A cup of coffee cools from 90°C to 80°C in t minutes when the room temperature is 20°C. The time taken by the similar cup of coffee to cool from 80°C to 60°C at the same room temperature is:

Solution

Physics sir · 5 steps

Use the average temperature in each interval: cooling rate is proportional to its excess over the room.

1. Average form. Over a short interval, Newton's law of cooling says the rate of fall of temperature is proportional to the average excess over the surroundings. Here the room is at θ₀ = 20°C.

   Δθ/(Δ t) = k((θ₁+θ₂)/2-θ₀)

2. First interval. From 90 to 80°C the drop is 10 and the average is 85, so 10/t = k(85-20) = 65k.
3. Second interval. From 80 to 60°C the drop is 20 and the average is 70, so 20/t' = k(70-20) = 50k.
4. Divide. Dividing the first equation by the second gives (10 t')/(20 t) = 65/50, so t' = (2 × 65)/50 t = 13/5 t.
5. So. The coffee takes 13/5t to cool from 80°C to 60°C, which is option (a).

The idea to keep. For Newton's cooling with two intervals, write drop over time equals k times (average temperature minus room) for each, then divide. Watch both the drop and the average change.

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