has angular momentum $mrv=(0.50)(0.80)(0.80\omega-6.0)=0.32\omega-2.4$ . Conservation gives $21.6=4.0\omega+0.32\omega-2.4$ , so $\omega=5.556\,\text{rad s}^{-1}=5.6\,\text{rad s}^{-1}$ to two significant figures.	5
04.1	Body: $-5.47\,\text{rad s}^{-1}$ ; wheel: $+235\,\text{rad s}^{-1}$	Let the body's angular velocity be $\Omega$ . The wheel's angular velocity in the inertial frame is then $\Omega+240$ . Initial angular momentum is zero, so $18.0\Omega+0.420(\Omega+240)=0$ . Hence $18.420\Omega=-100.8$ and $\Omega=-5.47231\,\text{rad s}^{-1}$ . The wheel's inertial-frame angular velocity is $-5.47231+240=234.52769\,\text{rad s}^{-1}$ . To three significant figures these are $-5.47\,\text{rad s}^{-1}$ and $+235\,\text{rad s}^{-1}$ ; the opposite signs are required by angular-momentum conservation.	5
05.1	$122\,\text{m s}^{-1}$ and $98.1\%$	Taking the turntable's initial rotation as positive, conservation of angular momentum gives $(0.720)(8.50)-(0.0600)(0.340)v=[0.720+(0.0600)(0.340)^2](5.00)$ . Thus $v=[6.120-3.63468]/0.02040=121.8294\,\text{m s}^{-1}$ . The initial kinetic energy is $\frac12(0.720)(8.50)^2+\frac12(0.0600)(121.8294)^2=471.282\,\text{J}$ . The final kinetic energy is $\frac12[0.720+(0.0600)(0.340)^2](5.00)^2=9.08670\,\text{J}$ . Hence the converted percentage is $100(471.282-9.08670)/471.282=98.0719\%$ , giving $122\,\text{m s}^{-1}$ and $98.1\%$ .	6

3.11.1.6 · Work and power

Tier 1 · Easy

Mark scheme for 3.11.1.6 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$63\,\text{J}$	Use $W=T\theta=(18)(3.5)=63\,\text{J}$ .	2
02.1	$23\,\text{rad}$	From $W=T\theta$ , $\theta=W/T=550/24=22.92\,\text{rad}$ , which is $23\,\text{rad}$ to two significant figures.	2

Tier 2 · Standard

Mark scheme for 3.11.1.6 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$5.3\,\text{kW}$	Convert the rotation rate: $1200\,\text{rev min}^{-1}=20\,\text{rev s}^{-1}$ , so $\omega=2\pi(20)=125.7\,\text{rad s}^{-1}$ . Then $P=T\omega=(42)(125.7)=5.28\times10^3\,\text{W}=5.3\,\text{kW}$ .	3
02.1	$4.8\,\text{kJ}$	At constant angular speed, the motor torque balances the load and friction torques: $T=35+5.0=40\,\text{N m}$ . The work supplied is $W=T\theta=(40)(120)=4800\,\text{J}=4.8\,\text{kJ}$ to two significant figures.	3
03.1	$1.55\,\text{kJ}$	Work is the area under the torque-angle graph. During the uniform increase, the mean torque is $(10+30)/2=20\,\text{N m}$ , so $W_1=(20)(40)=800\,\text{J}$ . During the constant-torque interval, $W_2=(30)(25)=750\,\text{J}$ . Thus $W=800+750=1550\,\text{J}=1.55\,\text{kJ}$ .	4

Tier 3 · Hard

Mark scheme for 3.11.1.6 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$2.4\,\text{kW}$ and $6.0\times10^4\,\text{J}$	The load torque is $T_{\text{load}}=84-9.0=75\,\text{N m}$ because the speed is constant. The useful power is $P=T_{\text{load}}\omega=(75)(32)=2400\,\text{W}=2.4\,\text{kW}$ . The useful energy is $E=Pt=(2400)(25)=60000\,\text{J}=6.0\times10^4\,\text{J}$ .	5
02.1	$5.0\,\text{kJ}$ and $0.99\,\text{kW}$	The motor work is $W_{\text{motor}}=T_{\text{mean}}\theta=(70)(80)=5600\,\text{J}$ . Friction removes $W_f=(8.0)(80)=640\,\text{J}$ . With no net change in rotational kinetic energy, the useful work is $5600-640=4960\,\text{J}=5.0\,\text{kJ}$ . The mean useful power is $4960/5.0=992\,\text{W}=0.99\,\text{kW}$ , to two significant figures.	5
03.1	$38\,\text{rad s}^{-1}$	The driving work is the torque-angle area $W_d=\frac12(50+20)(60)=2100\,\text{J}$ . Friction removes $W_f=(6.0)(60)=360\,\text{J}$ , so the rotational kinetic energy increases by $1740\,\text{J}$ . The initial kinetic energy is $\frac12(2.5)(8.0)^2=80\,\text{J}$ , giving $E_f=1820\,\text{J}$ . Hence $\omega_f=\sqrt{2E_f/I}=\sqrt{2(1820)/2.5}=38.16\,\text{rad s}^{-1}$ , or $38\,\text{rad s}^{-1}$ to two significant figures.	5
04.1	$100\,\text{rad s}^{-1}$ , $3.00\,\text{kW}$ and $135\,\text{kJ}$	The motor relation is $T_M=80-(60/120)\omega=80-0.500\omega$ . At steady speed the resultant torque is zero, so $80-0.500\omega=5.0+0.250\omega$ . This gives $\omega=75.0/0.750=100\,\text{rad s}^{-1}$ and $T_L=5.0+0.250(100)=30.0\,\text{N m}$ . The useful power is $P=T_L\omega=(30.0)(100)=3000\,\text{W}=3.00\,\text{kW}$ . In $45\,\text{s}$ the useful energy is $Pt=(3000)(45)=135000\,\text{J}=135\,\text{kJ}$ .	5
05.1	$9.14\,\text{kJ}$ and $1.14\,\text{kW}$	Uniform acceleration gives $\theta=\frac12(12+28)(8.0)=160\,\text{rad}$ . The motor does $W_M=(72)(160)=11520\,\text{J}$ and friction removes $W_f=(6.5)(160)=1040\,\text{J}$ . The rotational kinetic-energy increase is $\Delta E_k=\frac12(4.2)(28^2-12^2)=1344\,\text{J}$ . Energy conservation gives $W_{\text{useful}}=11520-1040-1344=9136\,\text{J}=9.14\,\text{kJ}$ . The mean useful power is $9136/8.0=1142\,\text{W}=1.14\,\text{kW}$ to three significant figures.	6

