Specific heat capacity and specific latent heatEdexcel International 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 (or 1 °C), without a change of state. Its unit is J kg⁻¹ K⁻¹.
The energy transferred when a mass m changes temperature by Δθ is
Worked example: 0.50 kg of water (c = 4180 J kg⁻¹ K⁻¹) is heated by 15 K. E = 0.50 × 4180 × 15 = 3.1 × 10⁴ J.
Do not forget the mass: c is per kilogram, so E = mcΔθ needs all three quantities.
Section 2
Specific latent heat
During a change of state the temperature stays constant even though energy is supplied or removed. The specific latent heat L is the energy needed to change the state of 1 kg of a substance without a change in temperature. Its unit is J kg⁻¹.
The specific latent heat of fusion is for melting or freezing, and the specific latent heat of vaporisation is for boiling or condensing. Vaporisation is much larger.
Worked example: melting 0.30 kg of ice with L = 3.34 × 10⁵ J kg⁻¹ needs E = 0.30 × 3.34 × 10⁵ = 1.0 × 10⁵ J.
Section 3
Combining the equations
A question may involve heating and then a change of state. Work in stages and add the energies.
Electrical heating supplies energy E = Pt = VIt. If no energy is lost, this equals the energy used for heating or for the change of state.
The efficiency can be found from
Energy lost to the surroundings makes the temperature rise smaller than expected.
Section 4
Core practical 13: specific latent heat
To determine the specific latent heat of vaporisation of water, boil water in an insulated container on a balance, using an electric heater.
- Measure the voltage and current to find E = VIt.
- Measure the fall in mass m on the balance over the time t.
- Calculate L = E / m.
Energy lost to the surroundings makes L too large, because less water is vaporised than the supplied energy should give. Lag the container, fit a lid, or repeat at a second power and use the difference between the two runs to cancel the loss.
Section 5
Core practical 12: the thermistor thermostat
An NTC thermistor's resistance falls as its temperature rises. In series with a fixed resistor across a constant supply, it forms a potential divider whose output depends on temperature.
To calibrate it as a thermostat:
- Place the thermistor in a water bath with a thermometer, and connect a voltmeter across the thermistor.
- Change the temperature in steps, stirring and waiting for the thermistor to reach the temperature of the bath.
- Record temperature and p.d. and plot a calibration graph.
- Read off the temperature that gives the p.d. at which the circuit should switch.
The p.d. across the thermistor is V = V_s × R_th / (R_th + R). The fixed resistor sets where the switching point lies.
For a calibration curve, give a reason for stirring and for waiting before each reading: both make the thermistor's temperature match the thermometer.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Specific heat capacity and specific latent heat
- A student heats 0.50 kg of water in a well-insulated beaker using a 60 W immersion heater. The specific heat capacity of water is 4180 J kg⁻¹ K⁻¹. The heater is switched on for 520 s.The student measures a temperature rise of 12 K, not 15 K. Calculate the percentage of the electrical energy supplied that was transferred to the water.2 marks
- A catering company uses ice to keep food chilled and steam to cook it. The specific latent heat of fusion of ice is 3.34 × 10⁵ J kg⁻¹ and the specific latent heat of vaporisation of water is 2.26 × 10⁶ J kg⁻¹. A cool box contains 0.30 kg of ice at 0 °C, which absorbs energy from the food at an average rate of 20 W.Calculate the energy released when 0.040 kg of steam at 100 °C condenses to water at 100 °C.2 marks
- A student determines the specific latent heat of vaporisation of water. A beaker of water is placed in an insulated container on a top-pan balance, and a 12.0 V electric heater with an ammeter in series is lowered into the water. Once the water is boiling steadily, the student records the balance reading and switches the heater on for a measured time, then records the balance reading again.The ammeter reads 4.20 A and the heater is on for 300 s. The balance reading falls by 6.0 g. Calculate the specific latent heat of vaporisation of water from these results.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).