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).
Mass of the copper block, m = 2.5 kg = 2500 g
Rise in the temperature of the copper block, Δθ = 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
= 487500 J
Let m1 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 = m1L
`:.m_1 =Q/L = 487500/335 = 1455.22 g`
Hence, the maximum amount of ice that can melt is 1.45 kg.
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`