The efficiency of a particular car engine is 27% when the enginedoes 8.3 kJ of w

Question

The efficiency of a particular car engine is 27% when the enginedoes 8.3 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 32%, what isQgain at the same workvalue?
3 kJ

(d) If a tune-up increases the efficiency to 32%, 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 32%, what isQgain at the same workvalue?
3 kJ

(d) If a tune-up increases the efficiency to 32%, what isQlost at the same workvalue?
4 kJ

Explanation / Answer

efficiency e = 27 % = 0.27 we know   W / Q (gain ) from this Q ( gain ) = W / e                            = 8.3 kJ / 0.27                            = 30.74 kJ heat lost Q ' = Q ( gain ) - W                     =22.44 kJ (b). efficency e = 32 % = 0.32 Q ( gain ) = W / e              = 8.3 kJ / 0.32             = 25.9375 kJ Q lost = Q ( gain ) - W            =17.6375 kJ

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