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

Options

A. `600 W`
B. `800 W`
C. `500 W`
D. `1200 W`

Answer

To find the rate at which the internal energy of the system increases, we can use the first law of thermodynamics, which states:

\[ \Delta U = Q - W \]

where:
- \(\Delta U\) is the change in internal energy,
- \(Q\) is the heat supplied to the system,
- \(W\) is the work done by the system.

Given:
- The rate of heat supplied to the system, \(Q = 1000 \, \text{W}\),
- The rate at which the system performs work, \(W = 200 \, \text{W}\).

Now, we can substitute the values into the equation:

\[ \Delta U = Q - W \]

Substituting the given values:

\[ \Delta U = 1000 \, \text{W} - 200 \, \text{W} \]

Calculating the right side:

\[ \Delta U = 800 \, \text{W} \]

Thus, the rate at which the internal energy of the system increases is \(800 \, \text{W}\).

### Final Answer:
The rate at which the internal energy of the system increases is \(800 \, \text{W}\).

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