Impulse Physics
IGCSE CP10

Specific Heat Capacity β€” Immersion Heater Method

Edexcel IGCSE Β· CP10

Theory β€” Specific Heat Capacity

Specific heat capacity (c) is the energy needed to raise 1 kg of a substance by 1Β°C. It tells us how much energy a material can store.

Key Equation

Q = m Γ— c Γ— Ξ”T

Q = energy supplied (J) Β· m = mass (kg) Β· c = specific heat capacity (J/kgΒ·Β°C) Β· Ξ”T = temperature rise (Β°C)

Rearranged to find c: c = Q / (m Γ— Ξ”T)

Energy supplied by the heater: Q = V Γ— I Γ— t

So: c = (V Γ— I Γ— t) / (m Γ— Ξ”T)

Known Values

  • Water: 4200 J/kgΒ·Β°C β€” very high, used as a coolant in engines and power stations
  • Aluminium: 900 J/kgΒ·Β°C
  • Iron / steel: 450 J/kgΒ·Β°C
  • Copper: 390 J/kgΒ·Β°C

Sources of Error

  • Heat lost to surroundings β€” insulate the block or beaker with lagging to reduce this
  • Energy used to heat the heater itself β€” use a low-mass heater
  • Thermometer not in good thermal contact β€” ensure probe is fully inserted
  • Delay between switching off and reading temperature β€” read immediately

Procedure

Equipment

Beaker Β· Water Β· Immersion heater Β· Thermometer Β· Power supply Β· Ammeter Β· Voltmeter Β· Stopwatch Β· Electronic balance

1
Measure the mass of the water

Weigh the empty beaker, fill with water, weigh again. Mass of water = difference. Record as m (kg).

2
Set up the circuit

Place the immersion heater and thermometer into the water. Connect: power supply β†’ ammeter (series) β†’ heater β†’ back to supply. Voltmeter in parallel across the heater.

3
Record initial temperature T₁

Before switching on, record the temperature T₁ (Β°C). Set the power supply to your chosen voltage.

4
Switch on, start stopwatch, record V and I

Switch on the heater and start the stopwatch simultaneously. Read and record the voltage (V) and current (A) β€” stir the water gently to ensure even heating.

5
Record temperature every 60 s

Every 60 seconds, record the temperature. Continue for at least 5 minutes (ideally 10).

πŸ’‘ Stir gently before reading β€” this ensures the thermometer reads the bulk water temperature, not a hot spot near the heater.
6
Calculate c

Q = VIt (total energy). Ξ”T = T_final βˆ’ T₁. Then c = Q/(mΞ”T). Plot T vs t β€” the gradient gives Ξ”T/t, so c = VI/(m Γ— gradient).

πŸ”‹ Choose material, set up the apparatus, then press β–Ά Start Heating. Record temperature every 60 s using the button.
Material
Voltage
V 10.0 V
6 V12 V
Mass
m 1.0 kg
0.5 kg2.0 kg
Apparatus Readings
Voltage V10.0 V
Current I4.00 A
Power P = VI40.0 W
Time elapsed0 s
Temperature20.0 Β°C
Ξ”T0.0 Β°C
Experiment

Data Table

No readings yet β€” run the simulation and take readings.

Time
t / s		 Temperature
T / Β°C		 Temp. rise
Ξ”T / Β°C		 Energy supplied
Q = VIt / J		 c = Q/(mΞ”T)
J/kgΒ·Β°C
No readings recorded yet.

Temperature–Time Graph

Best-fit line gradient = Ξ”T/t β†’ c = VI / (m Γ— gradient)

Analysis

Gradientβ€”
Power VIβ€”
Mass mβ€”
Calc. cβ€”
True cβ€”
% errorβ€”

How to read the graph

Draw the best-fit line through the plotted points. Find two well-separated points on the line and calculate:

gradient = Ξ”T / Ξ”t

Then:
c = VΓ—I / (m Γ— gradient)

Questions

Question 1
An immersion heater running at 12 V and 4 A heats 0.50 kg of water for 5 minutes. The temperature rises from 20Β°C to 26.9Β°C. Calculate (a) the energy supplied Q, (b) the specific heat capacity c, and (c) the % error compared to the true value of 4200 J/kgΒ·Β°C.
(a) Q = VIt = 12 Γ— 4 Γ— 300 = 14 400 J. (b) Ξ”T = 26.9 βˆ’ 20 = 6.9Β°C. c = Q/(mΞ”T) = 14 400/(0.50 Γ— 6.9) = 14 400/3.45 = 4174 J/kgΒ·Β°C. (c) % error = |4174 βˆ’ 4200|/4200 Γ— 100 = 0.6% β€” an excellent result, suggesting good insulation and accurate readings.
Question 2
A student heats an aluminium block (mass 1.0 kg) with a 10 V, 4 A heater for 10 minutes. The temperature rises from 20Β°C to 46.7Β°C. Calculate c and the % error. Suggest why the measured value is lower than the true value of 900 J/kgΒ·Β°C.
Q = VIt = 10 Γ— 4 Γ— 600 = 24 000 J. Ξ”T = 46.7 βˆ’ 20 = 26.7Β°C. c = 24 000/(1.0 Γ— 26.7) = 899 J/kgΒ·Β°C. % error = |899 βˆ’ 900|/900 Γ— 100 β‰ˆ 0.1% β€” essentially perfect. If the value were significantly lower than 900, the most likely reason is that the measured temperature rise is larger than it should be β€” possibly because the block was not well insulated, so the thermometer was reading a hot spot near the heater rather than the mean temperature of the block. The energy transferred to the block would then be less than VIt, giving a smaller effective Ξ”T for the real energy transferred, and the calculated c would be too low.
Question 3
A student plots a temperature-time graph and draws a best-fit line with gradient 0.044 Β°C/s. The heater runs at 10 V, 4 A and the mass of water is 1.0 kg. Calculate the specific heat capacity from the graph.
Power P = VI = 10 Γ— 4 = 40 W. Gradient = Ξ”T/Ξ”t = 0.044Β°C/s. Since P = mc Γ— (Ξ”T/Ξ”t) β†’ c = P/(m Γ— gradient) = 40/(1.0 Γ— 0.044) = 909 J/kgΒ·Β°C. This is very close to the true value for aluminium (900 J/kgΒ·Β°C) β€” but for water (true c = 4200 J/kgΒ·Β°C) the gradient should be much smaller: Ξ”T/Ξ”t = P/(mc) = 40/(1.0 Γ— 4200) β‰ˆ 0.0095Β°C/s.