Question

In the figure shown $AB$ is a rod of length $30\text{cm}$ and area of cross -section $1{\text{cm}}^2$ and thermal conductivity $336\text{W/K}$. The ends $A$ and $B$ are maintained at temperature ${20}^{\circ }\text{C}$ and ${40}^{\circ }\text{C}$ respectively. The point $C$ on this rod is connected to a box $D$ containing ice at $0^{\circ }\text{C}$, through a highly conducting wire of negligible heat capacity. The rate at which ice box melts is:

• A. $84\text{mg/s}$
• B. $84\text{g/s}$
• C. $20\text{mg/s}$
• D. $40\text{mg/s}$

Answer 1

The correct option is D $40\text{mg/s}$

Cross sectional area of rod $AB$,
$A=1{\text{cm}}^2={10}^{-4}{\text{m}}^2$
Thermal resistance of protion $AC$,
$R=\frac{L_{AC}}{KA}$
$R=\frac{(10\times {10}^{-2})}{336\times \left({10}^{-4}\right)}=\frac{1000}{336}\Omega$

Similarly the resistance of portion $BC$ of rod:
$R_{BC}=\frac{L_{BC}}{KA}$
$R_{BC}=\frac{0.20}{336\times {10}^{-4}}=\frac{2\times {10}^3}{336}\Omega$
$R_{BC}=2R$

The wire $CD$ connected the point $C$ on rod $AB$ and ice box $D$ $\left(0^{\circ }\text{C}\right)$ is highly conducting, hence temperature of point C will also be $0^{\circ }C$

The distribution of Heat current is shown in the figure.
$\Rightarrow H_1=\frac{\Delta T_1}{R}=\frac{20-0}{R}$
$H_1=\frac{20}{R}$

Similarly $H_2=\frac{\Delta T_2}{R_{BC}}=\frac{(40-0)}{2R}$
$H_2=\frac{40}{2R}=\frac{20}{R}$

Using conservation of energy at junction $C$,
$\Rightarrow H=H_1+H_2$
$H=\frac{20}{R}+\frac{20}{R}=\frac{40}{R}$
$H=\frac{40\times 336}{1000}=\frac{13440}{1000}=13.44\text{Watt}$

The heat current $H$ is the rate of heat transterred to the ice box $D$.
Let the rate of melting of ice is $m$
$\Rightarrow \frac{dQ}{dt}=m\times L_f$
$H=m\times L_f$

$m=\frac{H}{L_f}=\frac{\left(\frac{13.44}{4.2}\right)\text{cal/s}}{\left(80\text{cal/g}\right)}$

$m=0.04\text{g/s}$ $(\because 1\text{cal}\approx 4.2\text{J})$
$\Rightarrow m=0.04\times 1000=40\text{mg/s}$
$\begin{array}{c}\text{Why this question}?\\ \text{It uniquely tests your understanding of flow of heat current}\\ \text{based on temperature gradient.}\\ \text{Tip: In this problem the presence of highly conducting wire}\\ \text{between C}\&\text{D gives clue that both the points will}\\ \text{come to thermal equilibrium}(T_C=T_D=0^{\circ }\text{C})\\ \text{instantly due to negligible thermal resistance of wire.}\end{array}$