Thermodynamics Question 131
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
[Orissa PMT 2004]
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:
$ \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 $
