JEE Main 2023 · 6 April · morning · Single correct · +4 / −1

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

  1. (a) 600 W
  2. (b) 1200 W
  3. (c) 500 W
  4. (d) 800 W
Answer and solution

Answer (d) 800 W

Solution

Physics sir · 3 steps

Heat coming in pays for the work going out first, and whatever is left stays as internal energy.

  1. First law per second. The first law, Q = Δ U + W, holds every second too. So divide each term by time and work with rates.

    dU/dt = dQ/dt - dW/dt

  2. Signs. Heat is supplied to the system, so dQ/dt = +1000 W. Work is done by the system, so dW/dt = +200 W.
  3. So. dU/dt = 1000 - 200 = 800 W, which is option (d).

The idea to keep. Heat in is split two ways: some becomes internal energy, the rest leaves as work done by the system. With rates, use dQ/dt = dU/dt + dW/dt.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Thermodynamics

Asked about 150 times in JEE Main since 2020. This question is from 2023.