Specific Heat Capacity of Aluminium: Too High or Too Low?
In a Grade 10 thermal physics practical, students determine the specific heat capacity of an aluminium block. The block has mass 1.00 kg and contains a 24 W electrical heater and a thermometer. With the heater on for 300 s, the temperature rises from 20.0 °C to 26.5 °C. Using c = Pt / (mΔT), a group obtains 1110 J kg⁻¹ K⁻¹. The accepted value is 900 J kg⁻¹ K⁻¹. The students are confident because their temperature rise is clear and the stopwatch is digital. One student says the result is too high, so heat must have been gained from the room; another says heat loss would make the result too low. The teacher knows the heater is not fully embedded, the block is uninsulated, and the thermometer has a 0.5 °C resolution. The class must decide whether to repeat, correct, or evaluate the method.
1. Calculate the percentage difference and explain whether heat loss to the surroundings would make the measured specific heat capacity too high or too low. Show the reasoning with the equation.
Read the worked answer
The measured value is c = Pt / (mΔT) = (24 × 300) / (1.00 × 6.5) = 1108 J kg⁻¹ K⁻¹, approximately 1110. The percentage difference is (1110 - 900) / 900 × 100 ≈ 23%.
If heat is lost, the block receives less than Pt, so its temperature rise ΔT is smaller than it would be with perfect insulation. Since ΔT is in the denominator, a smaller ΔT makes the calculated c larger. The measured value is therefore too high, not too low. The same direction of bias follows if the heater is not fully embedded or if the thermometer lags: less energy reaches the block or the measured rise is too small, both increasing c.
2. The student says heat loss would make the result too low. Diagnose the reasoning error and propose a quick whole-class demonstration or thought experiment to correct it.
Read the worked answer
The student is confusing 'less energy supplied to the block' with 'less energy counted'. In the equation, the full electrical energy Pt is assumed to enter the block. If some is lost, the block's temperature rise is smaller, so dividing by a smaller ΔT increases the result. A quick numerical check: if c were 900 J kg⁻¹ K⁻¹, the expected rise would be Pt/(mc) = 7200/900 = 8.0 °C; the observed 6.5 °C is too small, which is why the calculated c is too large. Ask students to compare expected and observed ΔT, then change the equation by inserting a heat-loss term Q_lost on the left-hand side and solve for c. This makes the direction of bias explicit without a lengthy derivation.
3. Recommend an improved procedure that could realistically be done in one lesson, and explain how the class could use a graph to reduce the effect of heat loss.
Read the worked answer
Insulate the block, lag the heater leads, ensure the heater is fully embedded, and allow the block to reach a steady starting temperature before timing. Take readings every 30 s and plot temperature against time. At the start, heat loss is small because the block is near room temperature; the initial gradient is dT/dt = P/(mc), so using the initial gradient reduces the heat-loss effect. Repeat and average, and state uncertainties in mass, voltage, current, time and temperature. If time allows, heat to about 10 °C above room temperature and record a cooling curve to estimate the loss correction. Students should compare the corrected value with the accepted value and explain why the correction moves the result in the expected direction.