Newton's Law of Cooling (diff eq. - seperation of variables)

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SUMMARY

The discussion focuses on applying Newton's Law of Cooling to a problem involving the cooling of coffee. The initial temperature of the coffee is 90 degrees Celsius, and the room temperature is 25 degrees Celsius. The cooling rate is given as 1 degree per minute when the coffee temperature reaches 65 degrees, leading to the calculation of the cooling constant k, which is determined to be -0.025. The discussion also addresses the use of Euler's method to estimate the coffee temperature after five minutes, with a step size of h=1.

PREREQUISITES
  • Understanding of differential equations, specifically separation of variables.
  • Familiarity with Newton's Law of Cooling.
  • Knowledge of Euler's method for numerical approximation.
  • Basic calculus, including integration and differentiation.
NEXT STEPS
  • Study the derivation of Newton's Law of Cooling in detail.
  • Learn how to apply Euler's method for different types of differential equations.
  • Explore the implications of negative cooling constants in physical scenarios.
  • Practice solving similar problems involving temperature changes over time.
USEFUL FOR

This discussion is beneficial for students studying differential equations, particularly those interested in applications of Newton's Law of Cooling and numerical methods like Euler's method.

brusier
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Newton's Law of Cooling (diff eq. -- separation of variables)

Homework Statement



Fresh coffee sitting in a room cooling...you know the routine.
Anyhow T(0) = 90degreesCelcius.
Room temp=25degrees celsius

find k.
Then he asks us to use Euler's method to estimate coffee temp after five mins. (using step size h=1).

Homework Equations



T(t) satisfies the equation: dT/dt=k(T-Troom).
we know that at T=65 degrees (for coffee) dT/dt (or the rate of cooling, as the problem states)= 1 degree per minute

The Attempt at a Solution



Well thought hey look here the ole plug and chug MAY work, let's see...:
1=k(65-25) brings k=.025. but then I thought, hey for the crap to cool k needs to be negative (this is the only way the limit as t tends to infinity for e^(kt) to tend to 0). So this can't be the appropriate way of going about this. So, without thinking (again) I integrated to see if that would bring about new light. I got

T=65e^(kt)+25.

No real help even if the integration is right because I'm missing a value for t to help me solve for k. So I'm a little stuck here.

I'm confused even more in that the blank where the answer to the second part goes, the Euler's method answer, has a blank that reads T(10)=____________. But I know (thought) the question asked for T(5). Am I missing somthing here?? Thank you, Ian.
 
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brusier said:

Homework Statement



Fresh coffee sitting in a room cooling...you know the routine.
Anyhow T(0) = 90degreesCelcius.
Room temp=25degrees celsius

find k.
Then he asks us to use Euler's method to estimate coffee temp after five mins. (using step size h=1).

Homework Equations



T(t) satisfies the equation: dT/dt=k(T-Troom).
we know that at T=65 degrees (for coffee) dT/dt (or the rate of cooling, as the problem states)= 1 degree per minute

The Attempt at a Solution



Well thought hey look here the ole plug and chug MAY work, let's see...:
1=k(65-25) brings k=.025.

Since the temperature is decreasing 1 degree per minute, that means T' = -1, not +1.
 

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