A) 500; 7.5 %
B) 700; 10.5 %
C) 1000; 21 %
D) 1500; 15 %
Correct Answer: B
Solution :
From first law of thermodynamics \[Q=\Delta U+W\] or \[\Delta U=Q-W\] \[\therefore \] \[\Delta {{U}_{1}}={{Q}_{1}}-{{W}_{1}}=6000-2500=3500J\] \[\Delta {{U}_{2}}={{Q}_{2}}-{{W}_{2}}=5500+1000=-4500J\] \[\Delta {{U}_{3}}={{Q}_{3}}-{{W}_{3}}=-3000+1200=-1800J\] \[\Delta {{U}_{4}}={{Q}_{4}}-{{W}_{4}}=3500-x\] For cyclic process \[\Delta U=0\] \[\therefore \] \[3500-4500-1800+3500-x=0\] or \[x=700J\] Efficiency, \[\eta =\frac{output}{input}\times 100\] \[=\frac{{{W}_{1}}+{{W}_{2}}+{{W}_{3}}+{{W}_{4}}}{{{Q}_{1}}+{{Q}_{4}}}\] \[=\frac{(2500-1000-1200+700)}{6000+3500}\times 100\] \[=\frac{1000}{9500}\times 100\] \[\eta =10.5%\]You need to login to perform this action.
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