At 9 a.m., a thermometer reading 70F is taken outside where the temperature is 15F. At 9:05 a.m. the thermometer reading is at 45F. At 9:10 a.m. the thermometer is taken back indoors where the temperature is fixed at 70F find the reading 9:20 am.
The Answer is 58F
Newtons Law of Cooling is:
$\frac{dT}{dt} = -k(T-T_a)$
$\frac{dT}{T-T_a} = -k\cdot dt$
$T = Ce^{-kt} + T_a$
Given Conditions are:
Ta = 15F
T(0) = 70F
T(5) = 45F
T(10) = 31.363636C
T(20) = ?
At t=0; T= 70F
$70 = Ce^{-k*0} + 15$; C = 55;
Find k
$45 = 55e^{-k*5} + 15$; k = 0.1212271607
Find T(10)
$55e^{-0.1212271607*10}+15 = 31.3636367$
Find T(20); $T_a = 70$;
Use $T = T_a – Ce^{-kt}$ since colder to hotter ambient temp
$70 – 55e^{-0.1212271607*20} = 65.13148009$
I tried recomputing C and k when the thermometer was inside the room again. It just canceled out. What am I doing wrong?
Best Answer
We will solve this problem in two parts:
1) When it is taken outside:
$T_0=70^{\circ}F\hspace{10 mm}$$T_e=15^{\circ}F\hspace{10 mm}$$T(5)=45^{\circ}F\hspace{10 mm}$$T(10)=T^{\circ}F\hspace{10 mm}$
The general solution is: $\hspace{10 mm}T=15+55e^{-kt}$
Then, $\hspace{10 mm}T(5)=45=15+55e^{-k(5)}$;$\hspace{10 mm}e^{-5k}=\frac{30}{55}$
Let's substitute $\lambda=e^{-5k}$;$\hspace{10 mm}$then$\hspace{10 mm}\lambda^{2}=e^{-10k}$
Now,$\hspace{10 mm}T(10)=T=15+55e^{-k(10)}=10+55(\frac{36}{121})=26.36^{\circ}F$
2)When it is taken back inside:
$T_0=26.36^{\circ}F\hspace{10 mm}$$T_e=70^{\circ}F\hspace{10 mm}$
Ten minutes have gone by from 9:10 to 9:20 am, so:
$T(10)=70-43.64(\frac{36}{121})=57.02^{\circ}F$