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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`). |
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Answer» Correct Answer - D Let m be the mass of the container. Initial temperature of container, `T_i = (227 + 273) = 500 K` and final temperature of container, `T_f = (27 + 273) = 300 K ` Now, heat gained by the ice cube = heat lost by the container `:. (0.1)(8xx (10^4)) + (0.1)(10^3)(27) = -m (int_(500)^300)(A+BT)dT` or `10700 = -m [AT + (BT^2)/2]_(500)^(300)` After substituting the values of A and B and the proper limits, we get `m=0.495 kg` . |
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