Question

A source supplies heat to a system at the rate of \( 1000 W \). If the system performs work at the rate of \( 200 W \), the rate at which internal energy of the system increases is:

• A. \( 1200 W \)
• B. \( 600 W \)
• C. \( 500 W \)
• D. \( 800 W \)

Hint:

The rate of change of internal energy equals heat supplied minus work done by the system.

Answer 1

The Correct Option is D

Step 1: Apply the first law of thermodynamics:
\[ \frac{dQ}{dt} = \frac{dU}{dt} + \frac{dW}{dt} \] Step 2: Substitute the given values:
\[ 1000 = \frac{dU}{dt} + 200 \] Step 3: Solve for \( \frac{dU}{dt} \):
\[ \frac{dU}{dt} = 1000 - 200 = 800 W \] The internal energy change rate is \( 800 W \).