Thermodynamics

JEE Main level, single correct, medium

34 of 190

Question 34 of 190

A source supplies heat to a system at the rate of 1000W.1000W. If the system performs work at a rate of 200W.200W. The rate at which internal energy of the system increases is

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Answer: (B) 800W800W

According to the first law of thermodynamics, Q=W+UQ=W+U.
Upon differentiation, we get
dQdt=dWdt+dUdt⇒dUdt=dQdt−dWdt\dfrac{dQ}{dt}=\dfrac{dW}{dt}+\dfrac{dU}{dt} \Rightarrow \dfrac{dU}{dt}=\dfrac{dQ}{dt}-\dfrac{dW}{dt}
Given: dQdt=1000W\dfrac{dQ}{dt}=1000W
dWdt=200W\dfrac{dW}{dt}=200W
Therefore, dUdt=1000W−200W=800W\dfrac{dU}{dt}=1000W-200W=800W

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