Question

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 ${100}^{0}C$. At what point along its length should a temperature of ${200}^{0}C$ be maintained so that in steady state, the mass of ice melting be equal to that of the steam produced in the same interval of time? Assume that the whole system is insulated from the surroundings.

Answer 1

Suppose the temperature $\theta ={200}^{o}C$ be maintained at a point on the rod at a distance $x$ from the end at $0^{o}C$. As this temperature is higher, heat will flow from this point both toward ice and water ends.

Then, in an interval $t$, the amount of the heat flowing in ice $\left(at{0}^{o}C\right)$ and in water $(at{100}^{o}C$ are respectively.

$\frac{KA(\theta -{\theta }_{ice})t}{x}$ and $\frac{KA(\theta -{\theta }_{steam})t}{(1.5-x)}$

Let $m$ be the mass of ice melted and also of steam produced in time $4$, then

$m{L}_{ice}=\frac{KA(\theta -{\theta }_{ice})t}{x}$ and $m{L}_{steam}\frac{KA(\theta -{\theta }_{steam})t}{(1.5-x)}$

or $\frac{m{L}_{ice}}{m{L}_{steam}}=\frac{\theta -{\theta }_{ice}}{x}\times \frac{(1.5-x)}{\theta -thet{a}_{steam}}$

$\frac{L_{ice}}{L_{steam}}=\frac{\theta -{\theta }_{ice}}{\theta -{\theta }_{steam}}\times \frac{(1.5-x)}{x}$

Here, $L_{ice}=80cal/gm$

$L_{steam}=540cal/gm$

$\theta ={200}^{o}C$

${\theta }_{ice}=0^{o}C$

${\theta }_{steam}={100}^{o}C$

$\frac{80}{540}=\frac{200-0}{200-100}\times \frac{(1.5-x)}{x}$

$8\times 100x=54\times 200(1.5-x)$

$2x=40.5-27x$

$x=1.396m$

$x=139.6cm$