A body at `80" "^(@)C` cools down to `64" "^(@)C` in 5 minutes, and in 10 minutes it cools down to `52" "^(@)C`. What will be its temperature after 20 minutes ? What is the temperature of the environment ?

Explanation

For the first 5 minute, `Deltat=5min`.
`DeltaT=T_(2)-T_(1)=64-80=-16`
Now using `-(dQ)/(dt)=-k.[T-T_(s)]`
`(16)/(5)=+k.[(80+64)/(2)-T_(s)]` . .. . (1)
Here we have taken the average of initial and final temperature as the temperature of the body.
`T=(T_(1)+T_(2))/(2)`
For the next `Deltat=5` min.,
`(12)/(5)=k.((52+64)/(2)-T_(s))` . . . (2)
Dividing equation (1) by equation (2) `(16)/(5)xx(5)/(12)=(72-T_(s))/(58-T_(s))`
`:.4/3=(72-T_(s))/(58-T_(s))`
`:.232-4T_(s)=216-3T_(s)`
`:.T_(s)=16" "^(@)C`
From equation (1)
`(16)/(5)=k.(72-16)`
`=56k.`
`:.k.=(16)/(5xx56)=(2)/(32)`
For the last `Deltat=10` minutes,
`DeltaT=T-52`, where T = Final temperature `(52-T)/(10)=(2)/(35)((52+T)/(2)-16)(becauseTlt52)`
`:.52-T=(4)/(7)((52+T-32)/(2))(because(20)/(35)=(4)/(7))`
`:." "52-T=(2)/(7)(20+T)`
`:.364-7T=40+2T`
`:.324=9T`
`:.T=(324)/(9)=36" "^(@)C`