Question

A body cools from `80^(@)C` to `50^(@)C` in 5 min-utes Calculate the time it takes to cool from `60^(@)C` to `30^(@)C` The temperature of the surroundings is `20^(@)C` .

Answer 1

According to Newton’s law of cooling, we have:
`-(dT)/(dt)=K(T-T_(0))`
`(dT)/(K(T-T_(0))= -Kdt`…(i)
Where,
Temperature of the body = T
Temperature of the surroundings `= T_(0) = 20^(@)C`
K is a constant
Temperature of the body falls from `80^(@)C" to "50^(@)C` in time, t = 5 min = 300 s
Integrating equation (i), we get:
`underset(50)overset(80)int(dT)/(K(T-T_(0)))= -underset(0)overset(300)intKdt`
`[log_(e)(T-T_(0))]_(50)^(80)= -K[t]_(0)^(300)`
`(2.3026)/(K)"log"_(10) (80-20)/(50-20)= -300`
`(2.3026)/(K)"log"_(10)2= -300`
`(-2.3026)/(300)"log"_(10)2=K`...(ii)
The temperature of the body falls from `60^(@)C" to "30^(@)C` in time = t’
Hence, we get:
`(2.3026)/(K)"log"_(10)(60-20)/(30-20)= -t`
`(-2.3026)/(t)"log"_(10)4=(-2.3026)/(300)"log"_(10)2`
`:. t=300xx2=600s=10 min`
Therefore, the time taken to cool the body from `60^(@)C" to "30^(@)C` is 10 minutes.