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(D) According to the first law of thermodynamics,the heat supplied to a system $(dQ)$ is equal to the sum of the change in internal energy $(dU)$ and

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$800$

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(D) According to the first law of thermodynamics,the heat supplied to a system $(dQ)$ is equal to the sum of the change in internal energy $(dU)$ and the work done by the system $(dW)$: $dQ = dU + dW$.Dividing by the time interval $(dt)$,we get the rate equation: $\frac{dQ}{dt} = \frac{dU}{dt} + \frac{dW}{dt}$.Given:Rate of heat supply,$\frac{dQ}{dt} = 1000 \, W$.Rate of work done,$\frac{dW}{dt} = 200 \, W$.Rearranging the equation to find the rate of increase of internal energy:$\frac{dU}{dt} = \frac{dQ}{dt} - \frac{dW}{dt}$.Substituting the values:$\frac{dU}{dt} = 1000 \, W - 200 \, W = 800 \, W$.

$A$ 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 the internal energy of the system increases is $....... \, W$.

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