Question

A ‘thermocol’ ice box is a cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is $45^\circ C$, and the coefficient of thermal conductivity of thermocol is $0.01j{s^{ - 1}}{m^{ - 1}}{K^{ - 1}}$. [Heat of fusion of water =$335 \times {10^3}jk{g^{ - 1}}$]

Answer 1

The heat which will be conducted through the walls of the thermocol box in six hours the same heat will be used by ice to melt.
The conduction of heat through the wall of the thermocol box will be conducted as per the Fourier law of conduction

Complete step by step answer:
Here in this question, the box is of dimensions of side S= 30 cm and thickness = 5 cm is given. now 4 kg of ice is put in the box and the box is closed the ambient or outside temperature is around $45^\circ C$and now we have to calculate how much ice will be left after six hours
Firstly, here we will calculate the amount of heat which is conducted through the box in 6 hours
As per Fourier’s conduction law $Q = KA\dfrac{{\Delta T}}{L}$
Where Q is the amount of heat transferred
K is the coefficient of thermal conductivity
A is the are of a cross-section perpendicular to the flow of heat
L is the thickness of the wall
$\Delta T$ is the change in temperature
Now, Side of the given cubical ice box, S = 30cm =0.3m
Thickness of the ice box, L = 5.0cm =0.05m
Mass of ice kept in the icebox, m=4kg
Outside temperature, T= $45^\circ$C
So $\Delta T = 45^\circ C$ $\left( {\because {T_{ice}} = 0} \right)$
K=$0.01J{s^{ - 1}}{m^{ - 1}}{k^{ - 1}}$
Heat is being conducted from all 6 surfaces of the box so we will take the total area
A= Surface area of the box $6{S^2}$ $= 6{\left( {0.3} \right)^2}$ $= 0.54{m^3}$
Now Q will be $Q = \dfrac{{0.01 \times 0.54 \times 45}}{{0.05}}watts$
Now this will be the heat conducted through the walls of the cube per second we need total heat for 6 hours so
${Q_{total}} = Q \times 6 \times 60Joules$
${Q_{total}} = \dfrac{{0.01 \times 0.54 \times 45}}{{0.05}} \times 6 \times 60$
$\to {Q_{total}} = 104976J$
Now this heat will be taken up by ice to melt
As we know the amount of heat required to melt ice is calculated as
$Q = m(LH)$
Where LH is the latent heat of the ice and m is the mass of ice
Here LH is given as $LH = 335 \times {10^3}Jk{g^{ - 1}}$
So the mass of ice melted will be
$m = \dfrac{Q}{{LH}}$
$\to m = \dfrac{{104967}}{{335 \times {{10}^3}}}$
$\to m = 0.313kg$
This is the amount of ice melt so the remaining amount of ice will be $4 - 0.313 = 3.687kg$
Final Answer: Hence, the amount of ice remaining after 6 hours is 3.687 kg.

Note:
-The temperature in the Fourier should be kept in Kelvin only.
-All the units should be made dimensionally similar before performing calculations.
-The latent heat is the amount of heat required by the substance to phase change its 1 kg mass.
-Latent heat does not increase the temperature of the body it is utilized in the phase change.