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ExamsJEE MainPhysics

A hot liquid in a container of negligible heat capacity cools at a rate of 3 K/min just before it begins to solidify. While solidifying, its temperature stays constant for 30 min. Assuming the rate of heat loss stays constant, find the ratio of the specific heat capacity of the liquid to the specific latent heat of fusion (in K⁻¹).

  1. 1/30
  2. 1/60
  3. 1/90
  4. 1/120

Correct answer: 1/90

Solution

Let the constant rate of heat loss be P. In the liquid phase just before freezing, this power produces a 3 K/min temperature drop: P = m*s*3 (per minute). During solidification the temperature is constant, so all heat removed over 30 min is latent: m*L = P*30. Substituting P = 3*m*s gives m*L = 3*m*s*30 = 90*m*s, hence s/L = 1/90 K⁻¹.

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