Question

A cylindrical rod with one end in a steam chamber and the other end is in ice. It is found that 1gm of ice melts per second. If the rod is replaced by another one of same material and double the length and double area of cross section, the mass of ice that melts per second is:
(A) 2gm
(B) 4gm
(C) 1gm
(D) 0.5gm

Answer 1

The above problem is a case when the parameters of the metal body object are changed and its new conduction is needed. The conduction of heat through a rod of uniform cross section depends on its thermal conductivity, area, temperature gradient (difference in temperature upon length) and time.

Complete answer:
Let us first assign some useful terms before proceeding ahead.
Let the length of the rod initially be ‘l’ and afterwards it has been doubled to ‘2l’.
Let the area initially be ‘A’ and afterwards it has also been doubled to ‘2A’.
Let the thermal conductivity of the metal be ‘K’ and the temperature difference be $\vartriangle T$ .
Then, we can write:
$\Rightarrow Q=\dfrac{KA(\vartriangle T)t}{l}={{m}_{i}}{{L}_{i}}$
Where,
${{m}_{i}}$is the mass of ice melted per second and ${{L}_{i}}$ is the latent heat of fusion of ice.
From the above equation, it can be written as:
$\Rightarrow {{m}_{i}}\propto \dfrac{A}{l}$
Therefore, we can write:
$\Rightarrow \dfrac{{{m}_{i,0}}}{{{m}_{i}}}=\dfrac{\dfrac{{{A}_{i}}}{{{l}_{i}}}}{\dfrac{{{A}_{f}}}{{{l}_{f}}}}$
Where,
${{m}_{i,0}}$is the initial mass of ice that is melting per second. And,
${{m}_{i}}$ is the final rate of melt of ice.
Putting the values of known terms in above equation, we have:
$\begin{aligned} & \Rightarrow \dfrac{{{m}_{i,0}}}{{{m}_{i}}}=\dfrac{\dfrac{A}{l}}{\dfrac{2A}{2l}} \\\ & \Rightarrow \dfrac{{{m}_{i,0}}}{{{m}_{i}}}=1 \\\ & \therefore {{m}_{i}}={{m}_{i,0}} \\\ \end{aligned}$
Hence, the rate of melt of ice afterwards is the same as the initial rate of melt of ice, that is 1gm per second.

So, the correct answer is “Option C”.

Note: The relation of heat transfer used is only valid for a uniform object, that is whose area is uniform throughout its length. This is because for a non-uniform object the area will vary at certain points and hence the conduction at these points will also change, so we would need to use integration in order to derive the relation of conduction as a function of length.