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ICSE Class 10 Physics Question 9 of 25

Solved 2025 Specimen Paper ICSE Class 10 Physics — Question 22

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Question

Question 3(iv)

A solid of mass 60 g at 100°C is placed in 150 g of water at 20°C. The final steady temperature is 25°C. Calculate the heat capacity of solid.

[Specific heat capacity of water = 4.2 Jg⁻¹K⁻¹]

Answer

Given,

Mass of solid (ms) = 60 g

Mass of water (mw) = 150 g

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

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

Specific heat capacity of water (cw) = 4.2 Jg⁻¹K⁻¹

Let specific heat capacity of solid be cs.

Then,

Heat energy given by solid = mscs△ts = 60 x c x 75 = 4500 x c …..(1)

Heat energy taken by water = mwcw△tw = 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 × cs = 3150

cs=31504500\text c_\text s = \dfrac{3150}{4500}

cs = 0.7 J g⁻¹ K⁻¹

Now,

Heat capacity of solid = mass x specific heat capacity = 60 x 0.7 = 42 J K⁻¹

Hence, Heat capacity of solid = 42 J K⁻¹.