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
- 500 W
- 600 W
- 800 W
- 1200 W
The Correct Option is C
Approach Solution - 1
𝑑𝑄 = 𝑑𝑈 +𝑑𝑊
Also, we can write this as, 𝑑𝑄/𝑑𝑡 = 𝑑𝑈/𝑑𝑡 + 𝑑𝑊/𝑑𝑡
⟹ 1000𝑊 = 𝑑𝑈 𝑑𝑡 + 200𝑊
⟹ 𝑑𝑈 𝑑𝑡 = 800W
Approach Solution -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).
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