A calorimeter containing a hot lilquid is placed inside an enclosure whose walls are at `10^(@)C` and cools from `80^(@)C` to `60^(@)C` in 10 minutes. How long will it take to cool from `60^(@)C` to `40^(@)C`, if Newton's law of cooling holds good ?

Answer

`[15 min]`

Explanation

Here, `T_(0)=10^(@)C`,
In first case, `T_(1)=80^(@)C, T_(2)=60^(@)C`,
`Delta T=80-60=20^(@)C, Delta t=10 min= 10xx60 s`
Average temperature
`T=(T_(1)+T_(2))/(2)=(80+60)/(2)=70^(@)C`
Ac cording to Newton's law of cooling
`(Delta T)/(Delta t)= -K[T-T_(0)]`
or `(20)/(10xx60)= -K[70-10]= -Kxx60` ...(i)
In second case, `T_(1)=60^(@)C, T_(2)= 40^(@)C`,
`Delta T=60-40=20^(@)C, Delta t=?`,
Average Temperature, `T=(60+40)/(2)=50^(@)C`
`(Delta T)/(Delta t)= -K[T-T_(0)]` or `(20)/(Delta t)= -K[50-10]` .....(ii)
Dividing (i)by (ii), we get
`(Delta t)/(10xx60)= (60)/(40)= (3)/(2)` or `Delta t=(3)/(2)xx10xx60 s =15 min`