In a thermodynamic process, 50J of heat is going out of the system and 10J of work is done on the system. If the initial internal energy of the system was 60J, then the final internal energy of the system will be
A
Zero
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B
20J
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C
−20J
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D
120J
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Solution
The correct option is B20J dQ=−50J (negative because heat is going out of the system) W=−10J (work done on the system) Ui=60J dQ=dU+W −50=(Uf−60)+(−10) Uf=20J