Question

A $3\,{\text{cm}}$ cube of iron has one face at $100^\circ {\text{C}}$ and the other in a block of ice at $0^\circ {\text{C}}$ . If $k$ of iron $= 0.2$ CGS units and $L$ for ice is $80\,{\text{cal/g}}$ , then the amount of ice that melts in $10$ minutes is (assume steady state heat transfer).
(A) $450\,{\text{g}}$
(B) $900\,{\text{g}}$
(C) $350\,{\text{g}}$
(D) $500\,{\text{g}}$

Answer 1

First of all, we will find the amount of heat energy required to melt a certain amount of ice and then differentiate with respect to ice and then we will find an expression for steady state heat transfer. We will equate the two expressions and then substitute the values as required. We will manipulate accordingly and obtain the result.

Complete step by step answer:
The side of the square is $3\,{\text{cm}}$ .
One of the faces is at a temperature $100^\circ {\text{C}}$ and the other face is at a temperature of $0^\circ {\text{C}}$ .
$k$ of iron $= 0.2$ CGS units.
Latent heat of fusion for ice is, $L$ for ice is $80\,{\text{cal/g}}$ .
Time required is $10$ minutes i.e. $600\,{\text{s}}$ .
This problem is based on the principle of transfer of heat. As one side of the iron cube is under at higher temperature and another side is at freezing point, so the temperature flows from higher temperature to the lower temperature. The heat transferred melts the ice.
We know, the amount of heat energy required melt a certain amount of ice is given by:
$Q = Lm$ …… (1)
Where,
$Q$ indicates required heat.
$L$ indicates latent heat of fusion for ice.
$m$ indicates mass of ice.
Now we differentiate equation (1), with respect to time and we get:
$\dfrac{{dQ}}{{dt}} = \dfrac{d}{{dt}}\left( {Lm} \right)$
$\dfrac{{dQ}}{{dt}} = L\dfrac{{dm}}{{dt}}$ …… (2)
Here, $L$ is a constant term.
As, we are given that it is a steady state heat transfer, so we can write:
$\dfrac{{dQ}}{{dt}} = kA\dfrac{{dT}}{{dx}}$ …… (3)
Where,
$Q$ indicates required heat.
$A$ indicates the area of one face.
$T$ indicates temperature.
$x$ indicates small length.
Now, we equate the equation (2) and (3), and we get:

$$L\dfrac{{dm}}{{dt}} = kA\dfrac{{dT}}{{dx}} \\\ L \times \dfrac{m}{t} = kA \times \dfrac{{{T_1} - {T_2}}}{l} \\\$$

$m = \dfrac{t}{L} \times kA \times \dfrac{{{T_1} - {T_2}}}{l}$ …… (4)
Now we substitute the required values in the equation (4) and we get:

$$m = \dfrac{t}{L} \times kA \times \dfrac{{{T_1} - {T_2}}}{l} \\\ m = \dfrac{{600}}{{80}} \times 0.2 \times {3^2} \times \dfrac{{100 - 0}}{3} \\\ m = \dfrac{{600}}{{80}} \times 0.2 \times 3 \times 100 \\\ m = 450\,{\text{g}} \\\$$

Hence, the mass of ice that actually melts is $450\,{\text{g}}$ .
The correct option A.

Note: The given problem is based on calorimetry. Higher is the flow of heat from the higher temperature end to the lower temperature end, larger is the mass that melts. Steady-state conduction is the type of conduction that occurs where there is a steady temperature difference(s) that drives the conduction.