A2 Daily A Level Physics question
In a lab, a 500 W immersion heater warms 0.50 kg of water held in a 0.50 kg aluminium calorimeter. The heater runs for 2.0 minutes. Take c(water) = 4200 J kg⁻¹ °C⁻¹ and c(aluminium) = 900 J kg⁻¹ °C⁻¹. Assuming negligible heat losses, which estimate is closest to the temperature rise of the water-and-cup system?
Answer
The correct answer is D.
Correct: D — about 24 °C. Energy supplied is 500 × 120 = 60,000 J; total heat capacity is (0.50 × 4200) + (0.50 × 900) = 2550 J °C⁻¹, so the temperature rise is 60,000/2550 ≈ 23.5 °C. A uses 500 J instead of 60,000 J, confusing power (J s⁻¹) with energy delivered over 120 s. B treats the whole 1.0 kg as if it were water, using 4200 J kg⁻¹ °C⁻¹ for both water and aluminium, giving 60,000/(1.0 × 4200) ≈ 14 °C. C ignores the aluminium cup’s heat capacity, using only the water term 0.50 × 4200, giving 60,000/2100 ≈ 29 °C. D includes both water and aluminium, so it accounts for all the energy absorbed by the system.