A body cools from `80^(@)C` to `50^(@)`C in 5 minutes. Calculate the time it takes to cool from `60^(@)C` to `30^(@)C`. The temperature of the surroundings is `20^(@)C`

Explanation

`theta=(theta_1+theta_2)/(2)=(80+50)/2=65^(@)C" "dtheta=80-50=30^(@)C`
Using `(dtheta)/(dt)=K(theta-theta_0)" "i.e (30)/(5)=k(65-20)`
`:.K=(6/(45))`…………
`dtheta'=60-30=30^(@)C" "theta'=(60+30)/(2)=45^(@)C`
i.e dt'`=(dtheta')/(K(theta'-theta_0)=(30xx45)/(6xx25)`
`dt'=9`minutes.