Question

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.

Answer 1

T0=390 degF

Step-by-step 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

∫
`(dT)/(T-T0) =\int\limits^0.5_0 {-k} \, dt`

ln(T-T0), from 110 to 70=-kt, from 1/2 to 0

ln(110-T0)-ln(70-T0)=-k(1/2-0)

`2in(110-T0)/(70-T0) =-k\\k=-2ln(T0-110)/(T0-70) ...........................1`

integrating the LHS of the equation from 110 to 145F

∫∫∫
`(dT)/(T-T0) =\int\limits^1_0.5 {-k} \, dt`

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