A body cools from 60°C to 40°C in 6 minutes. If, temperature of surroundings is 10°C. Then, after the next 6 minutes, its temperature will be ________°C.
Show answer and solution
Answer: 28
Solution
Using Newton's Law of Cooling in average form:
(T₁ - T₂)/t = k·((T₁ + T₂)/2 - T₀)
First interval: (60 - 40)/6 = k·((60 + 40)/2 - 10)
20/6 = k·(50 - 10) = 40k
k = 1/12 per minute
Second interval: Let final temperature = T
(40 - T)/6 = (1/12)·((40 + T)/2 - 10)
(40 - T)/6 = (1/12)·((40 + T - 20)/2) = (20 + T)/24
24(40 - T) = 6(20 + T)
960 - 24T = 120 + 6T
840 = 30T
T = 28°C
(T₁ - T₂)/t = k·((T₁ + T₂)/2 - T₀)
First interval: (60 - 40)/6 = k·((60 + 40)/2 - 10)
20/6 = k·(50 - 10) = 40k
k = 1/12 per minute
Second interval: Let final temperature = T
(40 - T)/6 = (1/12)·((40 + T)/2 - 10)
(40 - T)/6 = (1/12)·((40 + T - 20)/2) = (20 + T)/24
24(40 - T) = 6(20 + T)
960 - 24T = 120 + 6T
840 = 30T
T = 28°C
[JEE Main 2023 · 29 Jan · Shift 1]