A bowl filled with very hot soup cools from 98°C to 86°C in 2 minutes when the room temperature is 22°C. How long it will take to cool from 75°C to 69°C?

Solution

Physics sir · 4 steps

Use the average temperature over each interval. A cooler bowl loses heat more slowly, so the time grows.

1. Averaged cooling law. For a small temperature drop, the rate of cooling is proportional to how much hotter the body is than the room. Use the average temperature over the interval.

   (Δ T)/t = k((T₁ + T₂)/2 - Tₛ)

2. First interval. The drop is 12 °C in 2 minutes, so the rate is 6 °C min⁻¹. The average temperature is 92 °C, which is 70 °C above the room. So 6 = 70k and k = 3/35 min⁻¹.
3. Second interval. The drop is 6 °C. The average temperature is 72 °C, which is 50 °C above the room. So 6/t = 50 × 3/35 = 30/7, which gives t = 42/30 = 1.4 minutes.
4. So. The soup takes 1.4 minutes to cool from 75 °C to 69 °C, which is option (c).

The idea to keep. In Newton's cooling problems, take the ratio of the two intervals so k cancels. The time depends on the temperature drop and on how far the average temperature sits above the surroundings.

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