Newton's Law of Cooling Problem

In summary, the problem involves finding the time it takes for a metal bar to reach certain temperatures when placed in boiling water. The temperature of the bar increases by 2 degrees per second and the water is at a constant boiling temperature of 100 degrees Celsius. By using the equation T(t)=Ce^{kt}+T_m and substituting in the given information, the correct function T(t) is found to be T(t)=-80e^{tln(.975)}+100. Solving for t=82.1 seconds gives a temperature of 90 degrees Celsius.
  • #1
AdkinsJr
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Homework Statement



A small metal bar, whose initial temperature was 20 degrees C is dropped into a large container of boiling water. How long will it take the bar to reach 90 degrees C if it is known that its temperature increases 2 degrees in 1 second. How long will it take the bar to reach 98 degrees C?

(the back of the book gives 82.1 seconds and 145.7 seconds)

Homework Equations



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

where [tex]T_m[/tex] is the temperature of the surroundings.

The Attempt at a Solution



[tex]\int{\frac{1}{T-T_m}\frac{dT}{dt}dt=\int kdt[/tex]

[tex]T(t)=Ce^{kt}+T_m[/tex]

I'm given the points [tex]T(0)=20[/tex] and [tex]T(1)=T(0)+2=22[/tex]

The problem is that this give two equations and three unknowns:

[tex]C+T_m=20[/tex]

[tex]Ce^k+T_m=22[/tex]

I'm given that the water is boiling, but that only gives a minimum temperature of 100 degrees celsius.
 
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  • #2
Actually if the water is boiling you can be sure what the temperature is, as it will be a function of the atmospheric pressure only. Assuming the experiment is performed at the sea level, the temperature of the boiling water shall be 100 C.
 
  • #3
Quiablo said:
Actually if the water is boiling you can be sure what the temperature is, as it will be a function of the atmospheric pressure only. Assuming the experiment is performed at the sea level, the temperature of the boiling water shall be 100 C.

Ok, you're right! I actually haven't taken chem yet. This is a differential equations application problem. Are you getting this from [tex]PV=nRT[/tex] ?I guess if you have [tex]T=\frac{PV}{nR}[/tex] I guess I can see how T is a function of pressure assuming constant volume.

I should have just tried using 100 C, because it solves the system and gives the correct function [tex]T(t)=-80e^{tln(.975)}+100[/tex] and t=82.1 gives 90 C.
 
  • #4
No, no, PV = nRT is used only for (ideal) gases, and we are talking about water, a liquid that is. Indeed, the fact that water boils at 100 degrees C and no other temperature, assuming one atmospheric pressure, has to do with the fact that any substance changes its state (solid to liquid or liquid to gas or gas to liquid or liquid to solid) at a specific temperature.
 
  • #5


I would first check the validity of the given information and equations. Is the rate of temperature increase truly 2 degrees per second? Is the initial temperature of the bar truly 20 degrees Celsius? These are important factors to consider in order to accurately solve the problem.

Assuming the given information is accurate, I would approach the problem by using Newton's Law of Cooling, which states that the rate of change of temperature of an object is proportional to the difference between its temperature and the temperature of its surroundings. In this case, the object is the metal bar and the surroundings are the boiling water.

Using the given equation, \frac{dT}{dt}=k(T-T_m), we can solve for k by plugging in the known values of T and T_m at t=0 and t=1. This will give us one equation and one unknown, allowing us to solve for k.

Once we have the value of k, we can use the equation T(t)=Ce^{kt}+T_m to solve for C. Plugging in the known values of T and T_m at t=0, we can solve for C.

Finally, we can use the equation T(t)=Ce^{kt}+T_m to solve for the time it takes for the bar to reach a specific temperature, such as 90 degrees Celsius or 98 degrees Celsius. This can be done by plugging in the known values of T, T_m, and C, and solving for t.

It is important to note that this solution assumes that the temperature increase of 2 degrees per second is constant and that the initial temperature of the bar is truly 20 degrees Celsius. Any deviations from these assumptions could result in a different solution. As a scientist, it is important to always question and verify the accuracy of given information before proceeding with a solution.
 

FAQ: Newton's Law of Cooling Problem

What is Newton's Law of Cooling Problem?

Newton's Law of Cooling Problem is a mathematical model that describes the rate at which an object cools down in a surrounding medium. It states that the rate of change of the temperature of an object is proportional to the difference between its temperature and the temperature of its surroundings.

What is the significance of Newton's Law of Cooling Problem?

This law is important in understanding the process of heat transfer and can be applied in various fields such as physics, engineering, and meteorology. It helps in predicting the cooling rate of objects and can aid in designing efficient cooling systems.

What are the assumptions made in Newton's Law of Cooling Problem?

The law assumes that the object and the medium are in thermal equilibrium, the medium has a constant temperature, and the temperature difference between the object and the medium is small. It also assumes that there is no heat transfer between the object and its surroundings other than through convection.

What is the formula for Newton's Law of Cooling Problem?

The formula is written as dT/dt = -k(T - Tm), where dT/dt is the rate of change of temperature, k is the cooling constant, T is the temperature of the object, and Tm is the temperature of the surrounding medium.

How is Newton's Law of Cooling Problem applied in real-life situations?

This law is used in various applications, such as predicting the cooling rate of food and beverages, designing heating and cooling systems for buildings, and understanding the cooling of electronic devices. It can also be applied in weather forecasting to predict the change in temperature over time.

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