InterviewSolution
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InterviewSolution
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Ten litres of water are boiled at 100°C, at a pressure of 1.013 × 105 Pa, and converted into steam. The specific latent heat of vaporization of water is 539 cal/g. Find (a) the heat supplied to the system (b) the work done by the system (c) the change in the internal energy of the system. 1 cm3 of water on conversion into steam, occupies 1671 cm3 (J = 4.186 J/cal) |
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Answer» Data : P = 1.013 × 105 Pa, V (water) = 10 L = 10-3 × 103 m , V(steam) = 1671 × 10-3 × 103 m , L = 539 cal/g = 539 × 103 \(\frac{cal}{kg}\) = 539 × 103 × 4.186 \(\frac{J}{kg}\) as J = 4.186 J/cal, mass of the water (M) = volume × density = 10 × 10-3 m3 × 103 kg/m3 = 10 kg (a) Q = ML = (10) (5.39 × 4.186 × 105) J = 2.256 × 107 J This is the heat supplied to the system (b) W = P∆V = (1.013 × 105) (1671 – 1) × 10-2 J = (1.013) (1670) × 103 J = 1.692 × 106 J This is the work done by the system. (c) ∆ U = Q – W = 22.56 × 106 – 1.692 × 106 = 2.0868 × 107J This is the change (increase) in the internal energy of the system. |
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