#### 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 of ice that can melt? (Specific heat of copper = 0.39 J g^{–1 }K^{–1}; heat of fusion of water = 335 J g^{–1}).

#### Solution 1

Mass of the copper block, *m *= 2.5 kg = 2500 g

Rise in the temperature of the copper block, Δ*θ* = 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* = *mC*Δ*θ*

= 2500 × 0.39 × 500

= 487500 J

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.

#### Solution 2

Here, mass of copper block, m = 2.5 kg = 2500 g

Fall in temperature, `triangle T = 500 - 0= 500^@C`

Specific heat of copper, `c = 0.39 J g^(-1) ""^@C^(-1)`

Latent heat of fusion, `L = 335 J g^(-1)`

Let the mass of ice melted be m'

As, Heat gained by ice = Heat lost by copper

`:. m'L = mc triangle T`

`m' = (mc triangleT)/L`

`m' = (2500xx0.39xx500)/335 =1500 g = 1.5 kg`