# Newton's law of cooling

1. Jul 29, 2011

### BeefBowl

1. The problem statement, all variables and given/known data
at t=0; thermometer reading (x) =80 F (taken outside where the air temp is 20 F)
at t=3; x=42 F
then the thermometer is brought inside where the air is at 80 F.
at t=10; x=71 F
When was the thermometer brought indoors?

2. Relevant equations
temp=ambient temperature; x= present temp of the body
x - temp=Ce-kt

3. The attempt at a solution

first I get the value of k: t=0; x=80
x=Ce-kt +20
C=60

x=60e-kt +20; t=3;x=42
k=0.334

Now, when brought inside:
at t=0, x=??
x-80=Ce-0.334t
C=x-80

And I am currently stuck up in here. If answered, could someone explain how does it happen? Thanks in advance!

2. Jul 29, 2011

### Dick

Call the temperature when it's brought in T0, and call the time it's brought in t0. Then t0 is the time to cool from 80F to T0 when the background is 20F, right? That means (80-T0)/60=e^(-k*t0), also right? Now call t1 the time to warm up from T0 to 71F when the background is 80F. Can you write a similar formula for t1? If you can do that then you've got two equations in the three unknowns t0, T0 and t1 (since you already know k). Now you need one more equation. How about t0+t1=10?

3. Aug 2, 2011

### BeefBowl

Thanks Dick! I get it now.