Question

A copper block of mass $2.5\text{kg}$ is heated in a furnace to a temperature of ${500}^{\circ }C$ and then placed on a large ice block. What is the maximum amount of ice that can melt? $\text{Specific heat of copper}=0.39\frac{J}{g{K}^{\prime }},\text{heat of fusion of water}=335J/g$

Answer 1

Given, mass of the copper block, $m=2.5\text{kg}=2500g$

Rise in temperature of the block, $\Delta T={500}^{\circ }C$

Specific heat of copper, $c=0.39J/gk$

Latent Heat of fusion of water, $L=335J/g$

Step 1: Find the lost by copper block.

The heat lost by the copper block,

$Q=mc\Delta T$

$=2500\times 0.39\times 500$

$=487500J$

Step 2: Find mass of ice that can be melt.

According to principle of calormetry. heat lost by the copper block will be used to melt the ice (when the copper block is placed on the ice block)

Let ${}^{\prime }{m}^{\prime }$ be the mass of the ice melted

Therefore, heat gained by ice = heat lost by copper

$Q=m^{\prime }L$

$m^{\prime }=\frac{Q}{L}$

$m^{\prime }=\frac{487500}{335}$

$m^{\prime }=1455g=1.455\text{kg}$