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Question

A slab of stone area 3600 cm2 and thickness 10 cm is exposed on the lower surface to steam at 1000C. A block of ice at 00C rest on upper surface of the slab. In one minute, 4.8 kg of ice is melted. The thermal conductivity of the stone in Js1m1k1 is (latent heat of ice =3.36×105J/kg)

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Solution

In Time t=60s,4.8 kg of ice in melted.
Heat consumed in process =mL=4.8×3.36×105
Q=1.6128×106 J
From Thermal transfer also,
Qt=KALT Here T= temp diff =100 K.
K=Qt.(LA).1T given,
=1.6128×10660×0.10.36×1100=74.67 J/smK

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