Newtons method of Cooling

[mattflint50]

The question concerns cooling soup:

Suppose that a cup of soup cooled from 90 degrees to 60 degrees in 10 minutes in a room whose temp was 20 degrees. Use Newton's method to evaluate the following questions.

a) How much longer would it take the soup to cool to 35 degrees?

I thought that this question was fairly straight forward. I simply set Ts=20, To=90, T=60 when t=10. I solved for k and got 0.05596. Then I went back and plugged this in.

b) Instead of being left to stand in the room, the cup of 90 degree soup is put into a freezer whose temperature is -15 degrees. How long will it take the temp to drop from 90 degrees to 35 degrees?

Please help me with (b) I am not sure which variables would change and to what. You help is certainly appreciated.

 

[stapel]

Does this exercise require Newton's "method" (an iterative process for finding zeroes) or Newton's "law" (a formula for determining cooling)?

Please reply with clarification, including definitions for the various variables you have used.

Thank you.

Eliz.

 

[galactus]

\(\displaystyle \L\\\frac{dT}{dt}=k(T-20); T(0)=90; T(10)=60\)

Separate variables:

\(\displaystyle \L\\\int\frac{dT}{T-20}=\int{k}dt\)

\(\displaystyle \L\\ln|T-20|=kt+C_{1}\)

\(\displaystyle \L\\T=C_{2}e^{kt}+20\)

When t=0, T=90:

\(\displaystyle \L\\90=20+C_{2}\)

Therefore, \(\displaystyle C_{2}=70\)

\(\displaystyle \L\\T=70e^{kt}+20\)

But, T(10)=60

\(\displaystyle \L\\60=70e^{k(10)}+20\)

Solve for k=-.05596178794

\(\displaystyle \H\\20+70e^{-.05596178794t}\)

Plug in t=35 and see.