Question

please solve this question​

Answer 1

Explanation:

In path

dQ=50050×4.2=210J

dW=W

AC

+W

CB

=50×10

3

×200×10

−6

=10J

dQ=dU+dW

⇒dU=dQ−dW=210−10=200J

In path ADB,dQ=?

du=200J (Internal energy change between 2 points is always same)

dW=W

AD

+W

DB

=0+155×10

3

×200×10

−8

=31J

dQ=dU+dW=200+31=231J=55cal

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Hope it help you

Answer 2

Answer:

this is answer of your question