An ice box used for keeping eatable cold has a total wall area of $1\;metr{e^2}$ and a wall thickness of $5.0cm$. The thermal conductivity of the ice box is $K = 0.01\;joule/metre{ - ^o}C$. It is filled with ice at ${0^o}C$ along with eatables on a day when the temperature is $30°C$ . The latent heat of fusion of ice is $334 \times {10^3}joules/kg$. The amount of ice melted in one day is ........ $gms$ ($1day = 86,400\;\sec onds$)
Medium
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(d) $\frac{{dQ}}{{dt}} = \frac{{KA}}{l}d\theta $ $ = \frac{{0.01 \times 1}}{{0.05}} \times 30$ $= 6J/sec$
Heat transferred in on day ($86400 sec$)
$\theta = 6 \times 86400 = 518400\,J$
Now $Q = mL$ $⇒$ $m = \frac{Q}{L}$$ = \frac{{518400}}{{334 \times {{10}^3}}}$
$= 1.552 kg = 1552g.$
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