A metal piece weighing 15 g is heated to 1000C and the immersed in a mixture of ice and water in thermal equilibrium. The volume of the mixture is found to be reduced by 0.15 cm3 with the temperature of the mixture remaining constant. Find the specific heat of the metal. Given specific gravity of ice = 0.92, specific gravity of water at 00C = 1.0, latent heat of fusion of ice = 80 cal g-1

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0.092 cal/g0C

0.124 cal/g0C

0.162 cal/g0C

0.242 cal/g0C

answer is A.

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Detailed Solution

mρice-mρH2O=0.15⇒m=0.15×0.920.08Hlost=Hgained⇒15×S×100=0.15×0.920.08×80 ⇒S=0.092 cal/g0C

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