Question

One end of a 2.35m long and 2.0cm radius aluminium rod $(235W.{{m}^{-1}}{{K}^{-1}})$ is held at ${{20}^{o}}C$. The other end of the rod is in contact with a block of ice at its melting point. The rate in $kg{{s}^{-1}}$ at which ice melts is:
(Latent heat of fusion $=336\times {{10}^{3}}K/kg$ )
$A.48\pi \times {{10}^{-6}}$
$B.24\pi \times {{10}^{-6}}$
$C.2.4\pi \times {{10}^{-6}}$
$D.4.8\pi \times {{10}^{-6}}$

Answer 1

Hint: We are using the thermodynamic properties to solve the problem. We will use the relationship of rate of melting in terms of length, temperature difference, area of cross-section and length. We know that in conduction heat flows from higher temperature to lower temperature.
Formula Used:
We will use the following formula to solve this thermodynamic problem:-
\[\dfrac{dm}{dt}=\dfrac{KA}{{{L}_{f}}}\left( \dfrac{dt}{dx} \right)\]

Complete step by step solution:
From the question we have the following parameters with us:-
Radius of aluminium rod, $r=2.0cm=2\times {{10}^{-2}}m$.
Length of the rod, $dx=2.35m$.
Thermal conductivity, $K=235W.{{m}^{-1}}{{K}^{-1}}$.
Temperature, $dt={{20}^{o}}C$
Latent heat of fusion, ${{L}_{f}}=336\times {{10}^{3}}K/kg$
We have to find the rate of melting, $\dfrac{dm}{dt}$.
We will use the following formula to find the correct answer:-
\[\dfrac{dm}{dt}=\dfrac{KA}{{{L}_{f}}}\left( \dfrac{dt}{dx} \right)\]………………. $(i)$
Putting the above given parameters in equation $(i)$we get
$\dfrac{dm}{dt}=\dfrac{235\times \pi {{(2\times {{10}^{-2}})}^{2}}\times 20}{2.35\times 336\times {{10}^{3}}}$……………. $(ii)$
We have used $A=\pi {{r}^{2}}$ in $(ii)$
Solving further we get
$\dfrac{dm}{dt}=2.4\pi \times {{10}^{-6}}kg{{s}^{-1}}$

Hence, option $(C)$ is correct.

Additional Information:
We should have some basic knowledge of conduction also.
Conduction is a mode of heat transfer in which particles at higher temperature transmit heat to the particles at lower temperature. The transfer of heat takes place through physical contacts and not by movement of particles. Conduction can’t take place in vacuum; it requires medium to transfer heat. The rate of conduction is fastest in solids and slowest in gases. In solids, metals are much better conductors than non-metals.

Note:
We should not get confused between the two similar formulae of \[\dfrac{dm}{dt}=\dfrac{KA}{{{L}_{f}}}\left( \dfrac{dt}{dx} \right)\] and $\dfrac{dQ}{dt}=KA\left( \dfrac{dt}{dx} \right)$. They are used for different calculations requirements. We should also take care of the fact that heat always flows from the region of higher temperature to the region of lower temperature.