A body cools from `70^(@)C` to `50^(@)C` in 6 minutes when the temperature of the surrounding is `30^(@)C`. What will be the temperature of the body after further 12 minutes if coolilng proseeds ac cording to Newton's law of cooling ?

Answer

`[34^(@)C]`

Explanation

Case (i) `dT =70-50=20^(@)C, dt=6` minute, `T_(0)=30^(@)C`
`T=(70+50)/(2)=60^(@)C`.
As, `(dT)/(dt)= -K[T-T_(0)]`
`:. (20)/(6)= -K[60-30]= -Kxx30` …(i)
Case (ii). Let `T_(3)` be the final temp. after further 12 minutes
`dT=(50-T_(3))^(@)C`, `dt=12 minute, T_(0)=30^(@)C`
`T=(50+T_(3))/(2)`
`:. (50-T_(3))/(12)= -K[(50+T_(3))/(2)-30]` .....(ii)
Dividing (i) by (ii), we get, `T_(3)=34^(@)C`