3.11.2.1 · First law of thermodynamics

Tier 1 · Easy

Mark scheme for 3.11.2.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$+340\,\text{J}$	Here $Q=+520\,\text{J}$ and $W=+180\,\text{J}$ . Therefore $\Delta U=Q-W=520-180=+340\,\text{J}$ .	2
02.1	$Q=-40\,\text{J}$	Here $\Delta U=-120\,\text{J}$ and $W=+80\,\text{J}$ . Therefore $Q=\Delta U+W=-120+80=-40\,\text{J}$ , so $40\,\text{J}$ leaves the gas by heating.	2

Tier 2 · Standard

Mark scheme for 3.11.2.1 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$+150\,\text{J}$	Energy leaving by heating gives $Q=-90\,\text{J}$ . Work done on the gas means work done by the gas is $W=-240\,\text{J}$ . Hence $\Delta U=Q-W=-90-(-240)=+150\,\text{J}$ .	3
02.1	$Q=+50\,\text{J}$ ; $50\,\text{J}$ is transferred to the gas by heating	The net work done by the gas is $W=+180-80=+100\,\text{J}$ . Its internal-energy change is $\Delta U=-50\,\text{J}$ . Therefore $Q=\Delta U+W=-50+100=+50\,\text{J}$ , so energy is transferred to the gas by heating.	3
03.1	$+260\,\text{J}$	For the first process, $Q_1=+600\,\text{J}$ and $W_1=+250\,\text{J}$ , so $\Delta U_1=Q_1-W_1=+350\,\text{J}$ . For the second, $Q_2=-180\,\text{J}$ and work done by the gas is $W_2=-90\,\text{J}$ , so $\Delta U_2=-180-(-90)=-90\,\text{J}$ . Therefore $\Delta U_{\text{total}}=350-90=+260\,\text{J}$ .	4

