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

According to Newton.s law of cooling,
`-(dT)/(dt)=K.(T-T_(s))`
`-((50-80)/(5))=K.[(50+80)/(2)-20]`
`6=K.(65-20)`
`:.K.=(6)/(45)`
Now, again,
`-(dT)/(dt)=K.(T-T_(s))`
`:.((30-60)/(t) )=(6)/(45)[(30+60)/(2)-20]`
`:.(30)/(t)=(2)/(15)[45-20]`
`:.(30)/(t)=(2)/(15)xx25`
`:.(30)/(t)=(10)/(3)`
`:.t=(90)/(10)`
`:.t=9" min"`