Newton's Law of Cooling air temperature

In summary, using the given temperatures at different time intervals and the equation u(t) = u_\infty + (u_0 - u_\infty) e^{-kt}, the temperature reading at 1:09PM can be calculated as u(4) = u_\infty + (u_0 - u_\infty) e^{-kt}. By substituting the given values, the temperature reading at 1:09PM is 58F.
  • #1
Mastur
41
0

Homework Statement


At 1:00PM, a thermometer reading 70oF is taken outside where the air temperature is -10oF. At 1:02PM, the reading is 26oF. At 1:05PM, the thermometer is taken back indoors, where the air is at 70oF. What is the temperature reading at 1:09PM?

Homework Equations


The Attempt at a Solution


At t=0, u=70F, temp.env.=-10F
At t=2, u=26F.
At t=5 u=? env=70F
At t=9, u=?

[itex]\frac{du}{dx}=k(u+10)[/itex]

u+10=cekt
at t=0, c=80F

u+10=80ekt
at t=2, u=26, k=-0.39925

Giving me the u value which is 0.86729

After that, I don't know what to do next. :|

I actually tried letting u=0.86729 and k=-0.39925 when t=0. And after doing the same process, I ended up having u=60.608F but the answer should be 58F.

Is there any shortcut or pattern in obtaining the answer to this kind problems?
 
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  • #2
Mastur said:

Homework Statement


At 1:00PM, a thermometer reading 70oF is taken outside where the air temperature is -10oF. At 1:02PM, the reading is 26oF. At 1:05PM, the thermometer is taken back indoors, where the air is at 70oF. What is the temperature reading at 1:09PM?

Homework Equations


The Attempt at a Solution


At t=0, u=70F, temp.env.=-10F
At t=2, u=26F.
At t=5 u=? env=70F
At t=9, u=?

[itex]\frac{du}{dx}=k(u+10)[/itex]

u+10=cekt
at t=0, c=80F

u+10=80ekt
at t=2, u=26, k=-0.39925

Giving me the u(5) value which is 0.86729
Ok so far so good. That's correct :)
After that, I don't know what to do next. :|

I actually tried letting u=0.86729 and k=-0.39925 when t=0. And after doing the same process, I ended up having u=60.608F but the answer should be 58F.

Is there any shortcut or pattern in obtaining the answer to this kind problems?

Your method seems ok. Treat the second calculation as a separate problem with a new time variable starting at t=0 when the thermometer is bought back indoors.

The easiest way to sort out the constants is to use

[tex]u(t) = u_\infty + (u_0 - u_\infty) e^{-kt}[/tex]

Where [itex]u_0=0.87[/itex] and [itex]u_\infty[/itex]=70. Just remember that you're trying to find u(4) because you've re-started the time variable.
 
Last edited:

Related to Newton's Law of Cooling air temperature

What is Newton's Law of Cooling air temperature?

Newton's Law of Cooling air temperature is a principle that explains the rate at which an object's temperature changes when it is placed in an environment with a different temperature. It states that the rate of change of an object's temperature is proportional to the difference between the object's temperature and the temperature of its surroundings.

How does Newton's Law of Cooling air temperature apply to everyday life?

This law is applicable in various situations, such as when a hot cup of coffee is left to cool down in a room or when a person steps out of a warm shower into a colder room. It also explains how the Earth's temperature changes as it receives heat from the sun and loses heat to outer space.

What factors affect the rate of cooling according to Newton's Law of Cooling air temperature?

The rate of cooling is affected by several factors, including the temperature difference between the object and its surroundings, the surface area of the object, the material of the object, and the presence of insulation or wind.

What is the mathematical formula for Newton's Law of Cooling air temperature?

The mathematical formula for Newton's Law of Cooling air temperature is dT/dt = -k(T-Ts), where dT/dt is the rate of change of temperature, k is the cooling constant, T is the temperature of the object, and Ts is the temperature of its surroundings.

Is Newton's Law of Cooling air temperature applicable to all objects?

Newtons's Law of Cooling air temperature is applicable to all objects, regardless of their size or material, as long as they are in an environment with a different temperature. However, the rate of cooling may vary depending on the specific properties of the object and its surroundings.

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