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A Carnot engine takes 3000 kcal of heat from a reservoir at 627^(@)C. The sink is at 27^(@)C. What is the amount of work done by the engine? |
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Answer» SOLUTION :Temperature of the reservoir, `T_(1) = (627 + 273)K = 900K` Temperature of the sink, `T_(2) = (27_273)K = 300K` `:. "EFFICIENCY", eta = 1 -(T_2)/(T_1) = 1 - (300)/(900) = 2/3` Again, `eta = W/(Q_1)` or, `W = etaQ_(1) = 2/3 xx 3000` `=2000 kcal = 2000 xx 4.2 xx 10^(3)J` `=8.4 xx 10^(6) J`. |
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