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 increases

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Correct answer: D. 800 W

Given, Rate of heat supplied, dQ dt =1000 ~W Rate of work performed, dw dt =200 ~W Using first law of thermodynamics d Q=d U+d w $ dU dt = dQ dt - dw dt dU dt =1000-200=800$

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