Solved 2024 Specimen Paper ICSE Class 10 Physics — Question 26

Given,

Mass of solid (m) = 60 g

Fall in temperature of solid = (100 - 25) = 75°C

Rise in temperature of water = (25 - 20) = 5°C

Let specific heat capacity of solid be c.

Heat energy given by solid = mc△t
= 60 x c x 75 = 4500 x c .....(1)

Heat energy taken by water = 150 × 4.2 × 5 = 3150 .....(2)

Assuming that there is no loss of heat energy,

Heat energy given by solid = Heat energy taken by water.

Equating equations 1 & 2, we get,

$4500 \times c = 3150 \\[0.5em] c = \dfrac{3150}{4500} \\[0.5em] c = 0.7 \text{ J g}^{-1} \text{K}^{-1} \\[0.5em]$

specific heat capacity of the solid = 0.7 J g^-1 K^-1

Heat capacity of solid = mass (m) x specific heat capacity = 60 x 0.7 = 42 J K^-1

Hence, Heat capacity of solid = 42 J K^-1