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In a heat engine, the temperature of the source and sink are are 500K and 375 K. If the engine consumes 25xx10^(5)J per cycle, the work done per cycle is |
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Answer» Solution :`T_(1) = 500 K, T_(2) = 375K,` Heat absorbed `= Q_(1) = 25 xx 10 ^(5) J` `ETA = (T_(1) -T_(2))/(T_(1)) = (500 - 375)/(500) = -0.25 = 25%` ` eta = (W)/(Q_(1))` `W = eta xx Q_(1) = 0.25 xx 25 xx 10 ^(5) J` `= 6. 25 xx 10 ^(5) J ` per cycle Heat REJECTED to the sink per cycle `=Q_(2) = ?` `Q_(2) = Q_(1) - W = 25 xx 10 ^(5) -6.25 xx 10 ^(5)` `= 18.75 xx 10 ^(5) J` |
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