A solid of mass 50 g at 150°C is placed in 100 g of water at 11°C, when the final temperature recorded is 20°C. Find the specific heat capacity of the solid. (Specific heat capacity of water = 4.2 J/g°C).

Given,

Mass of solid m₁ = 50 g,

temperature of solid t₁ = 150°C,

mass of water m₂ = 100 g

temperature of water t₂ = 11°C ,

temperature of mixture t = 20°C

Let specific heat capacity of the solid be c₁

Heat lost by the solid Q₁ = m₁ c₁ (t₁ - t)

= 50 x c₁ x (150 - 20) = 6500 c₁ J

Heat gained by water = m₂ c₂ (t - t₂ )

= 100 x 4.2 x (20 - 11) = 3780 J

If there is no loss of heat, by the principle of calorimetry

Heat lost by the solid = Heat gained by the water

6500 c₁ = 3780

c₁ = $\dfrac{3780}{6500}$ = 0.582 Jg^-1° C^-1