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Question

A slab of stone of area 0.36m^2 and thickness 0.1m is exposed to steam on lower surface at 373K. a block of ice at 273K rests on the upper surface of the slab. In one hour 4.8kg of ice is melted. The thermal conductivity of slab is(given latent heat of fusion of ice=3.36 * 10^5J/Kg)in J/m/s/C:1) 1.24 2) 1.29 3) 2.05 4) 1.02

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Solution

Qt= KATxthe above equation gives us the rate of flow of heat through the conductor.Q= KATxtwhere: A: areaK : conductivityThis should be equal to = mL , for the ice.KATxt = mLK(0.36)(373-273)0.160*60 = 4.8*3.36*105K = 1.244 J/ksec

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