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Newton's Law of Cooling differential equation

  • Thread starter Keldroc
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  • #1
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Homework Statement


Newton's Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings. Suppose t is time, T is the temperature of the object, and Ts is the surrounding temperature. The following differential equation describes Newton's Law
dt/dT=k(T−Ts)
where k is a constant.

Suppose that we consider a 96C cup of coffee in a 25C room. Suppose it is known that the coffee cools at a rate of 2C/min. when it is 70C Answer the following questions.
1. Find the constant k in the differential equation.
Answer (in per minute): k=

2. What is the limiting value of the temperature?
Answer (in Celsius): T=

3. Use Euler's method with step size h=2 minutes to estimate the temperature of the coffee after 10 minutes.
Answer (in Celsius): T(10)

Can someone help me get started with this problem? Thanks
 

Answers and Replies

  • #2
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Ok, so this seems to be a basic first order ODE problem.

Newton's Law of Cooling is as follows

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

For the given system (cup of coffee + room) we have

[tex]T[/tex] = temperature of the coffee

[tex]T_s[/tex] = temperature of the room, which we'll assume constant as the room temperature varies much slower than the cup of coffee, so significantly slower, that we'll assume it as constant for this problem.

[tex]\frac{dT}{dt}[/tex] states the rate of change of T, or the temperature of the cup of coffee.

[tex]k[/tex] is a constant which varies for every system, you can simply find the constant for this system by using the initial values given in the problem.

The limiting value for the temperature happens when the system reaches equilibrium, that is, when the coffee and the room temperature are equal, as we assumed the room temperature to be constant, that means the limiting value or the minimum value for the temperature of the coffee happens when [tex]T = T_s[/tex], that is, the coffee is at room temperature, in this case, we have [tex]\frac{dT}{dt} = 0[/tex] which means the heat exchange between the coffee and the room has ceased.

There's not much I can do for you as if I go any further I'll basically solve it for you (if I haven't already), this problem is a simple plug-in values for the variables.

Read the explanation above and try to solve it, if you still have any difficulties let me know.
 
  • #3
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Figured out the first two, but I don't understand how to use Euler's method on this problem. Can you help with that?
 
Last edited:
  • #4
HallsofIvy
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Do you know what Euler's method is? It's application here should be straightforward.
 

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