Question

A copper sphere of mass 500 g is heated to 100 °C and then introduced into a copper calorimeter containing 100 g of water at 20 °C. Find the maximum temperature of the mixture, if the mass of the calorimeter is 100 g and the specific heat of the calorimeter is 0.1 cal/(g.°C).

Answer 1

Data: m = 500 g, c = 0.1 cal/(g.°C), T’= 100 °C, m₁ = 100 g, c₁ = 1 cal/(g.°C), T₁ = 20°C, m₂ = 100 g, c₂ = 0.1 cal/(g.°C), T₂ = 20 °C, T= ?

Heat lost by the sphere = heat gained by the water and the calorimeter.

∴ mc (T’ – T) = m₁ c₁ (T – T₁ ) + m₂ c₂ (T – T₂ )

∴ 500 g × o.l cal/(g.°C) × (100 °C – T)

= 100 g × 1 cal/(g.°C) × (T – 20 °C) + 100 g × 0.1 cal/(g.°C) × (T – 20 °C)

∴ 50 (100 °C – T) = 100 × (T – 20 °C) + 10 × (T – 20 °C)

∴ 50 (100 °C – T) = 110 × (T – 20 °C) 

∴ 500 °C – 5T = 11T – 220 °C 

∴ 16T = 720 °C

∴ T = \(\cfrac{720^\circ C}{16}\) = 45 °C

Maximum temperature of the mixture = 45 °C.