Q141 / 180 · NEET 2026 (Cancelled) · Physics
An electric heater supplies heat to a system at a rate of $100$ W. If the system performs work at a rate of $75$ J/s, then the rate at which internal energy increases will be :
- A$75$ W
- B$100$ W
- C$125$ W
- Correct answerD$25$ W
Solution
According to the first law of thermodynamics:
$\dfrac{dQ}{dt} = \dfrac{dU}{dt} + \dfrac{dW}{dt}$
Given that the rate of heat supplied is $\dfrac{dQ}{dt} = 100$ W and the rate of work done by the system is $\dfrac{dW}{dt} = 75$ J/s = $75$ W.
Substituting the values:
$100 = \dfrac{dU}{dt} + 75$
$\dfrac{dU}{dt} = 100 - 75 = 25$ W
The rate at which internal energy increases is $25$ W.
Answer: $25$ W
$\dfrac{dQ}{dt} = \dfrac{dU}{dt} + \dfrac{dW}{dt}$
Given that the rate of heat supplied is $\dfrac{dQ}{dt} = 100$ W and the rate of work done by the system is $\dfrac{dW}{dt} = 75$ J/s = $75$ W.
Substituting the values:
$100 = \dfrac{dU}{dt} + 75$
$\dfrac{dU}{dt} = 100 - 75 = 25$ W
The rate at which internal energy increases is $25$ W.
Answer: $25$ W
Hence, the correct answer is option: 'D'.