The efficiency of a particular car engine is 24% when the enginedoes 9.6 kJ of w

Question

The efficiency of a particular car engine is 24% when the enginedoes 9.6 kJ of work per cycle. Assumethe process is reversible. (a) What is the energy the engine gains percycle as heat Qgain fromthe fuel combustion?
1 kJ

(b) What is the energy the engine loses per cycle as heatQlost?
2 kJ

(c) If a tune-up increases the efficiency to 29%, what isQgain at the same workvalue?
3 kJ

(d) If a tune-up increases the efficiency to 29%, what isQlost at the same workvalue?
4 kJ (a) What is the energy the engine gains percycle as heat Qgain fromthe fuel combustion?
1 kJ

(b) What is the energy the engine loses per cycle as heatQlost?
2 kJ

(c) If a tune-up increases the efficiency to 29%, what isQgain at the same workvalue?
3 kJ

(d) If a tune-up increases the efficiency to 29%, what isQlost at the same workvalue?
4 kJ

Explanation / Answer

We know that : .                  Efficiency  (E) = W / Q .         or             Qgain     =   W /E . (a) .                           Qgain    =   9.6 kJ / (24 / 100 ) .                                         =   40 kJ . (b) .                           Qloss      = Qgain   - W = 40 kJ - 9.6 kJ   = 30.4   kJ . (c) .                             Qgain    =   9.6 kJ / (29 / 100 ) .                                         =   33.10 kJ . (d) .                             Qloss     = Qgain   - W =  33.10 kJ - 9.6 kJ  =  23.5   kJ . Hope this helps u! .                                         =   33.10 kJ . (d) .                             Qloss     = Qgain   - W =  33.10 kJ - 9.6 kJ  =  23.5   kJ . Hope this helps u!

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