Newton's law of cooling and thermometer

[chrisdk]

Homework Statement

Hi, I need some help in solving Newton's law of cooling problem.

Use Newton’s law of cooling to determine the
reading on a thermometer 5 min after it is taken from an oven
at 72 ◦ C to the outdoors where the temperature is 20 ◦ C, if
the reading dropped to 48 ◦ C after one min.

Homework Equations

\frac{dT}{dt}=k(T-R)

The Attempt at a Solution

 

[Mark44]

What have you tried to do? You have to make an attempt before anyone will give you any help - them's the rules.

 

[chrisdk]

OK, we let T be the temperature of the object. Then, by Newton's Law of Cooling, \frac{du}{dt}=k(u-20), for some constant k. Let y=T-20. Then \frac{dy}{dt}=\frac{du}{dt}=ky. Thus we are now using the formula y=y_{0}e^{kt}. Since T is initially 72 ◦ C , y_{0}=72-20=52. So, y=52e^{kt}.
This is what I can do, so I'm asking you for some further help.

 

[Mark44]

OK, we let T be the temperature of the object. Then, by Newton's Law of Cooling, \frac{du}{dt}=k(u-20), for some constant k. Let y=T-20. Then \frac{dy}{dt}=\frac{du}{dt}=ky. Thus we are now using the formula y=y_{0}e^{kt}. Since T is initially 72 ◦ C , y_{0}=72-20=52. So, y=52e^{kt}.
This is what I can do, so I'm asking you for some further help.

You have thrown in an extra variable that you don't need.

Then, by Newton's Law of Cooling, \frac{dT}{dt}=k(T-20), for some constant k. Let u=T-20. Then \frac{du}{dt}=ku, and the solution to this differential equation is u=u_{0}e^{kt}. Since T is initially 72 ◦ C , u_{0}=72-20=52. So, u=52e^{kt}.

Reverting to T (we really want to know the temperature), we have T - 20 = 52e^kt, or T(t) = 20 + 52e^kt.

You're given that T(1) = 48, so use this fact to solve for k. After you have k, evaluate T(5) and you're done.

 

[chrisdk]

this helped me a lot, thank you! :)