Thermodynamics Question 344

Question: A system is provided with 200 cal of heat and the work done by the system on the surrounding is 40 J. Then its internal energy

Options:

A) Increases by 600 J

B) Decreases by 800 J

C) Increases by 800 J

D) Decreases by 50 J

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Answer:

Correct Answer: C

Solution:

[c] $ \Delta Q=\Delta U+\Delta W $

$ \because $ $ \Delta Q=200cal=200\times 4.2=840J $

and $ \Delta W=40J $

Therefore $ \Delta U=\Delta Q-\Delta W=840-40=800J $



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