Course: Combined Science + Separate Physics | Tier: Foundation + Higher | Linked equations: ΔE = mcΔθ, E = Pt
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Aim
To determine the specific heat capacity of a material, e.g. an aluminium block.
Method
- Measure the mass of the block with a balance.
- Put the immersion heater and thermometer into the holes in the block (a few drops of water in the thermometer hole improves contact).
- Wrap the block in insulation.
- Connect the heater to a power supply with a joulemeter (or an ammeter and voltmeter).
- Record the starting temperature, switch on, and record the temperature and energy transferred every minute for 10 minutes.
- Plot temperature against energy transferred. Specific heat capacity c = 1 ÷ (mass × gradient).
| Independent variable | Energy transferred (J) |
| Dependent variable | Temperature (°C) |
| Control variables | Mass and material of the block; amount of insulation |
Questions
Q1 (F) Name the independent and dependent variables.
Show answer
Independent: energy transferred to the block (or heating time). Dependent: temperature of the block.
Q2 (F) Why is the block wrapped in insulation?
Show answer
To reduce the energy transferred to the surroundings, so more of the energy supplied raises the temperature of the block. This makes the value of c more accurate.
Q3 (F) Why are a few drops of water put into the thermometer hole?
Show answer
It improves the thermal contact between the block and the thermometer, so the thermometer reads the block’s temperature accurately.
Q4 (F) State one hazard and how to reduce the risk.
Show answer
The heater and block get very hot and could cause burns. Don’t touch them while on, and let them cool before handling.
Q5 (F/H) A 1.0 kg block is given 18 000 J of energy. Its temperature rises from 20 °C to 40 °C. Calculate the specific heat capacity.
Show answer
F: ΔE = m c Δθ
I: 18 000 = 1.0 × c × 20
F: c = 18 000 ÷ 20
A: 900 J/kg °C
Q6 (F/H) The heater is connected to 12 V and draws 4.0 A for 5 minutes. The 1.0 kg block warms by 16 °C. Calculate the specific heat capacity.
Show answer
Step 1 – power: P = V I = 12 × 4.0 = 48 W
Step 2 – energy: 5 minutes = 300 s; E = P t = 48 × 300 = 14 400 J
Step 3: ΔE = m c Δθ, so 14 400 = 1.0 × c × 16
c = 14 400 ÷ 16 = 900 J/kg °C
Q7 (F/H) A student’s value of c is higher than the true value. Explain why.
Show answer
Some energy is transferred to the surroundings (and to heat the heater and thermometer), not the block. So the temperature rise is smaller than it should be for the energy measured, which makes the calculated c too high.
Q8 (H) Explain how to find c from a graph of temperature (y-axis) against energy transferred (x-axis).
Show answer
Find the gradient of the straight part of the line (Δθ ÷ ΔE). Since ΔE = mcΔθ, the gradient = 1 ÷ (mc), so c = 1 ÷ (m × gradient). Ignore the curved start, when the heater is still warming up.
