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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 100C. At what point along its length should a temperature of 200C be maintained so that in steady state, the mass of ice melting be equal to that the steam produced in same interval of time? Assume that the whole system is insulate from surroundings. Latent heat of fusion of ice and vaporization of water are 80 cal/g and 540 cal/g, respectively.


A

8.59 cm from ice end

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B

10.34 cm from water end

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C

10.34 cm from ice end

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D

8.76 cm from water end

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Solution

The correct option is B

10.34 cm from water end


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

Qice = KA(2000)(1.5x)×t

So ice melted by this heat

mice = QiceLF = KA80(2000)(1.5x)×t

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

Qwater = KA(200100)x×t

Steam formed by this heat

mstream = QwaterLv = KA(200100)500 × xt

According to given problem mice = msteam

i.e.,20080(1.5x) = 100540 × xx = 658m = 10.34cm

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


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