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A source supplies heat to a system at the rate of 1000 W. If the system performs work at a rate of 200 W. The
rate at which internal energy of the system increase is

Option: 1

500 W


Option: 2

600 W


Option: 3

800 W


Option: 4

1200 W


Answers (1)

best_answer

From Ist law of thermodynamics,
d Q=d U+d W
Also, we can write this as, \frac{d Q}{d t}=\frac{d U}{d t}+\frac{d W}{d t}
\begin{aligned} & \Rightarrow 1000 \mathrm{~W}=\frac{d U}{d t}+200 \mathrm{~W} \\ & \Rightarrow \frac{d U}{d t}=800 \mathrm{~W} \end{aligned}

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Divya Prakash Singh

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