One end of a copper rod of uniform cross-section and length 150 cm is in contact with ice and the other end with water at `100^(@)C` . At what point along its length should a temperature of `200^(@)C` be maintained so that in steady state, the mass of the ice melting is equal to that of steam produced in the same interval of time? Assume that whole system is insulated from the surroundings. Latent heat of steam =` 537 cal g^(-1)` and lalent heat of fusion of ice `= 80 cal g^(-1)`.
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Explanation

Let AB be a copper rod of length 150 cm. Its end A with water at `100^(@)C` and end B is with ice at `0^(@)C`. Let the point C of the rod be maintained at temperature `200^(@)C`, where AC = x cm and BC = (150-x) cm .
Let m (in gram) be the mass of water (converted into steam) or that of ice (melted into water) in the same time t. let A be the area of cross-section of the rod.
Amount of heat required to boil the water
`Q_(1) = mL = m xx 537 cal`
Since this amount of heat is conducted from the point C to the end A of the rod in time t, so
`Q_(1) = (KA(200-100)t)/(x) = m xx 537` ...(i)
The amount of heat required to melt the ice
`Q_(2) = mL' =m xx 80 cal`
Since this amount of heat is conducted from the point C to the end B of the rod in time t, so
`Q_(2) =(KA(200-0)t)/(150-x) = m xx 80` ..(ii)
Dividing (i) by (ii) we have
`(150-x)/(2x) = (537)/(80)`
or `80(150-x) = 2 x xx 537 or x = 10.4 cm`.