Question

A metal rod AB of length $10x$ has its one end $A$ in ice at $0^{\circ }\text{C}$ and the other end (B) in water at ${100}^{\circ }\text{C}$. If a point P on the rod is maintained at ${400}^{\circ }\text{C}$, then it is found that equal amounts of water and ice evaporate and melt per unit time. The latent heat of evaporation of water is $540{\text{cal g}}^{-1}$ and the latent heat of fusion of ice is $80{\text{cal g}}^{-1}$. If the point P is at a distance of $\lambda x$ from the end A, find the value of $\lambda$.
(Neglect any heat loss to the surroundings)

Answer 1

The rate of steady state heat transfer is given by
$\frac{\Delta Q}{\Delta t}=kA\left(\frac{\Delta T}{\Delta x}\right)$
The amount of heat flow from $P$ to $A$
$\frac{\Delta Q}{\Delta t}=\frac{kA(400-0)}{\lambda x}=L_f\frac{d{m}_1}{dt}$ ....(i)
where $m_1$ is the amount of ice melted.
The amount of heat flow from $P$ to $B$
$\frac{\Delta Q}{\Delta t}=\frac{kA(400-100)}{(10-\lambda )x}=L_v\frac{d{m}_2}{dt}$ ...(ii)
where $m_2$ is the amount of water evaporated.
Given, $m_1=m_2$
$\Rightarrow \frac{kA(400-0)}{\lambda x{L}_f}=\frac{kA(400-100)}{(10-\lambda )x{L}_v}$
$\Rightarrow \frac{400}{\lambda \times 80}=\frac{300}{(10-\lambda )540}$
$\Rightarrow 2160(10-\lambda )=240\lambda$
$\Rightarrow \lambda =9$