# Differential equation word problem

I'm doing a set of computer science programming questions and this was one, though I think it's simply solvable with differential equations, but I'm not too good with them. Any help?

Mr. Chickenorfish prepares a cup of coffee, then promptly forgets about it. Initially, the coffee is at $210$ F. Fifteen minutes later, the coffee is at $190$ F. If room temperature is $72$ F, when will the coffee be at $150$ F?

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## 2 Answers

You need a model for how the temperature decreases. I will guess that the heat loss is proportional to the temperature difference between the coffee and room temperature. Let the coffee temperature be $T$. My model says $\frac {dT}{dt}=k(T-72)$ and $T(0)=210, T(15)=190$. You need to solve the differential equation, use the two pieces of data to evaluate $k$ and the integration constant, then find the time when $T=150$

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I'm still not sure what to do from here –  Ryan Carter Mar 28 '13 at 23:49
@RyanCarter: copper.hat solved the differential equation for you. His $\tau$ is my $T$ and his $\lambda$ is my $k$. As we both said, plug the data you are given into the equation to determine $k$ or $\lambda$ –  Ross Millikan Mar 28 '13 at 23:52
@RyanCarter: Using copper.hat's notation, you are given $190=72+(210-72)e^{-(15\lambda)}$. Solve for $\lambda$ is the next step. –  Ross Millikan Mar 28 '13 at 23:55
Presumably the underlying model is an approximation of the Fourier equation in one dimension assuming the temperature gradient through the coffee cup wall is constant. Which would give the equations you mentioned above. –  copper.hat Mar 29 '13 at 0:30

It is just exponential decay. The temperature will be $\tau(t) = 72+(210-72)e^{-\lambda t}$. You are given $\tau(15) = 190$ which lets you compute $\lambda$, then you are asked to solve $\tau(t) = 150$ for $t$.

Solve for $\lambda$:

$\tau(15) = 190$ gives$\frac{190-72}{210-72} = e^{-\lambda 15}$, or equivalently $\ln(\frac{190-72}{210-72}) = -15\lambda$.

Solve for $t$:

Then solve for $t$ in $\frac{150-72}{210-72} = e^{-\lambda t}$, or equivalently $\ln(\frac{150-72}{210-72}) = -\lambda t$.

(Or, you could avoid computing $\lambda$ and just divide the above to get an expression for $\frac{t}{15}$.)

Pictures are more fun:

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Still not sure what to do –  Ryan Carter Mar 28 '13 at 23:52
@RossMillikan: Thanks! Just spotted that! –  copper.hat Mar 28 '13 at 23:53
@RyanCarter: Solve for $\lambda$, then solve for $t$ as above. –  copper.hat Mar 28 '13 at 23:53
I'm not great with logs –  Ryan Carter Mar 29 '13 at 0:01
You don't need to be a lumberjack to do the above. –  copper.hat Mar 29 '13 at 0:02