Specific Heat of Aluminium: Heat Loss and Cooling Correction
A student determines the specific heat capacity of aluminium using an electrical heater embedded in a 0.500 kg block. The heater is connected to a 12.0 V supply and draws 2.00 A for 180 s. A thermometer in the block reads 21.0 °C before and 28.5 °C after, with a maximum reading of 28.8 °C reached 30 s after the current is switched off. The block is wrapped in lagging, but the thermometer and heater leads pass through the lagging. The student calculates c = (12.0 × 2.00 × 180)/(0.500 × 7.5) = 1150 J kg⁻¹ K⁻¹. The accepted value is 900 J kg⁻¹ K⁻¹. The student says the value is too high because heat is lost to the surroundings, and that the result would be even higher without lagging.
1. Calculate the student's value and the percentage difference from the accepted value. Explain why heat loss during heating makes the calculated c too high, not too low.
Read the worked answer
c = E/(m ΔT) = (12.0 × 2.00 × 180)/(0.500 × (28.5 − 21.0)) = 4320/(0.500 × 7.5) = 4320/3.75 = 1152 J kg⁻¹ K⁻¹. The percentage difference from 900 is (1152 − 900)/900 × 100% = 28%. Heat loss during heating means that the electrical energy supplied is split: E_elec = m c ΔT + Q_loss. The student's calculation assumes all E_elec goes into the block, so c_calc = E_elec/(m ΔT) = c_true + Q_loss/(m ΔT). Since Q_loss is positive, c_calc is too high, not too low. Lagging reduces Q_loss, so without lagging the calculated value would be even higher. This direction is correct. However, the student's use of 28.5 °C rather than the maximum 28.8 °C also matters: using 28.8 °C gives ΔT = 7.8 K and c = 4320/(0.500 × 7.8) = 1108 J kg⁻¹ K⁻¹, which is still 23% high. So heat loss is a plausible major cause, but the value is not accurate enough to stop there.
2. The student uses 28.5 °C, not the maximum 28.8 °C. Explain how a cooling correction should be applied and whether it would reduce the discrepancy. Include the direction of the correction.
Read the worked answer
A cooling correction estimates the heat that would have been lost during the heating period by using the cooling curve after the heater is switched off. Plot temperature against time after switch-off, when the block is cooling under the same lagging conditions. Extrapolate this cooling curve back to the time of switch-off (or to the time when the heater current stops) to find the temperature the block would have had if no heat had been lost during heating. The difference between that extrapolated temperature and the measured temperature at switch-off is added to the measured temperature rise. Because the corrected ΔT is larger than 7.5 K, the corrected c = E/(m ΔT_corrected) is smaller than 1152 J kg⁻¹ K⁻¹, moving it towards 900 J kg⁻¹ K⁻¹. The maximum reading of 28.8 °C is not itself the cooling correction; it is the observed maximum after switch-off and may include heater lag. Using 28.8 °C reduces c to about 1108 J kg⁻¹ K⁻¹, but a proper cooling correction should reduce it further. The correction is therefore in the right direction and would reduce, but probably not eliminate, the discrepancy.
3. State two further improvements and explain which would have the larger effect on the result.
Read the worked answer
Two improvements: (i) include the heat capacity of the heater, thermometer and any container in the energy balance, or calibrate the apparatus with a material of known c; (ii) stir the block or use a better thermal contact so the thermometer reads the mean temperature, and use a smaller heating rate with a longer time to reduce temperature gradients. A third improvement is to insulate the block better and use a lid with minimal lead holes. The largest effect is likely the cooling correction and the heat capacity of the heater: the heater and thermometer have non-zero thermal capacity, so some of the 4320 J heats them rather than the aluminium. Ignoring them makes the calculated c too high. Better lagging reduces Q_loss but cannot eliminate the need for a cooling correction. Repeating with two different heating times and plotting E against ΔT would give a gradient m c and allow a correction for a constant heat loss per unit time; that is more reliable than a single measurement.