# Newton's Law of Cooling Example

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

• 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.? – D Tater Jul 14 '15 at 6:21

$T_a$ 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.

• Oh okay! That makes total sense. :) – D Tater Jul 14 '15 at 6:28

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.