Newton's law of cooling and thermometer

In summary, the thermometer reads 5 min after being taken from the oven at 72 ◦ C, and then 5 min after being taken outside the oven at 20 ◦ C, if the thermometer reading dropped from 48 ◦ C to 52 ◦ C after one minute. Using Newton's law of cooling, the temperature of the object is found to be 20 + 52ekt after 5 min.
  • #1
chrisdk
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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



[tex]\frac{dT}{dt}=k(T-R)[/tex]

The Attempt at a Solution

 
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  • #2
What have you tried to do? You have to make an attempt before anyone will give you any help - them's the rules.
 
  • #3
OK, we let T be the temperature of the object. Then, by Newton's Law of Cooling, [tex]\frac{du}{dt}=k(u-20)[/tex], for some constant k. Let [tex]y=T-20[/tex]. Then [tex]\frac{dy}{dt}=\frac{du}{dt}=ky[/tex]. Thus we are now using the formula [tex]y=y_{0}e^{kt}[/tex]. Since T is initially 72 ◦ C , [tex]y_{0}=72-20=52[/tex]. So, [tex]y=52e^{kt}[/tex].
This is what I can do, so I'm asking you for some further help.
 
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  • #4
chrisdk said:
OK, we let T be the temperature of the object. Then, by Newton's Law of Cooling, [tex]\frac{du}{dt}=k(u-20)[/tex], for some constant k. Let [tex]y=T-20[/tex]. Then [tex]\frac{dy}{dt}=\frac{du}{dt}=ky[/tex]. Thus we are now using the formula [tex]y=y_{0}e^{kt}[/tex]. Since T is initially 72 ◦ C , [tex]y_{0}=72-20=52[/tex]. So, [tex]y=52e^{kt}[/tex].
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, [tex]\frac{dT}{dt}=k(T-20)[/tex], for some constant k. Let [tex]u=T-20[/tex]. Then [tex]\frac{du}{dt}=ku[/tex], and the solution to this differential equation is [tex]u=u_{0}e^{kt}[/tex]. Since T is initially 72 ◦ C , [tex]u_{0}=72-20=52[/tex]. So, [tex]u=52e^{kt}[/tex].

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.
 
  • #5
this helped me a lot, thank you! :)
 

FAQ: Newton's law of cooling and thermometer

What is Newton's law of cooling?

Newton's law of cooling states that the rate of change of temperature of an object is directly proportional to the difference between its initial temperature and the temperature of its surroundings.

How does Newton's law of cooling apply to thermometers?

Thermometers work based on the principle of Newton's law of cooling. The change in temperature of a thermometer is measured by the movement of a liquid, such as mercury or alcohol, in a glass tube. As the temperature of the thermometer changes, the liquid expands or contracts, indicating the temperature on a scale.

What factors affect the accuracy of a thermometer?

The accuracy of a thermometer can be affected by factors such as the material used for the thermometer, the calibration of the scale, and the manufacturing process. Environmental factors, such as altitude and atmospheric pressure, can also affect the accuracy of a thermometer.

How does the shape and size of an object affect its rate of cooling?

The rate of cooling of an object is affected by its surface area and shape. An object with a larger surface area will cool faster than an object with a smaller surface area. Similarly, an object with a more streamlined shape will cool faster than an object with a more irregular shape.

Can Newton's law of cooling be used to predict the temperature of an object?

No, Newton's law of cooling can only be used to describe the rate of change of temperature of an object. It cannot predict the exact temperature of an object at a specific time, as it is affected by various external factors such as wind speed and humidity.

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