Heating Water with an Immersion Heater: An Impossible Efficiency
A student uses a 12.0 V, 2.0 A immersion heater to heat 250 g of water in an insulated polystyrene cup. A digital balance is used to measure the water, and a thermometer reads 18.0 °C before heating and 33.0 °C after 8.0 min. The student computes the energy supplied as (12.0 V)(2.0 A)(480 s) = 11 520 J and predicts a temperature rise of about 11.0 °C using c_water = 4.18 J g⁻¹ K⁻¹. The observed rise is 15.0 °C, about 36% larger. The student claims the heater must actually be delivering about 33 W rather than 24 W. A second student says the water mass must be wrong. The cup's heat capacity is small but not zero, and the thermometer has 1 °C graduations. The power supply displays 12.0 V throughout, but the current display flickers between 1.9 A and 2.1 A. The room is 22 °C.
1. Calculate the expected temperature rise from the electrical energy and the rise implied by the temperature data. What average power would be needed to explain the observation?
Read the worked answer
Electrical energy Q_el = VIt = 12.0 × 2.0 × 480 = 11 520 J. Predicted ΔT = Q/(mc) = 11 520/(250 × 4.18) = 11.0 K. The observed ΔT = 33.0 − 18.0 = 15.0 K. The heat actually absorbed by the water would be 250 × 4.18 × 15.0 = 15 675 J, implying an average power of 15 675/480 = 32.7 W. That is about 2.7 A at 12.0 V, not the measured 2.0 A. The 36% excess is far outside the current flicker and the thermometer's 1 °C resolution, which is only about 1 K in a 15 K rise.
2. Which explanation is more likely: the power is 33 W, or the water mass is about 184 g? Use the voltage and current data and identify a procedural error that would produce a 250 g reading.
Read the worked answer
The 33 W explanation requires 2.7 A, while the display flickers only between 1.9 and 2.1 A; a 35% current error is not plausible. A mass error is more likely: if the recorded 250 g included the cup, the water mass would be smaller. The required mass is m = Q/(cΔT) = 11 520/(4.18 × 15.0) = 184 g, so a cup of about 66 g would explain the reading. Decisive check: weigh the dry empty cup, then tare and weigh only the water. A thermometer calibration error would have to be large and would not be canceled if it is a scale error; the stated 1 °C graduations make that less likely.
3. If the actual water mass is 184 g, calculate the implied specific heat and assess whether the result is physically sensible.
Read the worked answer
Using 184 g: c = Q/(mΔT) = 11 520/(184 × 15.0) = 4.17 J g⁻¹ K⁻¹, within about 0.3% of 4.18. This is physically sensible for water and much more consistent than the 36% discrepancy. It supports the mass error, not a 33 W heater. A small heat loss to the cup would make the calculated c slightly low, but the cup is insulated and its heat capacity is small; the main correction is the mass. Repeating with a tared water mass would test the conclusion.