Question

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

• A. 776 gms
• B. 7760 gms
• C. 11520 gms
• D. 1552 gms

Answer 1

The correct option is D

1552 gms

$\frac{dQ}{dt}=\frac{KA}{l}d\theta =\frac{0.01\times 1}{0.05}\times 30=6J/sec$
Heat transferred in on day (86400 sec)
$Q=6\times 86400=518400J$
Now $Q=mL\Rightarrow m=\frac{Q}{L}=\frac{518400}{334\times {10}^3}$
$=1.552kg=1552g$.