Newton's law of cooling and thermometer

AI Thread Summary
The discussion focuses on applying Newton's law of cooling to determine the temperature of a thermometer after being removed from an oven. The initial temperature is 72 °C, and the outdoor temperature is 20 °C, with a recorded drop to 48 °C after one minute. Participants clarify the use of the differential equation and suggest simplifying variables to solve for the cooling constant k. Once k is determined using the temperature at one minute, the final temperature after five minutes can be calculated. The conversation emphasizes the importance of correctly applying the formula to reach the solution.
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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

 
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What have you tried to do? You have to make an attempt before anyone will give you any help - them's the rules.
 
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.
 
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chrisdk said:
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 = 52ekt, or T(t) = 20 + 52ekt.

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.
 
this helped me a lot, thank you! :)
 
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