Question

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 increase is

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

Answer 1

The Correct Option is C

From 1st law of thermodynamics,
𝑑𝑄 = 𝑑𝑈 +𝑑𝑊 
Also, we can write this as, 𝑑𝑄/𝑑𝑡 = 𝑑𝑈/𝑑𝑡 + 𝑑𝑊/𝑑𝑡 
⟹ 1000𝑊 = 𝑑𝑈 𝑑𝑡 + 200𝑊 
⟹ 𝑑𝑈 𝑑𝑡 = 800W

Answer 2

First Law of Thermodynamics Problem 

Step 1: First Law of Thermodynamics

The first law of thermodynamics states that:

\( \Delta Q = \Delta U + \Delta W \)

We can also express this in terms of rates as:

\( \frac{dQ}{dt} = \frac{dU}{dt} + \frac{dW}{dt} \)

Step 2: Given Values and Substitution

We are given the rate of heat transfer to the system as \( \frac{dQ}{dt} = 1000 \, \text{W} \) and the rate of work done by the system as \( \frac{dW}{dt} = 200 \, \text{W} \). Substituting these values into the equation:

\( 1000 \, \text{W} = \frac{dU}{dt} + 200 \, \text{W} \)

Step 3: Solve for \( \frac{dU}{dt} \)

Rearrange the equation to solve for the rate of change of internal energy:

\( \frac{dU}{dt} = 1000 \, \text{W} - 200 \, \text{W} = 800 \, \text{W} \)

Conclusion:

The rate at which the internal energy of the system increases is 800 W (Option 3).