Tier 3 · Hard

Mark scheme for 3.11.2.1 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$+1.8\,\text{kJ}$ , net work done by the gas	For A, $Q_A=+3.2\,\text{kJ}$ and $\Delta U_A=+1.1\,\text{kJ}$ , so $W_A=Q_A-\Delta U_A=+2.1\,\text{kJ}$ . Returning to the initial state requires $\Delta U_B=-1.1\,\text{kJ}$ . For B, $Q_B=-1.4\,\text{kJ}$ , so $W_B=Q_B-\Delta U_B=-1.4-(-1.1)=-0.3\,\text{kJ}$ . Thus $W_{\text{net}}=2.1-0.3=+1.8\,\text{kJ}$ .	5
02.1	$Q_C=-1.00\,\text{kJ}$ and $W_{\text{net}}=+0.10\,\text{kJ}$	For A, $\Delta U_A=Q_A-W_A=1.20-0.45=+0.75\,\text{kJ}$ . For B, $Q_B=-0.30\,\text{kJ}$ and $W_B=-0.60\,\text{kJ}$ , so $\Delta U_B=Q_B-W_B=+0.30\,\text{kJ}$ . The total change must be $-0.20\,\text{kJ}$ , hence $\Delta U_C=-0.20-0.75-0.30=-1.25\,\text{kJ}$ . Therefore $Q_C=\Delta U_C+W_C=-1.25+0.25=-1.00\,\text{kJ}$ . The net work is $0.45-0.60+0.25=+0.10\,\text{kJ}$ .	5
03.1	$Q_{BC}=+1.60\,\text{kJ}$	The direct route fixes the state-function change: $\Delta U_{AC}=2.40-0.90=+1.50\,\text{kJ}$ . From A to B, $Q_{AB}=-0.30\,\text{kJ}$ and $W_{AB}=-0.80\,\text{kJ}$ , so $\Delta U_{AB}=Q-W=-0.30-(-0.80)=+0.50\,\text{kJ}$ . Therefore $\Delta U_{BC}=1.50-0.50=+1.00\,\text{kJ}$ . Since $W_{BC}=+0.60\,\text{kJ}$ , $Q_{BC}=\Delta U_{BC}+W_{BC}=+1.60\,\text{kJ}$ .	5
04.1	$W_{AB}=+0.550\,\text{kJ}$ , $W_{BC}=+0.250\,\text{kJ}$ , $W_{CA}=-0.250\,\text{kJ}$ and $W_{\text{net}}=+0.550\,\text{kJ}$	For A to B, $\Delta U=3.05-2.20=+0.850\,\text{kJ}$ , so $W_{AB}=1.40-0.850=+0.550\,\text{kJ}$ . For B to C, $\Delta U=2.55-3.05=-0.500\,\text{kJ}$ , so $W_{BC}=-0.250-(-0.500)=+0.250\,\text{kJ}$ . For C to A, $\Delta U=2.20-2.55=-0.350\,\text{kJ}$ , so $W_{CA}=-0.600-(-0.350)=-0.250\,\text{kJ}$ . The net work is $0.550+0.250-0.250=+0.550\,\text{kJ}$ , consistent with zero net internal-energy change over a cycle.	6
05.1	$Q_{\text{compression}}=-430\,\text{J}$ , $0.900\,\text{kW}$ and $7.74\,\text{kW}$ rejected	During expansion, $\Delta U=Q-W=480-310=+170\,\text{J}$ . Returning to the initial state requires $\Delta U_{\text{compression}}=-170\,\text{J}$ , while work done by the gas during compression is $W=-260\,\text{J}$ . Hence $Q=\Delta U+W=-170-260=-430\,\text{J}$ . Net work per cycle is $310-260=50\,\text{J}$ , so the mechanical power is $(50)(18.0)=900\,\text{W}=0.900\,\text{kW}$ . The rejected-heat rate is $(430)(18.0)=7740\,\text{W}=7.74\,\text{kW}$ .	5

3.11.2.2 · Non-flow processes

Tier 1 · Easy

Mark scheme for 3.11.2.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$W=0\,\text{J}$ and $\Delta U=+450\,\text{J}$	The vessel is rigid, so the volume is constant and $W=0$ . From $Q=\Delta U+W$ , $\Delta U=Q-W=450-0=+450\,\text{J}$ .	2
02.1	$\Delta U=0$ and $Q=+210\,\text{J}$	The temperature of an ideal gas is constant, so $\Delta U=0$ . From $Q=\Delta U+W$ , $Q=0+210=+210\,\text{J}$ .	2

Tier 2 · Standard

Mark scheme for 3.11.2.2 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$96\,\text{kPa}$ and $\Delta U=0$	For an isothermal change, $p_1V_1=p_2V_2$ . Hence $p_2=p_1V_1/V_2=(240)(2.0/5.0)=96\,\text{kPa}$ . The temperature of an ideal gas is unchanged, so its internal energy is unchanged and $\Delta U=0$ .	3
02.1	$+2.8\times10^2\,\text{J}$	The work done is $W=p\Delta V=(2.00\times10^5)(2.2-1.5)\times10^{-3}=140\,\text{J}$ . Therefore $\Delta U=Q-W=420-140=+280\,\text{J}=+2.8\times10^2\,\text{J}$ to two significant figures.	3
03.1	$104\,\text{kPa}$	For the isothermal expansion, $p_1V_1=p_2V_2$ , so $p_2=260(1.5/4.5)=86.7\,\text{kPa}$ . During constant-volume heating, $p/T$ is constant. Hence $p_3=p_2(T_3/T_2)=86.7(348/290)=104\,\text{kPa}$ .	4

