A body cools from `80^(0)` C to `60^(0)` C in 2 minutes. In how much time it cools from `60^(0)` to `40^(0)` C ? The temperature of the surroundings is `10^(0)` C

Explanation

I case : Mean temperature of the body = `(80 + 60)/(2) = 70^(@)` C
Mean excess temperature = 70 - 10 = `60^(@)` C
`(d theta)/(dt) = K (theta - theta_(0) ) rArr (20)/(2) = K (60) rarr (1)`
II case : mean temperature of the body `(60+ 40)/(2) = 50^(@)` C
mean excess temperature = (50 - 10) = `40^(@)` C
let .t. minutes be the time to cool down from `60^(@)` C to `40^(@)` C
Then `(20)/(t ) = K (40) rarr (2)`
Dividing equation (1) by (2)
`(t)/(2) = (60)/(40) ` ie, t = 3 minutes