Question

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 the solid.

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

Answer 1

Given: Mass of hot solid (m₁) = 60 g

Mass of water (m₂) = 150 g 

Initial temperature of solid (t₁) = 100°C

Specific heat of solid (c₁) = ?

Water of mass (m₂) = 150 g

Initial temperature of water (t₂) = 20°C

Specific heat of water (c₂) = 4.2 J g⁻¹ K⁻¹

Final temperature of mixture (t_f) = 25°C = 298 K

According to principle of calorimetry,

Heat lost by solid = Heat gained by water

m₁ c₁ (t₁ − t_f) = m₂ c₂ (t_f − t₂)

60 × c₁ × (100 − 25) = 150 × 4.2 × (25 − 20)

c₁ × 60 × 75 = 150 × 4.2 × 5

c₁ = `(150 xx 4.2 xx 5)/(60 xx 75)`

c₁ = `3150/4500`

c₁ = 0.7 J g⁻¹ K⁻¹

Heat capacity of solid = Mass (m) × Specific heat of solid

= 60 × 0.7

= 42 J K⁻¹

Hence, the heat capacity of solid is 42 J K⁻¹.