Question

A copper block of mass `2.5 kg` is heated in a furnace to a temperature of `500^(@)C` and then placed on a large ice block. What is the maximum amount (approx.) of ice that can melt? (Specific heat copper `= 0.39 J//g^(@)C` heat of fusion of water `= 335 J//g`).

Answer 1

Mass of the copper block, m = 2.5 kg = 2500 g
Rise in the temperature of the copper block, `Delta theta = 500^(@)C`
Specific heat of copper, `C = 0.39 J g^(–1)"@C^(–1)`
Heat of fusion of water, `L = 335 J g^(–1)`
The maximum heat the copper block can lose, Q = mCΔθ
= 2500 × 0.39 × 500
= 2500 × 0.39 × 500
Let `m^(1) g` be the amount of ice that melts when the copper block is placed on the ice block.
The heat gained by the melted ice, `Q = m^(1)L`
`:.m_(1)=(Q)/(L)=(487500)/(335)=1455.22 g`
Hence, the maximum amount of ice that can melt is 1.45 kg