Vtyjkyuk

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:

$fracdTdt = -k(T-T_a)$
$fracdTT-T_a = -kcdot dt$
$T = Ce^-kt + T_a$

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?

We will solve this problem in two parts:

1) When it is taken outside:

$T_0=70^circFhspace10 mm$$T_e=15^circFhspace10 mm$$T(5)=45^circFhspace10 mm$$T(10)=T^circFhspace10 mm$

The general solution is: $hspace10 mmT=15+55e^-kt$

Then, $hspace10 mmT(5)=45=15+55e^-k(5)$;$hspace10 mme^-5k=frac3055$

Let's substitute $lambda=e^-5k$;$hspace10 mm$then$hspace10 mmlambda^2=e^-10k$

Now,$hspace10 mmT(10)=T=15+55e^-k(10)=10+55(frac36121)=26.36^circF$

2)When it is taken back inside:

$T_0=26.36^circFhspace10 mm$$T_e=70^circFhspace10 mm$

Ten minutes have gone by from 9:10 to 9:20 am, so:

$T(10)=70-43.64(frac36121)=57.02^circF$