Required practical 14 · Physics · Physics Paper 1, both tiers

Specific heat capacity

Determine the specific heat capacity of a material by linking the energy transferred to the temperature rise.

The exam trap. Use the temperature change, not a temperature. Energy lost to the air makes c come out too high.

Variables

Independent (you change): Energy transferred by the heater (power × time)
Dependent (you measure): Temperature of the block
Control (you keep the same): Mass of the block
Insulation around the block
Power of the heater

Measure the mass of the block and wrap it in insulation.
Why Insulation reduces energy lost to the surroundings.
Put the heater and thermometer in the holes; add a drop of oil to the thermometer hole.
Why Oil gives good thermal contact, so the reading is accurate.
Record the starting temperature, switch on and start the stopwatch.
Why You need both the energy supplied and the temperature change.
Record the temperature and the energy (or V, I and t) every minute for 10 minutes.
Why Energy = power × time; power = V × I.
Plot temperature against energy transferred and find the gradient of the straight part.
Why Gradient = 1 ÷ (m × c).
Calculate c = energy ÷ (mass × temperature change).
Why This rearranges ΔE = m c Δθ.

Hazard	How to reduce the risk
The heater and block become hot.	Do not touch them until cool.
Electrical equipment near water or oil.	Keep connections dry and use a low-voltage supply.

c = 15 000 ÷ (1.0 × 14.0) = 1070 J/kg °C for this block.
The value is higher than the accepted 900 J/kg °C because energy was lost, so the temperature rise was smaller.

Using a temperature instead of the temperature change in the calculation.
Instead Δθ = final − starting temperature.
Thinking specific heat capacity is the highest temperature a material can reach.
Instead It is the energy needed to raise 1 kg by 1 °C.
Measuring temperature but not energy.
Instead Record energy with a joulemeter, or V, I and t.