An experiment takes 10 minutes to raise the temperature of water in a container from 0 °C to 100 °C, and another 55 minutes to convert it totally into steam, by a heater supplying heat at a uniform rate. Neglecting the specific heat of the container and taking the specific heat of water to be 1 cal…

Question

An experiment takes 10 minutes to raise the temperature of water in a container from 0 °C to 100 °C, and another 55 minutes to convert it totally into steam, by a heater supplying heat at a uniform rate. Neglecting the specific heat of the container and taking the specific heat of water to be 1 cal g⁻¹ ° C⁻¹, the heat of vaporization according to this experiment will come out to be

Accepted Answer

(3) 550 cal g⁻¹. Idea: the heater delivers heat at a constant rate, so the heat supplied is proportional to the time, and the mass of water cancels between the two stages. Stage 1, 10 minutes: the heat per gram is cΔ T=1×100=100 cal g⁻¹. So the heater supplies (100)/(10)=10 cal g⁻¹ per minute. Stage 2, 55 minutes at 100 °C: all of that heat goes into the change of state, none into raising the temperature. L=10×55=550 cal g⁻¹ — a little above the accepted 540, which is what the experiment's losses look like.

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