Tier 3 · Hard

Mark scheme for 3.11.2.2 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$3.9\times10^2\,\text{kPa}$ and $\Delta U=+720\,\text{J}$	For an adiabatic change, $p_1V_1^\gamma=p_2V_2^\gamma$ , so $p_2=p_1(V_1/V_2)^\gamma=100(4.8/1.8)^{1.40}=394.8\,\text{kPa}$ , or $3.9\times10^2\,\text{kPa}$ to two significant figures. Adiabatic means $Q=0$ . Work done on the gas makes work done by the gas $W=-720\,\text{J}$ . Thus $\Delta U=Q-W=0-(-720)=+720\,\text{J}$ .	5
02.1	$\gamma=1.40$ , $\Delta U_{\text{adiabatic}}=-4.6\times10^2\,\text{J}$ and $\Delta U_{\text{total}}=-2.8\times10^2\,\text{J}$	For an adiabatic change, $p_1V_1^\gamma=p_2V_2^\gamma$ . Hence $\gamma=\ln(p_1/p_2)/\ln(V_2/V_1)=\ln(500/139)/\ln(3.00/1.20)=1.397$ , or $1.40$ to three significant figures. Adiabatic means $Q=0$ , so $\Delta U_{\text{adiabatic}}=-4.6\times10^2\,\text{J}$ . At constant volume the work is zero, so the later heating raises the internal energy by $1.8\times10^2\,\text{J}$ . The total change is therefore $-460+180=-280\,\text{J}=-2.8\times10^2\,\text{J}$ .	5
03.1	$V_B=9.40\times10^{-3}\,\text{m}^3$	Write volumes in units of $10^{-3}\,\text{m}^3$ and let $x=V_B/(10^{-3}\,\text{m}^3)$ . Isothermal expansion gives $p_B=300(1.20/x)=360/x\,\text{kPa}$ . The adiabatic compression gives $p_Bx^{1.40}=500(1.50)^{1.40}$ . Substitution gives $360x^{0.40}=500(1.50)^{1.40}$ , so $x=[500(1.50)^{1.40}/360]^{1/0.40}=9.397$ . Therefore $V_B=9.40\times10^{-3}\,\text{m}^3$ to three significant figures.	5
04.1	$453\,\text{kPa}$ , $-64.9\,\text{J}$ and $+585\,\text{J}$ (accept $-66.4\,\text{J}$ and $+586\,\text{J}$ from the rounded $453\,\text{kPa}$ )	For the adiabatic compression, $p_2=180(2.40/1.20)^{1.33}=452.5248\,\text{kPa}$ . Its work is $W_1=[(180\times10^3)(2.40\times10^{-3})-(452.5248\times10^3)(1.20\times10^{-3})]/(1.33-1)=-336.4539\,\text{J}$ . The constant-pressure work is $W_2=(452.5248\times10^3)(0.60\times10^{-3})=271.5149\,\text{J}$ , so $W_{\text{net}}=-64.93898\,\text{J}$ . Adiabatically, $\Delta U_1=-W_1=+336.4539\,\text{J}$ ; next, $\Delta U_2=520-271.5149=248.4851\,\text{J}$ . Thus $\Delta U_{\text{total}}=584.9390\,\text{J}$ , giving the stated rounded results.	6
05.1	$\gamma=1.55$ , theoretical work $=423\,\text{J}$ ; no, the sensor is $7.77\%$ low (accept $420\,\text{J}$ and $7.14\%$ from the rounded $\gamma=1.55$ )	From $p_1V_1^\gamma=p_2V_2^\gamma$ , $\gamma=\ln(640/150)/\ln(2.30/0.900)=1.5462857$ . The theoretical work is $[(640\times10^3)(0.900\times10^{-3})-(150\times10^3)(2.30\times10^{-3})]/(1.5462857-1)=422.8557\,\text{J}$ . The discrepancy relative to theory is $100(422.8557-390)/422.8557=7.76995\%$ . This exceeds $5.0\%$ , so the sensor value is not within the stated tolerance.	5

3.11.2.3 · The p-V diagram

Tier 1 · Easy

Mark scheme for 3.11.2.3 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$450\,\text{J}$	Use $W=p\Delta V$ . Here $p=1.80\times10^5\,\text{Pa}$ and $\Delta V=(3.7-1.2)\times10^{-3}=2.5\times10^{-3}\,\text{m}^3$ . Hence $W=(1.80\times10^5)(2.5\times10^{-3})=450\,\text{J}$ .	2
02.1	$-6.0\times10^2\,\text{J}$	$W=p\Delta V=(2.50\times10^5)(1.6-4.0)\times10^{-3}=-600\,\text{J}=-6.0\times10^2\,\text{J}$ to two significant figures. The work is negative because the path is a compression.	2

Tier 2 · Standard

Mark scheme for 3.11.2.3 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$6.3\times10^2\,\text{J}$	The net work is the rectangular loop area: $W=(p_{\text{high}}-p_{\text{low}})(V_{\text{high}}-V_{\text{low}})=(400-150)\times10^3(3.5-1.0)\times10^{-3}=625\,\text{J}$ . To two significant figures this is $6.3\times10^2\,\text{J}$ . The clockwise direction makes this work positive, done by the gas.	3
02.1	$-3.0\times10^2\,\text{J}$	The cycle has an isobaric path from A to B, a straight path from B to C and an isochoric return from C to A, so the enclosed triangular area is $\frac12\Delta p\Delta V=\frac12(2.00\times10^5)(3.0\times10^{-3})=300\,\text{J}$ . The stated order is anticlockwise, so the net work done by the gas is $W=-3.0\times10^2\,\text{J}$ to two significant figures.	3
03.1	$1.00\,\text{kJ}$	For the constant-pressure section, $W_1=(300\times10^3)(3.0-1.0)\times10^{-3}=600\,\text{J}$ . For the straight sloping section, the mean pressure is $(300+100)/2=200\,\text{kPa}$ , so $W_2=(200\times10^3)(5.0-3.0)\times10^{-3}=400\,\text{J}$ . Hence the total work is $600+400=1000\,\text{J}=1.00\,\text{kJ}$ .	4

