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Q. An ice box used for keeping eatables cool has a total wall area of $1 \,m ^{2}$ and a wall thickness of $5.0 \,cm$. The thermal conductivity of the ice box is $K=0.01 \,J / m ^{\circ} C$. It is filled with large amount of ice at $0^{\circ} C$ along with eatables on a day when the temperature is $30^{\circ} C$. The latent heat of fusion of ice is $334 \times 10^{3} J / kg$. The amount of ice melted in one day is ( $1$ day $=86,400\, s )$

Thermal Properties of Matter

Solution:

Quantity of heat transferred through wall will be utilized in melting of ice.
$ Q=\frac{K A \Delta \theta t}{\Delta x}=m L $
$\therefore $ Amount of ice melted $ m=\frac{K A \Delta \theta t}{\Delta x L} $
$\therefore m=\frac{0.01 \times 1 \times(30-0) \times 86400}{5 \times 10^{-2} \times 334 \times 10^{3}}$
$=1.552 \,kg $ or $ 1552 \,gm$