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 is (a) 600 W (b) 1200 W (c) 500 W (d) 800 W

Answer: (d). First law per unit time: dQ/dt = dU/dt + dW/dt dU/dt = 1000 - 200 = 800 W So the answer is (d) 800 W.

A source supplies heat to a system at the rate of 1000 W. If the… | JEE Main 2023 (6 Apr, Shift 1)

Q9JEE Main 2023 · 6 APR · SHIFT 1Easy

Heat, Work & First Law of Thermodynamics

A source supplies heat to a system at the rate of $1000\text{ W}$ . If the system performs work at a rate of $200\text{ W}$ . The rate at which internal energy of the system increases is

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Answer: (d)

First law per unit time: $\frac{dQ}{dt} = \frac{dU}{dt} + \frac{dW}{dt}$
$\frac{dU}{dt} = 1000 - 200 = 800\text{ W}$
So the answer is (d) 800 W.

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