Differential Equations: word problem

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Homework Help Overview

The problem involves a cooling cake and applies Newton's law of cooling to determine the time it takes for the cake's temperature to drop from 200 degrees F to 75 degrees F, given its initial temperature and the ambient temperature.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the formulation of the cooling equation and the implications of using a positive versus negative constant in the equation. There is an exploration of the mathematical steps taken to derive the cooling model and the interpretation of the results.

Discussion Status

Some participants express confidence in the mathematical approach, while others suggest reconsidering the sign of the cooling constant to ensure physical accuracy. There is an ongoing examination of the transient and steady-state behavior of the solution.

Contextual Notes

Participants note the importance of correctly interpreting the cooling constant and its effect on the solution, as well as referencing examples from textbooks that may present different conventions.

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Homework Statement



When a cake is removed from an oven, its temperature is measured at 300 degrees F, 3 min later its 200, how long will it take for it to cool off to 75? room temp is 70. Assume Newtons law of cooling applies. The rate of cooling is proportional to the difference between the current temp and the ambient temp.


Homework Equations



This is what i did:
dx/dt = k(x-70)

The Attempt at a Solution



dx/dt = k(x-70)
integral 1/(x-70) dx = integral k dt
ln|x-70| = kt+c
after some algebra...
x = C*e^(kt) +70 where C = + or - e^c
at x(o) = 300 so...
300 = C*e^(k*0) +70
which gives me C = 230

x(t) = 230*e^(kt) +70
x(3) = 200
200 =230*e^(k*3) +70
k = 1/3 * ln(13/23)

now that i have an equation i can solve for the time it takes to get to 75
i'll leave k as k since its confusing when put into the equation
75 = 230*e^(kt) +70
i solved and got
t = (3*ln(5/230))/(ln(13/23))
which is approximately 20 minutes

Did i do this right? or did i go wrong somewhere, i just want to make sure I am getting the right idea with these kinds of problems
 
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mathematically i find no mistake

but physically speaking we are formulating an equation for cooling so it is better to write the equation as
dx/dt = -k(x-70)
so that you will get a positive k value in the end of the calculation

because if k value is given as positive in the problem
your steady state result would tend to infinity instead of room temp 70.
x = C*e^(kt) +70
 
payumooli said:
mathematically i find no mistake

but physically speaking we are formulating an equation for cooling so it is better to write the equation as
dx/dt = -k(x-70)
so that you will get a positive k value in the end of the calculation

because if k value is given as positive in the problem
your steady state result would tend to infinity instead of room temp 70.
x = C*e^(kt) +70
No, it does not- his k is (1/3)ln(13/23), a negative number.
 
what i wanted to convey is these solutions have a steady state and transient state.

the transient state should tend to zero when time approaches infinty.(for this cooling problem)

x(t) = 230*e^(kt) +70, k<0 should be the solution if you have right idea about this problem
 
Thank you so much. I didn't think it would make a difference if it was -k or not because i saw an example somewhat similar to this in the book I'm using and it had a positive k. The answer just looked very confusing so i wanted to make sure i doing the question right
 

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