Specific heat capacity and latent heatEdexcel A-Level Physics: Revision notes
Section 1
Specific heat capacity
The specific heat capacity, c, of a substance is the energy needed to raise the temperature of 1 kg of the substance by 1 K without a change of state.
ΔE = mcΔθ
Units: J kg⁻¹ K⁻¹. For water c = 4180 J kg⁻¹ K⁻¹. A temperature change in °C has the same size as in K.
Worked example: the energy to heat 1.5 kg of water from 20 °C to 100 °C is 1.5 × 4180 × 80 = 5.0 × 10⁵ J. A 2.0 kW kettle transferring all its energy would take 5.0 × 10⁵ ÷ 2000 = 250 s.
A large c means that a lot of energy is needed per kelvin, so the temperature changes slowly.
Using the final temperature instead of the temperature change, or forgetting to convert grams to kilograms.
Section 2
Specific latent heat
During a change of state, energy is supplied but the temperature stays constant. The energy increases the potential energy of the molecules as the forces between them are overcome, and does not change the mean kinetic energy.
E = Lm
The specific latent heat, L, is the energy needed to change the state of 1 kg of a substance without changing its temperature. Fusion (melting) for ice is 3.34 × 10⁵ J kg⁻¹ and vaporisation for water is 2.26 × 10⁶ J kg⁻¹.
Worked example: to change 0.10 kg of ice at 0 °C into water at 20 °C: 0.10 × 3.34 × 10⁵ = 3.34 × 10⁴ J to melt, then 0.10 × 4180 × 20 = 8.4 × 10³ J to warm, a total of 4.2 × 10⁴ J.
Vaporisation needs far more energy than fusion because the molecules must be separated completely.
Section 3
Core practical: specific heat capacity
Heat a solid block of known mass with an electric heater of known power and measure the temperature rise.
Energy supplied = Pt (or VIt), so c = Pt ÷ (mΔθ).
Energy is lost to the surroundings, so the temperature rise is too small and the calculated c is too high. Improve the experiment by lagging the block, using a lid, putting oil in the thermometer hole, and plotting a graph of temperature against time or energy: the gradient gives 1/mc.
Section 4
Core practical: specific latent heat
Ice (fusion): place a heater in crushed ice in a funnel. After a time t, measure the mass of water collected. A second, identical funnel without a heater collects the water from ice melted by the room. Subtract it. L = Pt ÷ (m₁ − m₂).
Water (vaporisation): place a heater in boiling water on a balance and record the mass lost in time t. Energy losses make the result too high. Repeat with a different power for the same time and subtract: L = (P₁ − P₂)t ÷ (m₁ − m₂), which cancels the steady loss. Lagging also helps.
Section 5
Core practical: thermistor as a thermostat
An NTC thermistor has a resistance that falls as its temperature rises. Place it in series with a fixed resistor across a supply. The output p.d. across the fixed resistor is V_out = [R_fixed ÷ (R_thermistor + R_fixed)] × V_supply.
As the temperature rises, R_thermistor falls, so V_out across the fixed resistor rises.
To calibrate: place the thermistor in a water bath with a thermometer, raise the temperature in steps, wait for it to settle, and record the output p.d. at each temperature. Plot output p.d. against temperature. To use it as a thermostat, find the p.d. at the required temperature and use it to switch a relay, which turns the heater off.
Must know
- ΔE = mcΔθ and E = Lm; units J kg⁻¹ K⁻¹ and J kg⁻¹
- Temperature is constant during a change of state: the energy raises potential energy
- Energy losses make measured c and L too large; reduce them by insulation or by subtracting a control or second power
- A thermistor with a fixed resistor gives an output p.d. that rises with temperature across the fixed resistor
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Specific heat capacity and latent heat
- A 2.0 kW electric kettle contains 1.5 kg of water at 20 °C. The water is heated to 100 °C and the kettle is then switched off. The specific heat capacity of water is 4180 J kg⁻¹ K⁻¹. In practice the kettle takes 280 s to do this.Calculate the energy transferred to the surroundings and the kettle itself while the water is being heated.2 marks
- A student determines the specific latent heat of fusion of ice. A 60 W heater is placed in crushed ice at 0 °C in a funnel above a beaker. An identical funnel of ice without a heater is set up beside it. After 300 s the heated funnel has produced 62.0 g of melted water and the other funnel has produced 8.0 g.Calculate the specific latent heat of fusion of ice from these results.2 marks
- A student calibrates an NTC thermistor so that it can be used as the sensor in a thermostat for a greenhouse heater. The thermistor is in series with a 4.0 kΩ fixed resistor across a 6.0 V supply, and the output p.d. is taken across the fixed resistor. A relay switches the heater off when the output p.d. rises to a set value.Describe how the student could calibrate the thermistor circuit.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).