A) 1315J
B) 275J
C) 765J
D) 675J
Correct Answer: C
Solution :
\[\operatorname{In} a cyclic process \Delta U = 0\] 1st law of thermodynamics \[\operatorname{Q}=\Delta U+W\] \[{{\operatorname{Q}}_{1}}+ {{Q}_{2}}+{{Q}_{3}}+{{Q}_{4}}={{W}_{1}}+{{W}_{2}}+{{W}_{3}}+{{W}_{4}}\] \[5960-5585-2980+3645 =2200-825-1100+{{W}_{4}}\]\[{{\operatorname{W}}_{4}}= 1040-275\] \[{{\operatorname{W}}_{4}}=765J\]You need to login to perform this action.
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