Q164 / 180 · COMEDK 2025 (25 May Shift) · Physics
A thermal source supplies heat at a rate of $300 \mathrm{Js}^{-1}$. By using that heat a system performs work at a rate of $125 \mathrm{Js}^{-1}$. What is the rate at which the internal energy of the system increases?
- A$212.5 \mathrm{Js}^{-1}$
- Correct answerB$175 \mathrm{Js}^{-1}$
- C$87.5 \mathrm{Js}^{-1}$
- D$425 \mathrm{Js}^{-1}$
Solution
According to the first law of thermodynamics, the rate of heat supplied to the system is equal to the sum of the rate of work done by the system and the rate of change of internal energy of the system.
Let $\dfrac{dQ}{dt}$ be the rate of heat supply, $\dfrac{dW}{dt}$ be the rate of work done, and $\dfrac{dU}{dt}$ be the rate of change of internal energy.
The equation is given by $\dfrac{dQ}{dt} = \dfrac{dW}{dt} + \dfrac{dU}{dt}$.
Given values are $\dfrac{dQ}{dt} = 300 \text{ Js}^{-1}$ and $\dfrac{dW}{dt} = 125 \text{ Js}^{-1}$.
Substituting these values into the equation:
$300 = 125 + \dfrac{dU}{dt}$
$\dfrac{dU}{dt} = 300 - 125 = 175 \text{ Js}^{-1}$.
Answer: $175 \mathrm{Js}^{-1}$
Let $\dfrac{dQ}{dt}$ be the rate of heat supply, $\dfrac{dW}{dt}$ be the rate of work done, and $\dfrac{dU}{dt}$ be the rate of change of internal energy.
The equation is given by $\dfrac{dQ}{dt} = \dfrac{dW}{dt} + \dfrac{dU}{dt}$.
Given values are $\dfrac{dQ}{dt} = 300 \text{ Js}^{-1}$ and $\dfrac{dW}{dt} = 125 \text{ Js}^{-1}$.
Substituting these values into the equation:
$300 = 125 + \dfrac{dU}{dt}$
$\dfrac{dU}{dt} = 300 - 125 = 175 \text{ Js}^{-1}$.
Answer: $175 \mathrm{Js}^{-1}$
Hence, the correct answer is option: 'B'.