Tier 3 · Hard

Mark scheme for 3.11.2.3 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$480\,\text{J}$	For the straight expansion path, the area under the graph is average pressure times volume change: $W_1=\frac12(600+200)\times10^3(3.2-0.8)\times10^{-3}=960\,\text{J}$ . On the constant-pressure compression, $W_2=p\Delta V=(200\times10^3)(0.8-3.2)\times10^{-3}=-480\,\text{J}$ . The final constant-volume path does no work. Therefore $W_{\text{net}}=960-480=480\,\text{J}$ , positive because the loop is clockwise.	5
02.1	$-420\,\text{J}$ per cycle and $-7.56\,\text{kW}$ ; the negative power means a mechanical input of $7.56\,\text{kW}$ is required	The area represented by one grid square is $(25.0\times10^3)(0.20\times10^{-3})=5.00\,\text{J}$ . The magnitude of the enclosed area is therefore $84(5.00)=420\,\text{J}$ . An anticlockwise loop gives negative net work done by the gas, so $W=-420\,\text{J}$ per cycle. The signed mean mechanical power is $P=Wf=(-420)(18.0)=-7560\,\text{W}=-7.56\,\text{kW}$ . The negative sign means that the cycle requires a mean mechanical power input of $7.56\,\text{kW}$ .	5
03.1	$7.2\times10^2\,\text{J}$	For the curved expansion, the trapezium estimate is $W_{\text{exp}}=\frac12(500+360)(1.0\times10^{-3})+\frac12(360+240)(1.5\times10^{-3})+\frac12(240+180)(1.5\times10^{-3})=430+450+315=1195\,\text{J}$ . Both constant-volume processes do no work. The constant-pressure return does $W_{\text{comp}}=(120\times10^3)(1.0-5.0)\times10^{-3}=-480\,\text{J}$ . Thus $W_{\text{net}}=1195-480=715\,\text{J}=7.2\times10^2\,\text{J}$ to two significant figures.	5
04.1	$p_B=420\,\text{kPa}$	The constant-volume paths do no work. Along B to C, $W_{BC}=\frac12(p_B+260\times10^3)(4.80-1.20)\times10^{-3}$ . Along D to A, $W_{DA}=(140\times10^3)(1.20-4.80)\times10^{-3}=-504\,\text{J}$ . Hence $720=\frac12(p_B+260\times10^3)(3.60\times10^{-3})-504$ . The expansion work is therefore $1224\,\text{J}$ , so its mean pressure is $1224/(3.60\times10^{-3})=340\,\text{kPa}$ . Thus $\frac12(p_B+260)=340$ in kilopascals, giving $p_B=420\,\text{kPa}$ .	5
05.1	$W_{AB}=+700\,\text{J}$ , $W_{BC}=+700\,\text{J}$ , $W_{CD}=-280\,\text{J}$ , $W_{DA}=-280\,\text{J}$ ; net work $=+840\,\text{J}$ , so the cycle is clockwise	For each straight path, work is mean pressure multiplied by volume change. Thus $W_{AB}=\frac12(200+500)(3.0-1.0)=+700\,\text{J}$ and $W_{BC}=\frac12(500+200)(5.0-3.0)=+700\,\text{J}$ . Similarly, $W_{CD}=\frac12(200+80)(3.0-5.0)=-280\,\text{J}$ and $W_{DA}=\frac12(80+200)(1.0-3.0)=-280\,\text{J}$ , because $1\,\text{kPa}\,(10^{-3}\,\text{m}^3)=1\,\text{J}$ . Hence $W_{\text{net}}=700+700-280-280=+840\,\text{J}$ . Positive net work by the gas identifies a clockwise cycle.	6

3.11.2.4 · Engine cycles

Tier 1 · Easy

Mark scheme for 3.11.2.4 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$44\,\text{kW}$	$3000\,\text{rev min}^{-1}=50\,\text{rev s}^{-1}$ , so $\omega=2\pi(50)=314\,\text{rad s}^{-1}$ . Brake power is $P=T\omega=(140)(314)=4.40\times10^4\,\text{W}=44\,\text{kW}$ .	2
02.1	$30\,\text{cycles s}^{-1}$	$3600\,\text{rev min}^{-1}=60\,\text{rev s}^{-1}$ . A four-stroke cycle takes two crankshaft revolutions, so the cylinder completes $60/2=30\,\text{cycles s}^{-1}$ .	2

Tier 2 · Standard

Mark scheme for 3.11.2.4 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$49.6\,\text{kW}$	The crankshaft speed is $2400/60=40\,\text{rev s}^{-1}$ . A four-stroke cylinder completes one cycle every two revolutions, so each cylinder completes $20\,\text{cycles s}^{-1}$ . Thus $P_i=(620)(20)(4)=49600\,\text{W}=49.6\,\text{kW}$ .	3
02.1	$9.6\,\text{kW}$	The angular speed is $\omega=2\pi(2500/60)=261.8\,\text{rad s}^{-1}$ . Brake power is $P_b=T\omega=(200)(261.8)=52.36\,\text{kW}$ . Hence friction power is $P_i-P_b=62-52.36=9.64\,\text{kW}$ , or $9.6\,\text{kW}$ to two significant figures.	3
03.1	$28.6\%$	The angular speed is $\omega=2\pi(2400/60)=251.3\,\text{rad s}^{-1}$ . Hence the brake power is $P_b=T\omega=(150)(251.3)=37.70\,\text{kW}$ . The fuel input power is $(44\times10^6)(3.00\times10^{-3})=132\,\text{kW}$ . The overall efficiency is $P_b/P_{\text{in}}=37.70/132=0.2856=28.6\%$ .	4

