Thermal Properties of Matter

Question : 13

Total: 22

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 =0.39Jg–1K–1; heat of fusion of water =335Jg–1).

Solution:

Here, mass of copper block, m1=2.5 kg
Specific heat of copper,
C=0.39Jg−1K−1
=0.39×103J kg−1K−1
Temperature of furnace, ΔT=500°C
Latent heat of fusion,
L=335Jg−1
=335×103Jkg−1
If Q be the heat absorbed by the copper block, then
Q=m1CΔT... (i)
Let m2(kg) be the mass of ice melted when copper block is placed on it,∴Q=m2L... (ii)
∴ from (i) and (ii), we get, m1CΔT=m2L
m2=

m1CΔT

2.5×0.39×103×500

335×103

=1.155kg=1.5kg.

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