Answer:
T0=390 degF
Explanation:
The thermometer reading 70◦Fis placed in an oven preheated to a constant temperature. Through the glass window in the oven door, an observer records that the thermometer reads 110◦F after 12 minutes and 145◦F after 1 minute. How hot is the oven? Newton’s law of cooling yields the following differential equationdTdt=k(T−T0), whereT0 is the ambient temperature.
If T is temperature, then by Newton’s law
dTdt=-k(T−T0)
Where
T0− is temperature of oven
∫![\frac{dT}{T-T0} =\int\limits^0.5_0 {-k} \, dt](https://tex.z-dn.net/?f=%5Cfrac%7BdT%7D%7BT-T0%7D%20%3D%5Cint%5Climits%5E0.5_0%20%7B-k%7D%20%5C%2C%20dt)
ln(T-T0), from 110 to 70=-kt, from 1/2 to 0
ln(110-T0)-ln(70-T0)=-k(1/2-0)
![2in\frac{110-T0}{70-T0} =-k\\k=-2ln\frac{T0-110}{T0-70} ...........................1](https://tex.z-dn.net/?f=2in%5Cfrac%7B110-T0%7D%7B70-T0%7D%20%3D-k%5C%5Ck%3D-2ln%5Cfrac%7BT0-110%7D%7BT0-70%7D%20...........................1)
integrating the LHS of the equation from 110 to 145F
∫∫∫![\frac{dT}{T-T0} =\int\limits^1_0.5 {-k} \, dt](https://tex.z-dn.net/?f=%5Cfrac%7BdT%7D%7BT-T0%7D%20%3D%5Cint%5Climits%5E1_0.5%20%7B-k%7D%20%5C%2C%20dt)
ln(145-T0)-ln(110-T0)=-k(1-1/2)
2ln145-T0/(110-T0)=-k
k=-2lnT0-145/(T0-110).......................2
couplijng equatiuon 2 with 1
(T0-110)^2=(T0-70)(T0-145)
T0^2-220T0+12100=T0^2-215T0+10150
5T0=1950
T0=390 degF