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10 kilograms of water are boiled at \( 373 K \), at a pressure of \( 1.013 \times 10^{5} N / m ^{2} \) and converted into steam. The specific latent heat of vaporization of water is \( 539 kcal / kg .1 L \) of water, on conversion into steam, occupies \( 1671 L \). The mechanical equivalent of heat is \( 4186 J / kcal \). Calculate (a) the energy supplied to the system (water)(b) the work done by the system(c) the change in the internal energy of the system. |
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Answer» Given P = 1.013 x 105 N/ m2 M = 10 kg L = 539 kcal/kg V1 = 1 L V2 = 1671 L (a) Q = ML = 10 x 103 x 539 Q = 5390 kcal (b) Work done W = Pdv = 1.013 x 105 (V2- V1) = 1.013 x 105 (1671 -1) x 10-3 W = \(\frac {1691.71 \times 10^2\,J}{4.2}\) W = 402.7 x 102 W = 40.27 kcal (c) dθ = du + dW du = dθ - dW = 5390 - 40.27 du = 5349.73 kcal |
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