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# An ice box used for keeping eatables cool has a total wall area of $1m^{2}$ and a wall thickness of $\text{5}\text{.0cm}$. The thermal conductivity of the ice box is $K=0.01Jm^{\circ}C$. It is filled with ice along with eatables on a day when the temperature is $30^{\circ}C.$ The latent heat of fusion of ice is $334\times 10^{2} J/kg.$ The amount of ice melted in on day is $(1day=86,400sec)$\begin{align} & \text{A}\text{. 776g} \\ & \text{B}\text{. 7760g} \\ & \text{C}\text{.11520g} \\ & \text{D}\text{. 1552g} \\ \end{align}  Verified
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Hint: To find the amount or mass of the ice melted, we need to calculate the latent heat of fusion due ice , and the thermal conductivity needed to melt ice to water. We can equate them as the heat spent is used as the energy to melt the ice.

Formula used: $Q=mL$ and $Q=KA\dfrac{\Delta T}{d}t$

The latent heat of fusion, also called the enthalpy of fusion, is the energy required to convert a solid to liquid at constant pressure. Here, it is the energy needed to convert solid ice to liquid water at constant pressure or atmospheric pressure . The formula used is, $Q=mL$ where $Q$ is the energy(here, in the form of heat), $m$ is the mass of the substance and $L$ latent heat of fusion.
And thermal conductivity is the ability of a material to conduct heat. Here, since, we are converting ice to water, we need to exert energy in the form of heat. Then it is denoted mathematically as $Q=KA\dfrac{\Delta T}{d}t$ where, $K$ is the thermal conductivity of the material, here liquid, $A$ is the area , $d$ is the thickness of the walls, $\Delta T$ is the change in temperature and $t$ is the time taken.
Given that, $K=0.01 J/m^{\circ}C$, $\Delta T=30^{\circ}C.$, $A=1m^{2}$, $d=0.5cm$, $L=334\times 10^{2} J/kg$, $t=86,400sec$
$Q=mL=KA\dfrac{\Delta T}{d}t$
$m\times334\times 10^{2} J/kg=0.01 J/m^{\circ}C\times 1m^{2}\dfrac{30^{\circ}C.}{0.5\times 10^{-2}m}\times 86,400sec$
$m=\dfrac{518400}{33400}=1552g$