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A metallic ball of mass 200 g and at 200°C is placed over a block of ice, when 100 g of ice melts. If the specific latent heat of fusion of ice is 336 J g−1

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Question

A metallic ball of mass 200 g and at 200°C is placed over a block of ice, when 100 g of ice melts. If the specific latent heat of fusion of ice is 336 J g−1, find the specific heat capacity of the metallic ball.

Numerical
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Solution

Given: Mass of the metallic ball, m = 200 g

Initial temperature of the ball, θ1 = 200°C

Mass of ice melted, M =  100 g

Specific latent heat of fusion of ice, L = 336 J g−1

The ball cools to 0°C so final temperature θ2 = 0°C

Heat lost by the ball:

Qlost = mc(θ1 − θ2)

= 200 × c × (200 − 0)

= 40000c J

Heat gained by melting ice:

$$Q_{\text{gained}} = M L$$

= 100 × 336

= 33600 J

By the principle of calorimetry,

Heat lost = Heat gained

40000c = 33600

$$c = \frac{33600}{40000}$$

c = 0.84 J g−1°C−1

Therefore, the specific heat capacity of the metallic ball is 0.84 J g−1°C−1.

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Chapter 10: Calorimetry - EXERCISE-2 [Page 247]

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Goyal Brothers Prakashan A New Approach to ICSE Physics [English] Class 10
Chapter 10 Calorimetry
EXERCISE-2 | Q 9. (b) | Page 247
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