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A 1 kg ball of copper at 80⁰C is dropped into 500 g of water at 35⁰C. Assuming no heat is lost, what will be the equilibrium temperature? Assume specific heat of copper as 390 J/kg.K and specific heat of water as 4200 J/kg.K

A 1 kg ball of copper at 80⁰C is dropped into 500 g of water at 35⁰C. Assuming no heat is lost, what will be the equilibrium temperature? Assume specific heat of copper as 390 J/kg.K and specific heat of water as 4200 J/kg.K Correct Answer 42.05°C

Let the equilibrium temperature be T.

For thermal equilibrium:

Heat lost by copper ball from 80°C to T = heat gained by water from 35°C to T

Specific heat of copper = 390 J kg-1K-1 specific heat of water = 4200 J kg-1K-1

Specific heat = m × c × (T2 - T1)

(1) × (390) × (80 - T) = (0.5) × (4200) × (T - 35)

31200 - 390 T = 2100 T - 73500

2490 T = 104700

T = 42.05⁰C

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