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A metal piece weighing 15 g is heated to...

A metal piece weighing `15 g` is heated to `100^(@)C` and then immersed in a mixture of ice and water at the thermal equilibrium. The volume of the mixture is found to be reduced by `0.15 cm^(3)` with the temperature of mixture remaining constant. Find the specific heat of the metal. Given specific gravity of ice `= 0.92`, specific gravity of water at `0^(@)C = 1.0`, latent heat of fusion of ice `= 80 cal-g^(-1)`.

Text Solution

Verified by Experts

The correct Answer is:
`0.092 cal//gm"^(@)C`
`(m)/(rho_("ice")) - (m)/(rho_(H_(2)O)) = 0.15`
`m = (0.15 xx 0.92)/(0.08)`
`:. H_("lost") = H_("gained")`
`rArr 15 xx S xx 100 = (0.15 xx 0.92)/(0.008) xx 80`
`S = 0.092 cal//gm.^(@)C`.
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