Question

A thermocol ice box is an inexpensive and efficient way to store small quantities of cooked food particularly, in summer. A cubical icebox of a side $30{\text{cm}}$ has a thickness of $5{\text{cm}}$ . If $4{\text{kg}}$ of ice is put in the box, estimate the amount of ice remaining after $6{\text{h}}$ . The surrounding temperature is $45^\circ {\text{C}}$ and the coefficient of thermal conductivity of thermocol is $0.01{\text{ J}}{{\text{s}}^{ - 1}}{{\text{m}}^{ - 1}}{{\text{K}}^{ - 1}}$ . (Latent heat of fusion of water is $335 \times {10^3}{\text{Jk}}{{\text{g}}^{ - 1}}$ ).

Answer 1

By conduction, heat gets transferred from the surrounding to the box. This transferred heat melts the ice. So the heat transferred by conduction must be equal to the heat required to cause a change of state (i.e., melting of ice).

Formulas used: Area of a cube is $A = 6{a^2}$ , where $a$ is the side of the cube.
Heat transferred by conduction between a body and its surrounding is $Q = \dfrac{{\kappa A\Delta Tt}}{d}$ where, $\kappa$ is the coefficient of thermal conductivity, $A$ is the area of the body, $\Delta T$ is the difference in temperature between the body and surrounding, $t$ is the time for which the heat is transferred and $d$ is the thickness of the body.
The amount of heat required during a change of state is $Q = mL$ , where $m$ is the mass of the substance undergoing a change of state and $L$ is the amount of heat transferred per unit mass to the substance as it undergoes a change of state from solid to liquid and is called the latent heat of fusion.

Complete step by step answer.
Step 1: List the data given in the question.
Given, a cubical icebox of side $a = 30{\text{cm = 0}}{\text{.3m}}$ and thickness $d = 5{\text{cm = 0}}{\text{.05cm}}$ .
Then the area of the box will be, $A = 6 \times {\left( {0.3} \right)^2} = 0.54{{\text{m}}^2}$ .
A block of ice of mass $m = 4{\text{kg}}$ is put in the box.
The surrounding temperature is $45^\circ {\text{C}}$ . Since ice exists at $0^\circ {\text{C}}$ , the temperature difference between the box and surrounding will be $\Delta T = 45^\circ {\text{C}}$ .
The coefficient of thermal conductivity of thermocol is $\kappa = 0.01{\text{ J}}{{\text{s}}^{ - 1}}{{\text{m}}^{ - 1}}{{\text{K}}^{ - 1}}$ .
The latent heat of fusion of water is $L = 335 \times {10^3}{\text{Jk}}{{\text{g}}^{ - 1}}$ .
The time for which heat is transferred is $t = 6{\text{h = 6}} \times {\text{60}} \times {\text{60 = 21600s}}$ .
Step 2: Find the heat transferred to the box of ice by conduction.
The heat transferred to the box of ice by the surrounding is given by, $Q = \dfrac{{\kappa A\Delta Tt}}{d}$ , where $\kappa$ is the coefficient of thermal conductivity of thermocol, $A$ is the area of the box, $\Delta T$ is the difference in temperature between the box and its surrounding, $t$ is the time for which the heat is transferred and $d$ is the thickness of the box.
Substitute the values of $\kappa {\text{, }}A{\text{, }}\Delta T{\text{, t}}$ and $d$ in the above equation to obtain $Q$ .
Then we have, $Q = \dfrac{{0.01 \times 0.54 \times 45 \times 21600}}{{0.05}} = 104976{\text{J}}$ .
Step 3: Using the value of $Q = 104976{\text{J}}$ , find the amount of ice remaining in the box.
The heat transferred during a change of state is given by, $Q = mL$ .
Since we have $Q = 104976{\text{J}}$ and $L = 335 \times {10^3}{\text{Jk}}{{\text{g}}^{ - 1}}$ , we can obtain the amount of ice converted to water using the relation $m = \dfrac{Q}{L}$ .
Now, the amount of ice converted to water will be $m = \dfrac{{104976}}{{335 \times {{10}^3}}} = 0.313{\text{kg}}$ .
Therefore, the remaining mass of ice will be $4 - 0.313 = 3.687{\text{kg}}$ .

Note: In the calculation of heat transferred by conduction, the temperature difference is expressed in the units of degree Celsius. Even though the S. I. unit of temperature is Kelvin, a unit conversion was not done because a difference in temperature will be the same irrespective of the unit in which it is expressed.