An ice cube of mass 0.1 kg at `0^@C` is placed in an isolated container which is at `227^@C`. The specific heat s of the container varies with temperature T according to the empirical relation `s=A+BT`, where `A= 100 cal//kg-K and B = 2xx (10^-2) cal//kg-K^2`. If the final temperature of the container is `27^@C`, determine the mass of the container. (Latent heat of fusion for water = `8xx (10^4) cal//kg`, specific heat of water `=10^3 cal//kg-K`).

An ice cube of mass 0.1 kg at `0^@C` is placed in an isolated containe

1.	An ice cube of mass 0.1 kg at `0^@C` is placed in an isolated container which is at `227^@C`. The specific heat s of the container varies with temperature T according to the empirical relation `s=A+BT`, where `A= 100 cal//kg-K and B = 2xx (10^-2) cal//kg-K^2`. If the final temperature of the container is `27^@C`, determine the mass of the container. (Latent heat of fusion for water = `8xx (10^4) cal//kg`, specific heat of water `=10^3 cal//kg-K`).
Answer» Correct Answer - D<br>Let m be the mass of the container.<br> Initial temperature of container,<br> `T_i = (227 + 273) = 500 K`<br> and final temperature of container,<br> `T_f = (27 + 273) = 300 K `<br> Now, heat gained by the ice cube = heat lost by the container<br> `:. (0.1)(8xx (10^4)) + (0.1)(10^3)(27) = -m (int_(500)^300)(A+BT)dT`<br> or `10700 = -m [AT + (BT^2)/2]_(500)^(300)`<br> After substituting the values of A and B and the proper limits, we get<br> `m=0.495 kg` .

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