Question

A solid of mass 50g at 150°C is placed in 100g 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 .

Answer 1

m.c.t [where t is change in temperature]

Given :

hot body :                                  cold body :
m = 50g                                     m = 100g
c = ?                                           c = 4.2
t = 150 c                                      t = 11 c
            final temp = 20 c


Solution :

m.c.t = m.c.t
50 X c X (150 - 20) = 100 X 4.2 X (20-11)
          50 X c X 130 = 420 X 9
                             c = 420 X 9 / 50 X 130
                             c = 0.58 J/g/degree c


                                    Hope it helps................

Answer 2

Let subscript 1 denote solid (hot body) and subscript 2 denote water (cold body)
Initial temp. of solid T₁=150°C
Mass of solid m₁ = 50 g = 0.05 Kg
Initial temp. of water T₂=11°C
Mass of water m₂= 100 g = 0.1 Kg
Final temp of both T= 20°C
Specific heat capacity of water c₂=4.2 J/g°C 
Let Specific heat capacity of solid be c₁
now,
heat gained by water = heat lost by solid
heat gained/lost = mcΔT
∴ m₁c₁ΔT₁ = m₂c₂ΔT₂
⇒ 0.05*c₁*(150-20) = 0.1*4.2*(20-11)
⇒ 0.05*130*c₁ = 0.1*4.2*9
⇒ c₁ = 0.5815 J/g°C
Answer: Specific heat capacity of the solid is 0.5815 J/g°C