Tier 3 · Hard

Mark scheme for 3.11.2.4 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$P_i=38.0\,\text{kW}$ , $P_b=33.0\,\text{kW}$ , $P_f=5.0\,\text{kW}$ , thermal efficiency $=54.0\%$	The speed is $3000/60=50\,\text{rev s}^{-1}$ , so a four-stroke cylinder completes $25\,\text{cycles s}^{-1}$ . Hence $P_i=(380)(25)(4)=38000\,\text{W}=38.0\,\text{kW}$ . Also $\omega=2\pi(50)=314.16\,\text{rad s}^{-1}$ , so $P_b=T\omega=(105)(314.16)=32.99\,\text{kW}=33.0\,\text{kW}$ . Friction power is $P_f=P_i-P_b=38.0-32.99=5.01\,\text{kW}=5.0\,\text{kW}$ . Input power is $(44\times10^6)(1.60\times10^{-3})=70.4\,\text{kW}$ . Therefore thermal efficiency is $P_i/P_{\text{input}}=38.0/70.4=0.540=54.0\%$ .	6
02.1	$3.00\times10^{-3}\,\text{kg s}^{-1}$ and $83.8\%$	The speed is $30.0\,\text{rev s}^{-1}$ , so each cylinder completes $15.0\,\text{cycles s}^{-1}$ . The indicated power is $(450)(15.0)(6)=40500\,\text{W}$ . Since thermal efficiency is $P_i/P_{\text{in}}$ , $P_{\text{in}}=40500/0.300=135000\,\text{W}$ . The mass-flow rate is $P_{\text{in}}/q=135000/(45.0\times10^6)=3.00\times10^{-3}\,\text{kg s}^{-1}$ . The brake power is $P_b=T\omega=(180)[2\pi(30.0)]=33929\,\text{W}$ . Hence the mechanical efficiency is $P_b/P_i=33929/40500=0.8378=83.8\%$ .	6
03.1	The claim is incorrect: thermal efficiency increases from $26.0\%$ to $28.9\%$ , and mechanical efficiency increases from $72.5\%$ to $84.6\%$	Each cylinder completes $(2400/60)/2=20\,\text{cycles s}^{-1}$ , so the unchanged indicated power is $P_i=(390)(20)(4)=31.2\,\text{kW}$ . The thermal efficiencies are $31.2/120=26.0\%$ before and $31.2/108=28.9\%$ after. At $2400\,\text{rev min}^{-1}$ , $\omega=251.3\,\text{rad s}^{-1}$ . The brake powers are $(90)(251.3)=22.62\,\text{kW}$ and $(105)(251.3)=26.39\,\text{kW}$ , so the mechanical efficiencies are $22.62/31.2=72.5\%$ and $26.39/31.2=84.6\%$ . Both efficiencies improve, so the claim is incorrect.	6
04.1	$562\,\text{J}$ per cylinder per cycle and $3.50\times10^{-3}\,\text{kg s}^{-1}$	The angular speed is $2\pi(3600/60)=376.9911\,\text{rad s}^{-1}$ , so $P_b=(110)(376.9911)=41469.0\,\text{W}$ . Since mechanical efficiency is $P_b/P_i$ , $P_i=41469.0/0.820=50572.0\,\text{W}$ . Each cylinder completes $3600/(2\times60)=30.0$ cycles per second, giving $90.0$ cylinder-cycles per second. The area of the indicator diagram is $50572.0/90.0=561.911\,\text{J}$ . Thermal efficiency is $P_i/P_{\text{in}}$ , so $P_{\text{in}}=50572.0/0.340=148741\,\text{W}$ . Therefore $\dot m=148741/(42.5\times10^6)=3.49979\times10^{-3}\,\text{kg s}^{-1}$ , giving the stated results.	6
05.1	$11.0\,\text{kW}$ ; $3.87\times10^{-3}\,\text{kg s}^{-1}$ ; mechanical efficiency $79.0\%$	The angular speed is $\omega=2\pi(2775/60)=290.597\,\text{rad s}^{-1}$ , so the brake power is $P_b=T\omega=(143)(290.597)=41.5554\,\text{kW}$ . Friction power is $P_f=P_i-P_b=52.6-41.5554=11.0446\,\text{kW}$ , or $11.0\,\text{kW}$ to three significant figures. Thermal efficiency is $P_i/P_{\text{in}}$ , so $P_{\text{in}}=52.6/0.318=165.409\,\text{kW}$ . Hence the fuel mass-flow rate is $\dot m=(165.409\times10^3)/(42.7\times10^6)=3.87374\times10^{-3}\,\text{kg s}^{-1}$ , or $3.87\times10^{-3}\,\text{kg s}^{-1}$ . The mechanical efficiency is $P_b/P_i=41.5554/52.6=0.790$ , or $79.0\%$ .	6

3.11.2.5 · Second Law and engines

Tier 1 · Easy

Mark scheme for 3.11.2.5 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$0.36$ , or $36\%$	The work output is $W=Q_H-Q_C=5.0-3.2=1.8\,\text{kJ}$ . Therefore $\eta=W/Q_H=1.8/5.0=0.36=36\%$ .	2
02.1	$5.1\,\text{kJ}$	For a complete cycle, $Q_H=W+Q_C$ . Therefore $Q_C=Q_H-W=7.2-2.1=5.1\,\text{kJ}$ .	2

