Specific Heat of Water: Correcting for the Calorimeter and Heat Loss
A student determines the specific heat capacity of water using an electric heater. Water of mass 200 g is placed in a copper calorimeter of mass 80 g (c_Cu = 385 J kg⁻¹ K⁻¹). A 12.0 V supply drives 2.50 A through the heater for 300 s. The temperature rises from 20.1 °C to 30.3 °C. The student calculates c = IVt/(mΔT) = 4410 J kg⁻¹ K⁻¹ and compares this with 4180 J kg⁻¹ K⁻¹. The heater is switched off and the temperature continues to rise for a few seconds. The calorimeter is open, and the room is 2 °C cooler than the starting water. The student asks whether the result is acceptable, why it is high, and what correction would bring it closer.
1. Calculate the uncorrected specific heat capacity and identify whether the error is random or systematic.
Read the worked answer
Q = IVt = (2.50)(12.0)(300) = 9000 J. ΔT = 30.3 − 20.1 = 10.2 K, m = 0.200 kg. Uncorrected c = 9000/[(0.200)(10.2)] = 4.41 × 10³ J kg⁻¹ K⁻¹. This is about 5.5% above 4180. The error is systematic, not random: the calculation assumes all 9000 J entered the water, but some heated the calorimeter and some was lost to the cooler room. Both omissions make ΔT smaller than it should be for the water alone, so the calculated c is too high. The post-switch-off rise partly offsets this, because it increases the recorded ΔT, but it is not controlled.
2. Apply the calorimeter correction and calculate the corrected value. Is the remaining discrepancy acceptable?
Read the worked answer
Heat absorbed by the copper calorimeter: Q_cal = m_Cu c_Cu ΔT = (0.080)(385)(10.2) = 314 J. Energy to water: 9000 − 314 = 8686 J. Corrected c = 8686/[(0.200)(10.2)] = 4.26 × 10³ J kg⁻¹ K⁻¹. This is only 1.9% above 4180. The remaining excess is consistent with heat loss from the open calorimeter to a room 2 °C cooler; correcting for that would lower the value further. It could also include thermometer calibration error. The corrected result is therefore acceptable within about 2%, but the uncorrected 4410 should not be reported as the specific heat capacity of water.
3. Design an improved procedure to reduce both errors and explain how to treat the post-switch-off temperature rise.
Read the worked answer
Use a lagged, covered calorimeter with a stirrer and a calibrated thermometer; measure the water mass by difference. Start with water slightly below room temperature and finish slightly above, so the average temperature is near room temperature and net heat exchange is small. Record temperature every 15–30 s, switch off at a known time, and extrapolate the cooling curve back to the switch-off time to correct for the heater's thermal lag and for cooling. Alternatively use a joulemeter to measure electrical energy directly. Include the calorimeter heat capacity in the calculation and repeat to estimate uncertainty.