Question

One end of a copper rod of uniform cross-section and of length $1.5\text{m}$ is kept in contact with ice and the other end with water at ${100}^{\circ }\text{C}$. At what point along its length should a temperature of ${200}^{\circ }\text{C}$ be maintained so that in steady state, the mass of ice melting be equal to that of the steam produced in same interval of time? Assume that the whole system is insulated from surroundings. Latent heat of fusion of ice and vapourization of water are $80\text{cal/g}$ and $540\text{cal/g}$, respectively.

• A. $10.34\text{cm}$ from water at ${100}^{\circ }\text{C}$.
• B. $10.34\text{cm}$ from ice at $0^{\circ }\text{C}$.
• C. $4.66\text{cm}$ from water at ${100}^{\circ }\text{C}$.
• D. $\text{None of these}$

Answer 1

The correct option is A $10.34\text{cm}$ from water at ${100}^{\circ }\text{C}$.


If the point is at a distance $x$ from water at ${100}^{\circ }\text{C}$, heat conducted to ice in time $t$,
$Q_{\text{ice}}=KA\frac{(200-0)}{(1.5-x)}\times t$
So ice melted by this heat
$m_{\text{ice}}=\frac{Q_{\text{ice}}}{L_F}=\frac{KA}{80}\frac{(200-0)}{(1.5-x)}\times t$
Similarly heat conducted by the rod to the water at ${100}^{\circ }\text{C}$ in time $t$,
$Q_{\text{water}}=KA\frac{(200-100)}{x}t$
Steam formed by this heat
$m_{\text{system}}=\frac{Q_{\text{water}}}{L_V}=KA\frac{(200-100)}{(540\times x)}t$
According to given problem $m_{\text{ice}}=m_{\text{steam}}$
i.e.,
$\frac{200}{80(1.5-x)}=\frac{100}{540\times x}$
$\Rightarrow x=\frac{6}{58}\text{m}=10.34\text{cm}$
i.e., ${200}^{\circ }\text{C}$ temperature must be maintained at a distance $10.34\text{cm}$ from water at ${100}^{\circ }\text{C}$.