Answer in Molecular Physics | Thermodynamics for Giselle Labagan #144533
February 21st, 2023
i) “Q_1 = 5000 ;J”
“Q_2 = 4500 ;J”
“u03b7 = 1 – frac{Q_2}{Q_1} = 1 – frac{4500}{5000} = 0.1”
ii) “Q_1 = 25000 ;J”
“W = 2000 ;J”
“W = Q_1 u2013 Q_2”
“Q_2 = 25000 u2013 2000 = 23000 ;J”
“u03b7 = 1 – frac{Q_2}{Q_1} = 1 – frac{23000}{25000} = 0.08”
iii) “W = 400 ;J”
“Q_2 = 2800 ;J”
“W = Q_1 u2013 Q_2”
“Q_1 = W + Q_2 = 400 + 2800 = 3200 ;J”
“u03b7 = 1 – frac{Q_2}{Q_1} = 1 – frac{2800}{3200} = 0.125”
Engine (iii) has the highest thermal efficiency 1.25.
Engine (ii) has the lowest thermal efficiency 0.08.