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Q.
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
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Solution:
$ dQ=+\,200\,cal $ $ =200\times 4.2=840\,J $ $ dW=+\,40\,J $ From the first law of thermodynamics $ dQ=dU+dW $ $ dU=dQ-dW $ $ =840-40=800\,J $ So, the internal energy of the .system increased by 800 J.