A bodycools from 80^(@)C to 60^(@)C in 2 mintues .In how time it cools from 60^(@) to 40^(@)C ? The temperatureof the surroudingis 10^(@)C

A bodycools from 80^(@)C to 60^(@)C in 2 mintues .In how time it cools

1.	A bodycools from 80^(@)C to 60^(@)C in 2 mintues .In how time it cools from 60^(@) to 40^(@)C ? The temperatureof the surroudingis 10^(@)C
Answer» Solution :I CASE : Meantemperatureof the body `= (80 + 60)/(2) = 70^(@)C`. Meanecess temperature `= 70 -10=60^(@)C` `(d theta)/(dt) = K ( theta - theta_(0))rArr (20)/(2) = K(60)…….(1)` II case : Meantemperature of body`(60+40)/(2) =50^(@)C` Meanexcess temperature ` = (50-10) = 40^(@)C`. Let.t. minutesbe thetime periodto COOL down `60^(@)C` to `40^(@)C`. The`(20)/(t) = k (40).........(2)` `(1)+ (2) , (t)/(2) = (60)/(40) i.e., t = 3 ` minutes.