question_answer

One end of a copper rod of uniform cross-section and of length 3.1 m is kept in contact with ice and the other end with water at 100°C. At what point along it's length should a temperature of 200°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. Latent heat of fusion of ice and vaporisation of water are 80 cal/gm and 540 cal/gm respectively

A)            40 cm from 100°C end      

B)            40 cm from 0°C end

C)            125 cm from 100°C end    

D)            125 cm from 0°C end

Correct Answer: A

Solution :

                   Rate of flow of heat is given by \[\frac{dQ}{dt}=\frac{\Delta \theta }{l/KA}\]also \[\frac{dQ}{dt}=L\frac{dm}{dt}\] (where L = Latent heat) Þ \[\frac{dm}{dt}=\frac{KA}{l}\,\left( \frac{\Delta \theta }{L} \right)\]. Let the desire point is at a distance x from water at 100°C. Q Rate of ice melting = Rate at which steam is being produced Þ \[{{\left( \frac{dm}{dt} \right)}_{Steam}}={{\left( \frac{dm}{dt} \right)}_{Ice}}\]Þ \[{{\left( \frac{\Delta \theta }{Ll} \right)}_{Steam}}={{\left( \frac{\Delta \theta }{Ll} \right)}_{Ice}}\] Þ \[\frac{(200-100)}{540\times x}=\frac{(200-0)}{80\,(3.1-x)}\]Þ x = 0.4 m = 40 cm