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