1.

Calculate the amount of heat required in calorie to change 1 g of ice at `-10^(@)C` to steam at `120^(@)C`. The entire process is carried out at atmospheric pressure. Specific heat of ice and water are `0.5 cal g^(-1) .^(@)C^(-1)` and `1.0 cal g^(-1) .^(@)C^(-1)` respectively. Latent heat of fusion of ice and vaporization of water are `80 cal g^(-1)` and `540 cal g^(-1)` respectively. Assume steam to be an ideal gas with its molecules having 6 degrees of freedom. Gas constant `R = 2 cal mol^(-1) K^(-1)`.

Answer» Correct Answer - 733.8 cal


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