Question

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 $(1day=86,400sec)$
\[\begin{align}
  & \text{A}\text{. 776g} \\
 & \text{B}\text{. 7760g} \\
 & \text{C}\text{.11520g} \\
 & \text{D}\text{. 1552g} \\
\end{align}\]

Answer 1

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 $Q=KA\dfrac{\Delta T}{d}t$

Complete step by step answer:
The question has two terms, which are the latent heat of fusion and thermal conductivity. To begin with, let us understand the terms.
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, $K=0.01 J/m^{\circ}C$, $\Delta T=30^{\circ}C.$, $A=1m^{2}$, $d=0.5cm$, $L=334\times 10^{2} J/kg$, $t=86,400sec$
Since in both the cases the energy used in the form of heat is the same, we can equate them.
$Q=mL=KA\dfrac{\Delta T}{d}t$
Substituting, we get
$m\times334\times 10^{2} J/kg=0.01 J/m^{\circ}C\times 1m^{2}\dfrac{30^{\circ}C.}{0.5\times 10^{-2}m}\times 86,400sec$
$m=\dfrac{518400}{33400}=1552g$

So, the correct answer is “Option D”.

Note: The latent heat of fusion is the enthalpy which is a state function, depending on internal energy, pressure and volume. Students tend to confuse between enthalpy and entropy, where entropy is freedom or disorder of the system, and is also a state function which depends on temperature and pressure.