Tier 2 · Standard

Mark scheme for 3.11.2.5 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$0.569$ , or $56.9\%$ ; practical irreversibilities such as friction and unwanted heat transfer make the actual efficiency lower.	Use $\eta_{\max}=(T_H-T_C)/T_H=(720-310)/720=0.569=56.9\%$ . This limit assumes an ideal reversible engine. A real engine has friction and transfers heat across finite temperature differences and to its surroundings, so less of $Q_H$ becomes useful work.	3
02.1	$2.7\,\text{kJ}$	The maximum theoretical efficiency is $\eta_{\max}=(800-320)/800=0.600$ . The actual efficiency is $0.75(0.600)=0.450$ . Since $\eta=W/Q_H$ , the work output is $W=(0.450)(6.0)=2.7\,\text{kJ}$ to two significant figures.	4
03.1	$77.7\,{}^\circ\text{C}$	The hot-source temperature is $T_H=650+273=923\,\text{K}$ . Using $\eta_{\max}=(T_H-T_C)/T_H$ , $0.620=(923-T_C)/923$ , so $T_C=350.74\,\text{K}$ . Converting back gives $T_C=350.74-273=77.74\,{}^\circ\text{C}$ , or $77.7\,{}^\circ\text{C}$ to three significant figures.	4

Tier 3 · Hard

Mark scheme for 3.11.2.5 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Engine efficiency $=37.5\%$ , useful-energy fraction $=87.5\%$ , maximum theoretical efficiency $=64.7\%$	The engine efficiency counts work output: $\eta=W/Q_H=45/120=0.375=37.5\%$ . The CHP useful-energy fraction counts electrical work plus useful heating: $(45+60)/120=0.875=87.5\%$ . The maximum theoretical engine efficiency is $(850-300)/850=0.647=64.7\%$ . The $87.5\%$ figure is not a heat-to-work efficiency: it includes lower-grade heat that the Second Law requires the engine to reject, while the Carnot figure limits only the fraction convertible to work.	5
02.1	$8.00\,\text{kJ}$ , $6.60\,\text{kJ}$ and $67.5\%$	The maximum theoretical efficiency is $\eta_{\max}=(T_H-T_C)/T_H=(900-300)/900=0.6667$ . The corresponding maximum work is $(0.6667)(12.0)=8.00\,\text{kJ}$ . The actual work is $(0.450)(12.0)=5.40\,\text{kJ}$ , so the actual rejected energy is $12.0-5.40=6.60\,\text{kJ}$ . The actual work is $100(5.40/8.00)=67.5\%$ of the maximum theoretical work.	5
03.1	The claim is not forbidden at $300\,\text{K}$ but is impossible at $360\,\text{K}$ ; the minimum rejected energies are $7.50\,\text{kJ}$ and $9.00\,\text{kJ}$ respectively	The claimed efficiency is $9.60/18.0=0.533$ . With the $300\,\text{K}$ sink, $\eta_{\max}=(T_H-T_C)/T_H=(720-300)/720=0.583$ , so the maximum work is $(0.583)(18.0)=10.5\,\text{kJ}$ . The claim is below this limit and is not forbidden by the Second Law. With the $360\,\text{K}$ sink, $\eta_{\max}=(720-360)/720=0.500$ , so the maximum work is $9.00\,\text{kJ}$ . The claimed $9.60\,\text{kJ}$ exceeds this limit and is impossible. At the theoretical limits, the minimum rejected energies are $18.0-10.5=7.50\,\text{kJ}$ and $18.0-9.00=9.00\,\text{kJ}$ respectively.	5
04.1	$787\,\text{K}$ and $6.30\,\text{kJ}$ rejected; a lower source temperature gives an insufficient maximum theoretical efficiency	The actual efficiency is $4.20/10.5=0.400$ . Therefore the required maximum theoretical efficiency is $0.400/0.640=0.625$ . Using $0.625=(T_H-295)/T_H$ gives $0.625T_H=T_H-295$ , so $T_H=295/0.375=786.667\,\text{K}$ , or $787\,\text{K}$ . Energy conservation gives $Q_C=10.5-4.20=6.30\,\text{kJ}$ . Reducing $T_H$ would reduce $(T_H-T_C)/T_H$ below $0.625$ , so an engine operating at only $64.0\%$ of that limit could not reach the required $0.400$ actual efficiency.	5
05.1	Measured work $=33.9\,\text{kW}$ and efficiency $=40.1\%$ ; maximum theoretical work $=32.4\,\text{kW}$ , so the reported sink temperature is impossible	Energy conservation gives $P_W=84.6-50.7=33.9\,\text{kW}$ . The measured efficiency is $\eta=33.9/84.6=0.400709$ , or $40.1\%$ . If the sink were at $460\,\text{K}$ , the maximum theoretical efficiency would be $(745-460)/745=0.382550$ . The corresponding maximum work output is $(0.382550)(84.6)=32.3638\,\text{kW}$ , or $32.4\,\text{kW}$ . The measured $33.9\,\text{kW}$ exceeds this maximum theoretical value, so the reported sink temperature is impossible.	6

