Specific heat capacity
dE = m c d(theta)
The temperature CHANGE trips everyone — d(theta) is the difference, not the final temperature.
Know the equation
| Symbol | Quantity | Unit |
|---|---|---|
| dE | change in thermal energy | J |
| m | mass | kg |
| c | specific heat capacity | J/kg C |
| d(theta) | temperature change | C |
Rearrangements
- c = dE / (m d(theta))
- d(theta) = dE / (m c)
- m = dE / (c d(theta))
Apply it — mark your own working
Work each one out on paper first, then reveal the mark scheme and tick the marks you actually earned. That is exactly how you should mark past papers.
0.50 kg of water is heated from 20 C to 80 C. The specific heat capacity of water is 4200 J/kg C. Calculate the energy transferred to the water.
Do the calculation on paper first — then mark it.
A 2.0 kg metal block is given 23 000 J of energy and its temperature rises by 25 C. Calculate the specific heat capacity of the metal.
Do the calculation on paper first — then mark it.
Where the marks get lost
- Putting the final temperature into d(theta) instead of the temperature change (here 60, not 80).
- Forgetting the mass must be in kg.
- When rearranging for c, dividing by only m or only d(theta) instead of by both (m x d(theta)).
Exam tip: Work out d(theta) as a separate first line and label it. That single step earns a mark and stops the whole calculation going wrong.
Still losing marks on the calculations?
I'll go through your working line by line and show you exactly where the marks are — your first lesson is free.