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

Since `(d theta)/(dt)=-K(theta-theta_0` where `theta=(theta_1+theta_2)/2" "theta_1=373K" "theta" "theta_2=273K`
`(80-50)/(5)=-K((80+50)/(2)-20)" "6=-K(45)`
`:.-K=6/(45)=2/(15)`
`(60-30)/t=(2/(15))((60+30)/2-20)` i.e, `((30)/t)=2/(15)(25)`
`:.t=(30xx15)/(50)=9` minutes