A body cools from `80^(@)C` to `60^(@)C` in 2 mintues . In how time it cools from `60^(@)` to `40^(@)C` ? The temperature of the surrouding is `10^(@)C`

Explanation

I case :
Mean temperature of the body `= (80 + 60)/(2) = 70^(@)C`. Mean ecess temperature `= 70 -10=60^(@)C`
`(d theta)/(dt) = k ( theta - theta_(0)) rArr (20)/(2) = K(60)…….(1)`
II case :
Mean temperature of body `(60+40)/(2) = 50^(@)C` Mean excess temperature ` = (50-10) = 40^(@)C`.
Let .t. minutes be the time period to cool down `60^(@)C` to `40^(@)C`.
The `(20)/(t) = k (40).........(2)`
`(1)+ (2) , (t)/(2) = (60)/(40) i.e., t = 3 ` minutes.