Why the specific heat capacity came out 13% too high
A student is asked to measure the specific heat capacity of water. She pours 0.400 kg of water at 20.4 °C into a copper can of mass 0.085 kg (specific heat capacity 385 J kg⁻¹ K⁻¹), lowers a 120 W immersion heater into it, switches on for exactly 240 s while stirring, and records a final temperature of 35.4 °C. She ignores the can and obtains 4800 J kg⁻¹ K⁻¹. The accepted value is 4180 J kg⁻¹ K⁻¹. The can is lagged at the sides but open at the top. The room is at 21 °C. The teacher says the error is not simply random and asks her to account for it quantitatively and to redesign the experiment.
1. Reproduce the student's calculation and explain, with reasons, the direction in which each of her two approximations (ignoring the can, and heat loss) pushes her answer.
Read the worked answer
Energy supplied: E = Pt = 120 x 240 = 28 800 J. Temperature rise: Δθ = 35.4 - 20.4 = 15.0 K. Student's result: c = E/(mΔθ) = 28 800/(0.400 x 15.0) = 28 800/6.00 = 4800 J kg⁻¹ K⁻¹.
Ignoring the can: some of the electrical energy went into the copper, not the water. Crediting all of it to the water inflates the numerator, so c comes out too high. Size of the effect: 0.085 x 385 x 15.0 = 491 J, which corresponds to 491/(0.400 x 15.0) ≈ 82 J kg⁻¹ K⁻¹.
Heat loss to the room: the water is warmer than the 21 °C room for most of the run, so energy escapes. The measured Δθ is therefore smaller than the energy input would produce, the denominator is too small, and again c comes out too high. Both errors lie on the same side, which is why the total is a 15% overshoot rather than a cancellation.
2. Correct the calculation for the can and estimate how much energy was lost to the surroundings. Comment on whether that loss rate is plausible.
Read the worked answer
Correcting for the can: energy to the copper = 0.085 x 385 x 15.0 = 491 J. Energy remaining for the water = 28 800 - 491 = 28 309 J, giving c = 28 309/6.00 = 4720 J kg⁻¹ K⁻¹ - still 13% above 4180.
Energy the water should have received: 4180 x 0.400 x 15.0 = 25 080 J. Total useful energy = 25 080 + 491 = 25 571 J. Energy lost ≈ 28 800 - 25 571 = 3 229 J ≈ 3.2 kJ, about 11% of the input.
Plausibility: spread over 240 s this is an average loss of about 13 W from an open-topped, part-lagged can with a hot heater and a thermometer lead protruding. That is entirely reasonable for apparatus of this kind; a loss four or five times larger would suggest a missing lid or a badly lagged can, and a much smaller one would suggest the discrepancy lies elsewhere, for example a miscalibrated thermometer.
3. The student proposes simply 'heating for longer to reduce the percentage error'. Evaluate this and suggest two changes that would genuinely improve the result.
Read the worked answer
Heating longer does not help. The loss is roughly proportional to the time the can is above room temperature, and so is the energy input. Extending the run by an hour at about 35 °C would lose an even larger fraction, and the water would approach boiling, where evaporation becomes dominant. The percentage error in c does not fall; it may grow.
What does help:
- Start below room temperature and finish above it. Cool the water to about 15 °C, run to about 27 °C. Heat gained from the room while the water is cooler roughly cancels heat lost while it is warmer, removing the systematic error rather than merely shrinking it.
- Use a lid, better lagging and a proper stirrer, so the loss coefficient itself is smaller, and take the temperature with a 0.1 °C thermometer (a 0.2 °C reading error alone is worth about 1.3% in c).
A stronger method is to plot temperature against time before, during and after heating: the pre-heating gradient measures the cooling correction directly, and the gradient during heating gives the corrected Δθ for the energy input, turning an uncontrolled error into a measured one.