One end of a copper rod of uniform cross section and of length 1.5 m is kept in contact with ice and the other end with water at 100^oC. At what point along its length should a temperature of 200^oC 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 cal/g and 540 cal/g, respectively.

If the point is at a distance x from water at 100^oC, heat conducted to ice in time t,

${\mathrm{Q}}_{\mathrm{ice}}=\mathrm{KA}\frac{(200-0)}{(1.5-\mathrm{x})}\times \mathrm{t}$

So ice melted by this heat

${\mathrm{m}}_{\mathrm{ice}}=\frac{{\mathrm{Q}}_{\mathrm{ice}}}{{\mathrm{L}}_{\mathrm{F}}}=\frac{\mathrm{KA}}{80}\frac{(200-0)}{(1.5-\mathrm{x})}\times \mathrm{t}$

Similarly heat conducted by the rod to the water at 100^oC in time t,

${\mathrm{Q}}_{\text{water }}=\mathrm{KA}\frac{(200-100)}{\mathrm{x}}\mathrm{t}$

Steam formed by this heat

${\mathrm{m}}_{\text{steam }}=\frac{{\mathrm{Q}}_{\text{water }}}{{\mathrm{L}}_{\mathrm{V}}}=\mathrm{KA}\frac{(200-100)}{(540\times \mathrm{x})}\mathrm{t}$

According to given problem ${\mathrm{m}}_{\text{ice }}={\mathrm{m}}_{\text{steam }}$

$\text{ i.e., }\frac{200}{80(1.5-\mathrm{x})}=\frac{100}{540\times \mathrm{x}}\Rightarrow \mathrm{x}=\frac{6}{58}\mathrm{m}=10.34\mathrm{cm}$

i.e., 200^oC temperature must be maintained at a distance 10.34 cm from water at 100^oC.