JEE Main 2021 · 22 July · evening · Numerical answer · +4 / −0

In 5 minutes, a body cools from 75°C to 65°C at room temperature of 25°C. The temperature of body at the end of next 5 minutes is _________° C.

Answer and solution

Answer 57

Solution

Physics sir · 4 steps

Track the excess over the room, not the temperature itself: it shrinks by the same factor every 5 minutes.

  1. Excess temperature. Newton's law says the excess θ - θ₀ over the surroundings decays exponentially. So in equal time intervals, the excess is multiplied by the same factor each time.

    θ - θ₀ = (θᵢ - θ₀) e⁻ᵏᵗ

  2. First 5 minutes. The excess goes from 75 - 25 = 50°C to 65 - 25 = 40°C. The factor for 5 minutes is e⁻⁵ᵏ = 40/50 = 0.8.
  3. Next 5 minutes. The same factor applies again: the new excess is 40 × 0.8 = 32°C.
  4. So. The temperature is 25 + 32 = 57 °C. The average-temperature form of Newton's law gives the same 57.

The common slip: 55. Assuming the body drops another 10°C in the next 5 minutes gives 55. The fall is smaller now because the body is closer to room temperature.

The idea to keep. In Newton's cooling, equal times multiply the excess over the surroundings by the same factor. Cooling always slows down as the body nears room temperature.

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

Study Heat Transfer

Asked about 30 times in JEE Main since 2020. This question is from 2021.