A cup of warm water at $46$ degrees is placed into a refrigerator. 10 minutes later, the water is $39$ degrees, and another 10 minutes later, the water is $33$ degrees.
Use Newton's law of cooling to find the ambient temperature in the refrigerator.
Newton's law of cooling is $\frac{dH}{dt}=-k(H-H_a)$ where $H$ is the temperature of an object and $H_a$ is the ambient temperature. This condenses into $y=y_0e^{-kt}$, where $y$ is $H-H_a$ and $y_0$ is $y$ when $t=0$.
If $t$ is diven in units of 10 minutes, we get $y_0-7=y_0e^{-k}$ and $y_0-13=y_0e^{-2k}$, which can be converted to
$$\ln (y_0-7)-\ln y_0=-k$$
$$\ln (y_0-13)-\ln y_0=-2k$$
But I can't get any further.
Wolfram says $y_0$ is $49$ degrees, which means the refrigerator is $-3$ degrees.
How do I get this result?
Best Answer
Since, as you said, the solution of the heat equation is: $$y(t)=y_0e^{-kt}$$ with $y_0$ and $k$ unknown and $y(1) = y_0 - 7$, and $y(2) = y_0 - 13$.
Hence $$ y_0 - 7 = y_0 e^{-k} \\ y_0 - 13 = y_0 e^{-2k} $$
Let $w = e^{-k}$, thus, $w^2 = e^{-2k}$. With this substitution, the equations become $$ y_0 - 7 = y_0 w \\ y_0 - 13 = y_0 w^2 $$
Solving the first equation for $w$ and plugging the solution into the second we get: $$ \frac{y_0 - 7}{y_0} = w \\ y_0 - 13 = y_0 \left( \frac{y_0 - 7}{y_0} \right) ^2 $$
To solve the second equation for $y_0$ we can multiply both sides by $y_0$ and we get: $$ y_0^2 - 13 y_0 = y_0^2 - 14 y_0 + 49 $$
I.e.: $ y_0 = 49 $ which makes
$$ H_a = 46 - 49 = -3 $$ (the final solution should have units).
Just for fun, we get that $w = \frac{49 - 7}{49} = \frac{6}{7}$ and $k = -\ln w = 0.15\ldots$.
Remark: I only applied the logarithm in the end because we aimed to find $y_0$ to get $H_a$. We do not need to compute $w$ nor $k$ to solve this problem.