Question

A cubical thermocol ice box of side $20\,cm$ has a thickness $5\,cm$ . If $5\,kg$ of ice is put in the box, the amount of ice remaining after $10\,hours$ is:
(The outside temperature is $50$ and the coefficient of thermal conductivity of thermocol $= 0.01\,J{s^{ - 1}}{m^{ - 2}}$ , latent heat of fusion of ice $= 335 \times {10^3}\,Jk{g^{ - 1}}$ )
A. $3.7\,kg$
B. $3.9\,kg$
C. $4.7\,kg$
D. $4.9\,kg$

Answer 1

Here we have to first find the total surface area of the box, then the total heat entering the box through all sides. Then we have to use the relation between heat and thermal conductivity to find the heat.
At last we have to find the mass of ice through heat and latent heat of fusion.

Complete step by step answer:
Given,
Length of each side, $l = 20\,cm = 20 \times {10^{ - 2}}\,m$
Thickness, $x = 5\,cm = 5 \times {10^{ - 2}}\,m$
Since, the thermocole is cubical,
So total surface area will be:
$\Rightarrow a = 6{l^2} \\\ \Rightarrow a = 6 \times {\left( {20 \times {{10}^{ - 2}}\,m} \right)^2} \\\ \Rightarrow a = 24 \times {10^{ - 2}}\,{m^2} \\\$
The outside temperature $= 50$
So, change in temperature,
$\Delta T = 50 - 0 = 50$
Thermal conductivity, $K = 0.01\,J{s^{ - 1}}{m^{ - 2}}$
Time,
$t = 10\,hr \\\ \Rightarrow t= 36 \times {10^3}\,s \\\$
Latent heat of fusion, ${L_f} = 335 \times {10^3}\,Jk{g^{ - 1}}$
So, total heat entering the box is:
$Q = \dfrac{{K \times a \times \Delta T \times t}}{x} \\\ \Rightarrow Q = \dfrac{{0.01\,J{s^{ - 1}}{m^{ - 2}} \times 24 \times {{10}^{ - 2}}{m^2} \times 50 \times 36 \times {{10}^3}\,s}}{{5 \times {{10}^{ - 2}}\,m}} \\\ \Rightarrow Q = 86400\,J \\\$

Let the ice that melts be $m\,kg$
Relationship between heat and latent heat is:
$Q = m{L_f} \\\ m = \dfrac{Q} {{{L_f}}} \\\ \Rightarrow Q = \dfrac{{86400\,J}} {{335 \times {{10}^3}\,Jk{g^{ - 1}}}} \\\ \Rightarrow Q = 0.258\,kg \\\$

Therefore, amount of ice left
$m = \left( {5 - 0.258} \right)\,kg \\\ \Rightarrow m = 4.742\,kg \\\ \Rightarrow m = 4.7\,kg \\\$
Hence, option C is correct.

Additional information: Thermal conductivity: The thermal conductivity of a material is an estimation of its capacity to conduct heat. It is usually denoted by $K$ . Heat transfer happens at a lower rate in materials of low thermal conductivity than in materials of high thermal conductivity.
Latent heat of fusion: The latent heat of fusion is the enthalpy change of some quantity of the material when it is melted. When the heat of the fusion is referred to as the unit of mass, it is generally known as basic heat of fusion, while the molar heat of the fusion corresponds to the shift of the enthalpy by the quantity of the material in moles.

Note: Here we should be careful with the units. If they are not in standard form we need to convert them. Also at last we need to subtract $5$ from the result because it is mentioned in the question that $5\,kg$ of ice is put in the box.