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JEE Advanced 2019 · Paper 1, Physics question 17

A liquid at 30°C is poured very slowly into a Calorimeter that is at temperature of 110°C. The boiling temperature of the liquid is 80°C. It is found that the first 5 gm of the liquid completely evaporates. After pouring another 80 gm of the liquid the equilibrium temperature is found to be 50°C. The ratio of the Latent heat of the liquid to its specific heat will be ______ °C.

[Neglect the heat exchange with surrounding]

Answer and worked solution

Official answer: 270

Let the calorimeter's heat capacity be CC, and the liquid's specific heat ss and latent heat LL.

First 5 g. It evaporates completely only while the calorimeter is above the boiling point, so the calorimeter cools from 110 °C to 80 °C:
30C=5(50s+L).30C=5(50s+L).

Next 80 g. The calorimeter cools from 80 °C to 50 °C while the liquid warms from 30 °C to 50 °C:
30C=80s(20)=1600s.30C=80s(20)=1600s.

Combining.
250s+5L=1600s  ⟹  Ls=270 ∘C.250s+5L=1600s\implies\frac Ls=270\ ^\circ\text{C}.
Answer: 270

Topics:Calorimetry and specific heatLatent heat and phase change
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