1
$\begingroup$

A $200°F$ cup of tea is left in a $65°F$ room. At time $t=0$ the tea is cooling at $5°F$ per minute. Write an initial-value problem (differential equation with an initial condition) that models the temperature $T$ of the tea. Assume Newton's Law of Cooling. (Not required to solve the differential equation)

I'm struggling with this problem but I know newton's law of cooling is: $$\frac{dT}{dt}=-k(T-Ta)$$

$\endgroup$
1
  • $\begingroup$ Oh okay, also back to the answer, wouldn't you assume T=200 since its the and -k(200-65) since T should the the initial temperature and Ta the ambient? I was thinking that t=0 where its cooling at 5F per minute. Wouldn't it look like T(0)=200, T(1)=195, t(2)=190. Or am I wrong.? $\endgroup$ – D Tater Jul 14 '15 at 6:21
2
$\begingroup$

$T_a$ in Newton's law is a temperature of room; $T_a = 65$. So, equation for modeling is $$ \frac{dT}{dt} = -k(T-65). $$ Now we should to determine $k$. "At time $t=0$ the tea is cooling at $5^\circ$F per minute". Ok, we have $$ \left.\frac{dT}{dt}\right|_{t=0} = -k(200-65)=135k = 5\Longrightarrow k =\frac{1}{27} $$ (time in minutes, of course).

UPDATE (Answer to OP question in comment)

No, $5^\circ$F per minute is a speed; it is derivative of $T$ (LHS of Newton's law). And by Newton's law, it's not a constant. You can solve equation above: $$ T = 65 + 135e^{-t/27}. $$ It's exponent, not a linear function (as you assummed).

And how you can see from it (or from law directly), if $T=65$, $dT/dt$ is zero; if the tea has cooled, it is no longer cool.

$\endgroup$
1
  • $\begingroup$ Oh okay! That makes total sense. :) $\endgroup$ – D Tater Jul 14 '15 at 6:28
0
$\begingroup$

The heat transfer equation

$$\frac{dT}{dt}=-k (T-T_a)$$

has a well known (and easy to derive) solution $$T=T_a+(T_0-T_a)e^{-kt}$$

Your initial conditions let you obtain $k$ directly from the first equation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.