3.11.2.6 · Reversed heat engines

Tier 1 · Easy

Mark scheme for 3.11.2.6 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$\mathrm{COP}_{\text{ref}}=3.0$	For a refrigerator, the desired transfer is $Q_C$ . Hence $\mathrm{COP}_{\text{ref}}=Q_C/W=360/120=3.0$ .	2
02.1	$1.20\times10^3\,\text{J}$	Energy conservation for a reversed heat engine gives $Q_H=Q_C+W=900+300=1200\,\text{J}=1.20\times10^3\,\text{J}$ .	2

Tier 2 · Standard

Mark scheme for 3.11.2.6 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$\mathrm{COP}_{\text{hp}}=3.5$ and $10\,\text{kW}$ extracted	For a heat pump, $\mathrm{COP}_{\text{hp}}=Q_H/W=14/4.0=3.5$ . Energy conservation gives $Q_C=Q_H-W=14-4.0=10\,\text{kW}$ extracted from outside.	3
02.1	$10.5\,\text{kW}$ and $62.0\%$	Energy conservation gives $Q_H=Q_C+W=8.40+2.10=10.5\,\text{kW}$ . The actual refrigerator COP is $Q_C/W=8.40/2.10=4.00$ . The maximum theoretical value is $T_C/(T_H-T_C)=258/(298-258)=6.45$ . Therefore the actual COP is $100(4.00/6.45)=62.0\%$ of the maximum theoretical value.	4
03.1	$0.188\,\text{kW}$	The temperatures are $T_C=255\,\text{K}$ and $T_H=295\,\text{K}$ . The maximum refrigerator COP is $T_C/(T_H-T_C)=255/40=6.375$ . Since $\mathrm{COP}_{\text{ref}}=Q_C/W$ , the minimum input power is $W=1.20/6.375=0.1882\,\text{kW}$ , or $0.188\,\text{kW}$ to three significant figures.	4

Tier 3 · Hard

Mark scheme for 3.11.2.6 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$3.7\,\text{kW}$	For an ideal heat pump, $\mathrm{COP}_{\text{hp,max}}=T_H/(T_H-T_C)=293/(293-268)=11.72$ . The actual COP is $0.42(11.72)=4.922$ . Since $\mathrm{COP}_{\text{hp}}=Q_H/W$ , the electrical input is $W=Q_H/\mathrm{COP}_{\text{hp}}=18/4.922=3.66\,\text{kW}$ , giving $3.7\,\text{kW}$ to two significant figures.	5
02.1	$5.78$ , $8.67\,\text{kW}$ and $\pounds3.36$	The maximum theoretical refrigerator COP is $T_C/(T_H-T_C)=263/(296-263)=7.970$ . The actual refrigerator COP is $0.60(7.970)=4.782$ . For the same device, $\mathrm{COP}_{\text{hp}}=\mathrm{COP}_{\text{ref}}+1=5.782$ , or $5.78$ . The heating rate is $Q_H=(5.782)(1.50)=8.673\,\text{kW}$ , or $8.67\,\text{kW}$ . The electrical energy used is $(1.50)(8.0)=12.0\,\text{kW h}$ , so the cost is $(12.0)(\pounds 0.28)=\pounds 3.36$ .	5
03.1	$278\,\text{K}$	The actual COP is $0.50T_C/(300-T_C)$ , so the cooling rate is $200[0.50T_C/(300-T_C)]$ . At steady temperature this equals the leakage rate: $100T_C/(300-T_C)=60(300-T_C)$ . Let $x=300-T_C$ . Then $100(300-x)=60x^2$ , or $3x^2+5x-1500=0$ . The positive root is $x=[-5+\sqrt{18025}]/6=21.54\,\text{K}$ , so $T_C=300-21.54=278.46\,\text{K}=278\,\text{K}$ to three significant figures.	5
04.1	$k=15.0\,\text{W K}^{-1}$ and $919\,\text{W}$	The ideal refrigerator COP is $258/(303-258)=5.73333$ , so the actual COP is $(0.480)(5.73333)=2.75200$ . While running, the cooling rate is $(2.75200)(350)=963.200\,\text{W}$ ; its time average is $(0.700)(963.200)=674.240\,\text{W}$ . Hence $k=674.240/(303-258)=14.9831\,\text{W K}^{-1}$ , or $15.0\,\text{W K}^{-1}$ . The mean electrical input is $(0.700)(350)=245.0\,\text{W}$ , so the mean transfer to the room is $674.240+245.0=919.240\,\text{W}$ , or $919\,\text{W}$ .	5
05.1	Actual COP $=4.56$ , maximum theoretical COP $=8.00$ , outside temperature $=260\,\text{K}$ and $11.4\,\text{kW}$ extracted	The actual heat-pump coefficient of performance is $\mathrm{COP}_{\text{hp}}=Q_H/W=14.6/3.20=4.5625$ , or $4.56$ . Therefore $\mathrm{COP}_{\text{hp,max}}=4.5625/0.570=8.00439$ , or $8.00$ . Using $\mathrm{COP}_{\text{hp,max}}=T_H/(T_H-T_C)$ gives $8.00439=297/(297-T_C)$ , so $T_C=259.895\,\text{K}$ , or $260\,\text{K}$ . Energy conservation gives $Q_C=Q_H-W=14.6-3.20=11.4\,\text{kW}$ extracted from outside.	6