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. The maximum amount of ice that can be melt, will be :

Given :

Specific heat of copper = 0.4 Jg⁻¹ °C⁻¹

Heat of fusion of water = 400 Jg⁻¹

Concept:

When copper block after getting heated by a furnace is kept in a block of ice, the heat is liberated by copper block and recieved by block of ice i.e.

Heat lost by copper block = Heat gain by ice

Calculation:

Given:

mc = mass of copper block = 2.5 kg, tc = 500°C, Ccopper = 0.4 Jg−1°C−1

L = heat of fusion of water = 400 Jg−1

Heat lost by copper block = Heat gain by ice

m_cC_coppert_c = m_iceL

2.5 × 0.4 × 500 = m_ice × 400

mice = 1.25 kg