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:

Given: Heat is supplied to the system at the rate of 1000 W and the system performs work at the rate of 200 W. Find: The rate at which the internal energy increases. Using the first law of thermodynamics, Q = dUdt + W Substituting the given values, 1000 = dUdt + 200 Therefore, dUdt = 800 W Therefore, the rate of increase of internal energy is 800 W. The correct option is C.

A source supplies heat to a system at the rate of 1000 W. If the… — Solution | JEE Physics

MCQEasyJEE 2023First Law & Internal Energy

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

A
$500 \, \text{W}$
B
$600 \, \text{W}$
C
$800 \, \text{W}$
D
$1200 \, \text{W}$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Heat is supplied to the system at the rate of $1000 \, \text{W}$ and the system performs work at the rate of $200 \, \text{W}$ .

Find: The rate at which the internal energy increases.

Using the first law of thermodynamics,

\dot{Q} = \frac{dU}{dt} + \dot{W}

Substituting the given values,

1000 = \frac{dU}{dt} + 200

Therefore,

\frac{dU}{dt} = 800 \, \text{W}

Therefore, the rate of increase of internal energy is $800 \, \text{W}$ . The correct option is C.

Common mistakes

Using the wrong sign convention for work. If the system performs work, that work is subtracted from the heat supplied. Use $\dot{Q} = \frac{dU}{dt} + \dot{W}$ , not addition of both given rates.
Confusing power with energy. Here $1000 \, \text{W}$ and $200 \, \text{W}$ are rates, so the answer must also be a rate in $\text{W}$ , not in joules.
Relying on the incorrect extracted solution statement. The numerical substitution there is inconsistent with the question values. Always match the formula with the actual heat input and work output given in the question.

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Answer

Step-by-step solution

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