Question

Hi! Any help with this ODE word problem will be appreciated!

I'll simplify it to the important data:
We are keeping track of the temperature of coffee. Well, we have the temperature at two times and want a third. I would like to be able to conceptualize what's going on, if anyone can help!

7:30am \(200^\circ\text F\)
7:45am \(120^\circ\text F\)
8:00am \(?^\circ\text F\)

The rate of change of heat is a simplified version of Newton's Law of Cooling, that \(\dfrac{dT}{dt}=k(T-T_r)\) where \(k\) is a proportionality constant and \(T_r\) is the room temperature (\(70^\circ\text F\)).

Answer 1

*

Answer 2

hM?

Answer 3

Hm?*

Answer 4

I figure that I probably need to find out what \(k\) is...

Answer 5

@sauravshakya

Answer 6

Can I use \(\Large{\int\limits_{7.5}^{7.75}}\normalsize\dfrac{dT}{dt}dt=\Large{\int\limits_{7.5}^{7.75}}\normalsize k(T-T_r)\ dt=200-120\)?

I'll think on that...

Answer 7

I haven't done anything like it in class, that I recall.

Answer 8

But I don't have \(T(t)\) to do integration...

Answer 9

You can say nothing about 8:00 AM here. Because We don't know how the temperature function is changing over time.

Answer 10

That's what I was thinking as well... But we know the rate-of-change function, two points on the \(T\) vs \(t\) curve, which is one solution to the rate-of-change function...

Answer 11

I know that initial value problems require one point...

Answer 12

I think I would solve the equation then use the given data to find the constants

\[ \frac{dT}{dt}= k(T -T_r) \\ \frac{dT}{(T -T_r)}= k \ dt \\\ln(T -T_r)= k \ t + C
\]
where t is time in minutes from 7:30
make each side the exponent of e
\[ T- T_r= A e^{k\ t} \]
now use 7:30 t=0 T= 200 to find A
and 7:45 t=15 to find k

Answer 13

Thank you very much! I'm working on digesting it now.

Answer 14

\(A\) is an arbitrary constant like \(C\), right?

Answer 15

yes A = e^C
we don't care about C per se.

Answer 16

This is somewhat related, @phi , but do we always need to look at things like \(t=0\) for initial value problems? Or can we use non-zero values for the independent variable?

Answer 17

Thank you!

Answer 18

you can use non zero values of t. But obviously t=0 makes it especially easy to find A

Answer 19

Right, thank you very much! :)

Answer 20

In case it is of any viewer's concern:
\(A=130^\circ\text F\)
\(k=-0.0637\ [s^{-1}]\)
and the coffee will be about \(89.2^\circ\text F\) at 8:00am.

:)

The problem was about a professor with an 8am class, who never drinks coffee unless it is at \(90^\circ\text F\) or cooler. Good news for him - he can drink the coffee in class!

I might show off and solve for the time \(t\) at which he can take his first sip. :P But not until I get the real problems done.

Thank you!