Question

A thermocol box has a total wall area (including the lid) of $1{{ }}{{{m}}^2}$ and wall thickness of 3 cm. it is filled with ice at $0^\circ C$. If the average temperature outside the box is $30^\circ C$ throughout the day, the amount of ice that melts in one day is (Use ${K_{thermocol}} = 0.03{{ W/mK}}$, ${L_{fusion(ice)}} = 3 \times {10^5}{{ J/kg}}$
A) 1kg
B) 2.88kg
C) 25.92kg
D) 8.64kg

Answer 1

Hint: In this question, we are going to use the relation of heat with the change in temperature and area of the box. In this question the ice is changing into water, so there is a phase change. Hence we will equate the heat to \[mL\].

Complete step by step solution:
Given,
Area of the thermocol box $A = 1{{ }}{{{m}}^2}$
Latent heat ${L_{fusion(ice)}} = 3 \times {10^5}{{ J/kg}}$
${K_{thermocol}} = 0.03{{ W/mK}}$
Temperature inside the box ${T_1} = 0^\circ {{C}}$
Temperature outside the box ${T_2} = 30^\circ {{C}}$
Thickness of the box $d = 3{{ cm}}$
Thickness of the box $d = 3 \times {10^{ - 2}}{{m}}$
Difference between temperatures of thermocol box and outside the box
$\Rightarrow \Delta T = {T_2} - {T_1}$
$\Rightarrow \Delta T = 30 - 0$
$\Rightarrow \Delta T = 30^\circ {{C}}$
Time $t = 24 \times 60 \times 60{{ s}}$
Time $t = 86400{{ s}}$
Using the formula of heat,
$\Rightarrow \dfrac{Q}{t} = \dfrac{{KA}}{d}\Delta T$
Or
$\Rightarrow \dfrac{{mL}}{t} = \dfrac{{KA}}{d}\Delta T$
Putting the values of all the variables \[Q,t,K,A,d{{ and }} \Delta T\],
$\Rightarrow \dfrac{{m \times 3 \times {{10}^5}}}{{86400}} = \dfrac{{0.03 \times 1}}{{3 \times {{10}^{ - 2}}}} \times 30$
$\Rightarrow m = \dfrac{{0.03 \times 1 \times 30 \times 86400}}{{3 \times {{10}^{ - 2}} \times 3 \times {{10}^5}}}$
$\Rightarrow m = \dfrac{{77760}}{{9 \times {{10}^3}}}$
$\Rightarrow m = 8640 \times {10^{ - 3}}$
$\Rightarrow m = 8.64{{ kg}}$

The amount of ice melted in one day is $m = 8.64{{ kg}}$.

Note: In this type of questions, we have to be very careful about the units of all the variables. The units of all the variables should be in the same system of units. In this question we have seen that the ice is changing into water. So, this is the problem of phase change. The phase change of ice to water is a first order phase transition.
We have to remember that in problems of phase change we have to use the latent heat of the first phase because the latent heat is the energy absorbed or released by a body during a thermodynamic process. Latent heat is also called the heat of fusion. We will equate the product of mass and latent heat with the heat by the given equation $Q = mL$. Some solids exist in crystalline forms. The transition between these involves absorption or release of latent heat.