A metal rod AB of length 10x has its one end A in ice: Heat Conduction


Problem: A metal rod AB of length 10x has its one end A in ice at 0°C and the other end B in water at 100°C. If a point P on the rod is maintained at 400°C, then it is found that equal amounts of water and ice evaporates and melts per unit time. The latent heat of evaporation of water is 540 cal/gram and latent heat of melting of ice is 80cal/gram. If the point P is at a distance of λx from the ice end A, find the value of λ. (Neglect any heat loss to the surroundings.)  (IIT JEE 2009)

Answer: The answer is λ = 9.

Solution: The situation is shown in the figure.

A Metal Rod AB of length 10x

Let $A$ be the cross-sectional area of the rod and $k$ be the thermal conductivity of the rod's material. The rates of heat flow from the point P towards the end A and the end B, are given by \begin{align} \frac{\mathrm{d}Q_A}{\mathrm{d}t}&=kA \frac{T_P-T_A}{\lambda x}\\ &=\frac{400kA}{\lambda x},\nonumber\\ \frac{\mathrm{d}Q_B}{\mathrm{d}t}&=kA \frac{T_P-T_B}{10x-\lambda x} \nonumber\\ &=\frac{300kA}{10 x-\lambda x}.\nonumber \end{align} Let the rate of melting of ice be $\mathrm{d}m_i/\mathrm{d}t$ and the rate of evaporation of water be $\mathrm{d}m_w/\mathrm{d}t$. The rates of heat required for the ice and the water are \begin{align} &\mathrm{d}Q_A/\mathrm{d}t=(\mathrm{d}m_i/\mathrm{d}t)L_i,\nonumber\\ &\mathrm{d}Q_B/\mathrm{d}t=(\mathrm{d}m_w/\mathrm{d}t) L_w, \nonumber \end{align} where $L_i=$ 80 cal/gram and $L_w=$ 540 cal/g are the latent heats of fusion and evaporation, respectively. Use $\mathrm{d}m_i/\mathrm{d}t=\mathrm{d}m_w/\mathrm{d}t$ and simplify to get, \begin{align} \left(\frac{400}{\lambda}\right)\cdot \left(\frac{10-\lambda}{300}\right)=\frac{80}{540},\nonumber \end{align} which gives, $\lambda=9$.

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