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)`.

Options

A. `0.092 cal//gm^(@)C`
B. `0.124 cal//gm^(@)C`
C. `0.162 cal//gm^(@)C`
D. `0.242 cal//gm^(@)C`

Answer

A

Explanation

`(m)/(rho_(ice))-(m)/(rho_(H_(2)O)) = 0.15`
`m=(0.15xx0.92)/(0.08)`
`:. H_("lost") =H_("gained") rArr 15xxSxx100=(0.15xx0.92)/(0.08)xx80`
`S = 0.092 cal//gm^(@)C`.