In a cyclic process, heat transfers are +34.7kJ , −15.2kJ, −13.56kJ and +21.5kJ. What is the net work done in the process?
A
+17.44kJ
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B
+27.44kJ
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C
−17.44kJ
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D
−27.44kJ
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Solution
The correct option is D−27.44kJ In the cyclic process, △Ucyc=0 as internal energy is state function.
as w+q=△U
Therefore, wcyc+qcyc=0 wcyc=−qcyc
so, the net work done is wcyc=−[34.7−15.2−13.56+21.5]kJ wcyc=−27.44kJ