Question

a solid of mass 150g is heated to 100c and dropped into a 20c 500ml of water and what is the equilibrium temperature of the mixture?​

Answer 1

Answer:

Explanation:

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

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Answer 2

A solid of mass 150g is heated to 100°C and dropped into a 20°C , 500ml of water. the specific heat capacity of the solid is 0.5J/g/°C.

We have to find the equilibrium temperature of the mixture.

concept :

• heat always flows from higher temperature to lower temperature.
• heat lost by the object of higher temperature equals heat gained by the object of lower temperature.

here, heat flows from the solid of mass 150g to water. hence heat lost by solid equals heat gained by water.

i.e., heat lost by solid = heat gained by water

⇒m₁s₁(T₁ - T) = m₂s₂(T - T₂)

• here mass of solid = m₁ = 150g
• specific heat capacity of solid, s₁ = 0.5 J/g/°C
• Temperature of solid, T₁ = 100°C
• mass of water , m₂ = 500ml × 1g/ml = 500g [ ∵ density of water is 1 g/ml ]
• temperature of water , T₂ = 20°C
• specific heat capacity of water , s₂ = 4.2 J/g/°C

⇒150 × 0.5 × (100 - T) = 500 × 4.2 × (T - 20)

⇒75(100 - T) = 2100(T - 20)

⇒(100 - T) = 28(T - 20)

⇒100 - T = 28T - 560

⇒660 = 29

⇒T = 660/56 ≈ 12 °C

Therefore the equilibrium temperature would be 12°C.