Physics · Thermal Properties of Matter

JEE Advanced 2019 — Paper 1 — Question 15

A liquid at $30^{\circ} \mathrm{C}$ is poured very slowly into a Calorimeter that is at temperature of $110^{\circ} \mathrm{C}$ . The boiling temperature of the liquid is $80^{\circ} \mathrm{C}$ . 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 $50^{\circ} \mathrm{C}$ . The ratio of the Latent heat of the liquid to its specific heat will be $\_\_\_\_$ ${ }^{\circ} \mathrm{C}$ . [Neglect the heat exchange with surrounding]

Answer: 270

Numerical answer — enter this value.

Step-by-step solution

Let specific heat of liquid = $c$ , latent heat = $L$ , heat capacity of calorimeter = $C$ .

First 5 g completely evaporates:

Q = 5c(80-30)+5L = 250c+5L

Calorimeter cools from $110^\circ$ C to $80^\circ$ C:

250c+5L = 30C \qquad (1)

After pouring another 80 g, equilibrium temperature = $50^\circ$ C. (The first 5 g has evaporated.)

Heat gained by liquid:

Q = 80c(50-30) = 1600c

Calorimeter cools from $80^\circ$ C to $50^\circ$ C:

1600c = 30C \qquad (2)

From (2): $C=\dfrac{1600c}{30}$

Substitute in (1):

250c+5L = 1600c \Rightarrow 5L = 1350c

\dfrac{L}{c} = 270^\circ\text{C}

Answer key and solution verified before publishing.

Practise Thermal Properties of Matter

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Advanced 2019
Paper: Paper 1
Subject: Physics
Chapter: Thermal Properties of Matter
Topic: Thermometry and Calorimetry