Question

One end of a copper rod of uniform cross-section and length $1.5$ meters is in contact with melting ice and the other end with boiling water. At what point along its length should a temperature of ${200}^{o}C$ be maintained, so that in steady state, the mass of ice melting is equal to that of steam produced in the same interval of time? Assume that the whole system is insulated from the surroundings.

Answer 1

Melting ice at 0°

Boiling water at 100°C

Let the distance be x from point A

Where the temperature is 200°C

We know that,

$\frac{d\theta }{dt}=\frac{KA\Delta T}{l}$ (1)

Where,

$\frac{d\theta }{dt}$= Rate of flow of heat.

K=Thermal conductivity

A=Area

$\Delta$T=Temperature difference

l= Length of material

We know,

$\theta$=mL [heat=$\theta$,m=mass,L=latent heat]

Differentiating on both sides

$\frac{d\theta }{dt}=\frac{dm}{dt}L$ (2)

From 1 and 2

$\frac{dm}{dt}=\frac{KA\Delta T}{Ll}$ (3)

At steady state

Rate at which ice is melting= Rate at which steam is produced

${\left(\frac{dm}{dt}\right)}_{ice}={\left(\frac{dm}{dt}\right)}_{steam}$

Using 3

$\frac{KA(200-0)}{L_{ice}x}=\frac{KA(200-100)}{(1.5-x)L_{steam}}$

$\frac{2}{x\times 80}=\left(\frac{1}{(1.5-x)540}\right)$

$\frac{1}{4x}=\left(\frac{1}{54(1.5-x)}\right)$

$54\times 1.5-54x=4x$

$54\times 1.5=58x$

$x=\frac{54\times 1.5}{58}$

=$1.4$ meters