Specific heat capacity (required practical)

Energy

GCSE Physics (8463)

ORIGINAL

Energy from E = Pt, the temperature change, then c = ΔE ÷ (mΔθ) — and why the result comes out too high.

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Question

In the specific heat capacity required practical, a student heats a

1.0\,\text{kg}

aluminium block with a

50\,\text{W}

heater for

6.0

minutes. The temperature of the block rises from

21.0\,{}^\circ\text{C}

40.5\,{}^\circ\text{C}

. (a) Calculate the specific heat capacity of aluminium given by these results. (b) The accepted value is about

900\,\text{J/kg}\,{}^\circ\text{C}

. Explain why the student's value is higher.

Every step worked, with the reasoning.

1
$t = 6.0 \times 60 = 360\,\text{s}$
Power is in watts (joules per second), so the time must be in seconds.
2
$E = Pt = 50 \times 360 = 18\,000\,\text{J}$
The energy transferred by the heater is power × time.
3
$\Delta\theta = 40.5 - 21.0 = 19.5\,{}^\circ\text{C}$
$\Delta E = mc\Delta\theta$ uses the temperature change, not the final temperature.
4
$c = \dfrac{\Delta E}{m\Delta\theta} = \dfrac{18\,000}{1.0 \times 19.5} = 923\ldots \approx 920\,\text{J/kg}\,{}^\circ\text{C}$
Rearrange $\Delta E = mc\Delta\theta$ for $c$ , then round to 2 significant figures to match the mass and power.
5
Some of the $18\,000\,\text{J}$ is transferred to the surroundings, so the block's temperature rise is smaller than if all of it went into the block. A smaller $\Delta\theta$ in the denominator gives a larger $c$ .
The calculation assumes every joule heats the block. Insulating the block and making good thermal contact with the heater reduce the error.

Answer: (a) $920\,\text{J/kg}\,{}^\circ\text{C}$ (2 s.f.); (b) energy is lost to the surroundings, so the measured $\Delta\theta$ is too small and the calculated $c$ is too large.

Question

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