A physics student heats 2.0 kg of water from 25°C to 75°C using an immersion heater. If the specific heat capacity of water is 4200 J/(kg·°C), how much heat energy was supplied to the water?

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Question

A physics student heats 2.0 kg of water from 25°C to 75°C using an immersion heater. If the specific heat capacity of water is 4200 J/(kg·°C), how much heat energy was supplied to the water?

Accepted Answer

420 000 J — Award 1 mark for correct substitution into Q = mcΔθ = 2.0 × 4200 × (75 – 25) = 2.0 × 4200 × 50 = 420 000 J. A is incorrect — uses temperature difference of 25°C only. C is incorrect — uses final temperature of 75°C instead of temperature change. D is incorrect — uses sum of temperatures.

Answer

210 000 J

Answer

420 000 J

Answer

840 000 J

Answer

630 000 J

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