Q: A body is originally at 80 C and cools down to 60 degrees Centigrade  in 20 minutes. If the temperature of the air is 40 degree centigrade, find the temperature of the body after 40 minutes.

 

Q: A body is originally at 80 C and cools down to 60 degrees Centigrade in 20 minutes. If
the temperature of the air is 40 degree centigrade, find the temperature of the body after 40
minutes.
Let be the temperature of the body at time t By Newton’s law of cooling, we have d k 0 dt d k 40 dt
d kdt 40 Integrating on both sides, we get
d 40 kdt
ln 40 kt ln C
ln 40 ln C kt
ln 40 kt
C 40 kt
e
C 40 Ce kt Ce kt 40 When t 0, 80C Substitute t 0, 80C in Ce kt 40
80 Ce k 0 40 80 40 C
C 40 Since where 0 40 0 is the temperature of the air Hence 40e kt 40 Substitute t 20, 60C in 40e kt 40 60 40e k 20 40
60 40 40e k 20
20 40e 20 k
20 1
e 20 k 40 2
1
ln e 20 k ln ln1 ln 2 ln 2
2
20k ln 2
1
k
ln 2
20
Substitute 1
k
ln 2
20 in the equation 40e kt 40 40e When 1 ln 2 t 20 . Then 40 t 40 40e 1 ln 2 40 20 40 40e ln 2 2 40 40e 2ln 2 40
2 40eln 2 40 40 22 40
1 40 40
4 10 40 50C Hence the temperature of the body after 40 minutes is 50 degree centigrade

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