A metal rod $A B$ of length $10 x$ has its one end $A$ in ice at $0^{\circ} \mathrm{C}$ and the other end $B$ in water at $100^{\circ} \mathrm{C}$. If a point $P$ on the rod is maintained at $400^{\circ} \mathrm{C}$, then it is found that equal amounts of water and ice evaporate and melt per unit…

Question

A metal rod $A B$ of length $10 x$ has its one end $A$ in ice at $0^{\circ} \mathrm{C}$ and the other end $B$ in water at $100^{\circ} \mathrm{C}$. If a point $P$ on the rod is maintained at $400^{\circ} \mathrm{C}$, 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 $540 \mathrm{calg}^{-1}$ and latent heat of melting of ice is $80 \mathrm{calg}^{-1}$. If the point $P$ is at a distance of $\lambda x$ from the ice end $A$, find the value of $\lambda$. [Neglect any heat loss to the surrounding.]

Aicharya · Accepted Answer

Heat will flow both sides from point $P$. $ \begin{aligned} L_1 \frac{d m_1}{d t} & =\left(\frac{\text { Temperature difference }}{\text { Thermal resistance }}\right)_1 \\ & =\frac{400}{(\lambda x) / k A} \\ L_1 \frac{d m_2}{d t} & =\frac{400-100}{(100-\lambda) x / k A} \end{aligned} $ In about two equations, $ \begin{aligned} & \frac{d m_1}{d t}=\frac{d m_2}{d t} \quad \text { (given) } \\ & L_1=80 \mathrm{calg}^{-1} \text { and } L_2=540 \mathrm{calg}^{-1} \end{aligned} $ Solving these two equations we get $\lambda=9$. `

A metal rod $A B$ of length $10 x$ has its one end $A$ in ice at $0^{\circ} \mathrm{C}$ and the other end…

Solution