Question

A cylindrical rod with one end in a steam chamber and the other end in ice results in melting of 0.1 gm of ice per second. If the rod is replaced by another rod with half the length and double the radius of the first and if the thermal conductivity of the material of the second rod is 0.25 times that of first, the rate at which ice melt in gm/second will be
A. 0.1
B. 0.2
C. 1.6
D. 3.2

Answer 1

If one end of metal rod is heated, it will transfer heat to the other end through the process called conduction. In the study state, the rate of flow of heat or heat current is proportional to area of cross section, A, temperature difference (θ1-θ2) and inversely proportional to the length of the rod. The thermal conduction in metal bar is given by,
$Q=\dfrac{KA(\theta_1-\theta_2)t}{L}$
Here,
Q = Total heat
K=Thermal conductivity of the material
A= Area of cross section
L= Length of the rod
t = time

Complete step by step answer:
At steady state heat lost transmitted through rod = heat gained by ice
By taking the ratio of heat transfer in two metal rods, we obtain
$Q=\dfrac{KA({\theta_1}-{\theta_2})t}{L}=mLwt$
m = mass of ice
$\text{Lw}$= Latent heat of water
t = time
Therefore,
$m\propto \dfrac{KA}{L}$
Let Q1 and Q2 be the heat transferred by the two rods. By taking the ratio of heat transfer in two metal rods, we obtain
$\dfrac{{{m}_{2}}}{{{m}_{1}}}=\left( \dfrac{K_2A_2}{L_2} \right)/\left( \dfrac{K_1A_1}{L_1} \right)$
The rate at which ice melt in gm/second will be
${{m}_{2}}=\dfrac{K_2A_2}{K_1A_1}\times \dfrac{L_1}{L_2}\times {{m}_{1}}$
Substituting, ${{K}_{2}}=0.25{{K}_{1}}$, ${{A}_{2}}=4{{A}_{1}}$, ${{L}_{2}}=0.5{{L}_{1}}$ and ${{m}_{1}}=0.1\text{ g}$, we get
${{m}_{2}}=\dfrac{0.25K1\times 4A1}{K1A1}\times \dfrac{L1}{0.5L1}\times 0.1$
Cancelling the common terms and simplifying,
${{m}_{2}}=\dfrac{0.25 {K_1} \times 4 {A_1}}{{ {K_1} {A_1}}}\times \dfrac{ {L_1}}{{0.5 {L_1}}}\times 0.1$
$\text{ }=\dfrac{0.25\times 4\times 0.1}{{0.5}}$
$\text{ }=0.2$
Hence, the rate at which ice melts is 0.2 gm/second.
The correct answer is (B).

Note:
The area of the second rod becomes four times the area of the first rod as radius doubles. That is, if ${{A}_{1}}=\pi {{r}^{2}}$, ${{A}_{2}}$ can be calculated as
$\begin{gathered} & {{A}_{2}}=\pi {{\left( 2r \right)}^{2}} \\\ & \text{ }=4\pi {{r}^{2}} \\\ & \text{ }=4{{A}_{1}} \\\ \end{gathered}$
Assuming that if radius is doubled, the area will also be doubled will give us wrong results, that is, ${{A}_{2}}=2{{A}_{1}}$ is incorrect.
We can calculate the increase in conductance by multiplying the increase in area and conductivity and dividing by increase in length as follows:
$\dfrac{4\times 0.25}{0.5}=2\text{ times}$
Since, the rate of heat transfer varies directly as thermal conductance, it also doubles, that is, $0.1\times 2=0.2\text{ g/s}$.
The thermal conductivity normally varies with temperature. But it can be considered constant over a normal temperature range.