Question

A metallic rod of cross-sectional area $9.0$$c{m^2}$ and length $0.54$$m$, with the surface insulated to prevent heat loss, has one end immersed in boiling water and the other in ice-water. The heat conducted through the rod melts the ice at the rate of $1$$gm$ for $33$$sec$. Thermal conductivity of the rod is
(A) $330$ $W{m^{ - 1}}{K^{ - 1}}$
(B) $60$ $W{m^{ - 1}}{K^{ - 1}}$
(C) $600$ $W{m^{ - 1}}{K^{ - 1}}$
(D) $33$ $W{m^{ - 1}}{K^{ - 1}}$

Answer 1

Hint: As one end of the rod is immersed in boiling water which is hot and the other end is immersed in a mixture of ice-water which is cold, a heat transfer will occur and the heat current will flow from hot end to cold end. Now, this heat will be used to melt the ice. Find this rate of heat transferred in terms of the given quantities and then find the thermal conductivity.

Complete step by step answer:
As the two ends are immersed in two different containers, the temperature of these two ends will be different. The difference between the ends will be $\Delta T = 100 - 0 = 100^\circ C$ (since the temperature of boiling water is $100^\circ C$ and that of ice is $0^\circ C$).
If an amount of $\Delta Q$ heat flows through a cross section of area $A$ in time $\Delta t$, then the rate of heat flow is $\dfrac{{\Delta Q}}{{\Delta t}}$. This rate is found to be proportional to the cross section $A$, the temperature difference between the ends $\Delta T$ and inversely proportional to the length $l$ that the heat current covers. Mathematically,
$
 \dfrac{{\Delta Q}}{{\Delta t}} \propto \dfrac{{A\Delta T}}{l} \\
 \dfrac{{\Delta Q}}{{\Delta t}} = K\dfrac{{A\Delta T}}{l} \\
 $
where, $K$ is a constant and called as the thermal conductivity.
So, the rate of heat flowing through the rod will be $\dfrac{{\Delta Q}}{{\Delta t}} = K\dfrac{{A\Delta T}}{l}$
Given that $A = 9 \times {10^{ - 4}}$${m^2}$, $l = 0.54$$m$ and having found out that $\Delta T = 100^\circ C$
$
 \dfrac{{\Delta Q}}{{\Delta t}} = K\dfrac{{\left( {9 \times {{10}^{ - 4}}} \right)\left( {100} \right)}}{{\left( {0.54} \right)}} \\
 \dfrac{{\Delta Q}}{{\Delta t}} = K\left( {\dfrac{1}{6}} \right) = \dfrac{K}{6} \\
 $
As this is the heat which is being transferred across the rod and then used to melt the ice, the rate of melting of ice is supposed to be equal to this heat current.
The latent heat of fusion ice is $3.36 \times {10^5}Jk{g^{ - 1}}$. Therefore, the rate at which the ice melts is $\left( {3.36 \times {{10}^5}\dfrac{J}{{kg}}} \right) \times \left( {\dfrac{{1g}}{{33s}}} \right) = \left( {3.36 \times {{10}^5}\dfrac{J}{{kg}}} \right) \times \left( {\dfrac{{1 \times {{10}^{ - 3}}kg}}{{33s}}} \right) = 10.182J{s^{ - 1}}$
Therefore, we will $\dfrac{K}{6} = 10.182$$W$.
\[K = 10.182 \times 6\]
$\therefore K = 61.092$$W{m^{ - 1}}{K^{ - 1}}$
The most appropriate value from the options is $60$ $W{m^{ - 1}}{K^{ - 1}}$.
Hence, the thermal conductivity of the rod is $60$ $W{m^{ - 1}}{K^{ - 1}}$.

So, the correct answer is “Option B”.

Note:
Here, we should keep in mind the units of each quantity and convert them appropriately when required. Also remember the value of latent heat of fusion of ice used in this question, it will help you to solve these kinds of questions. The instantaneous rate of heat flowing through a cross section is given as $\dfrac{{dq}}{{dt}} = K\dfrac{{AdT}}{{dl}}$.