# Knowing T-T_s=(T_0 - T_s)e^(kt), A pan of warm water (46dgC) was put in a refrigerator. 10 minutes later, the water's temperature was 39dgC; 10 minutes after that, it was 33dgC. Use Newton's law of cooling to estimate how cold the refrigerator was?

## The question is asking for ${T}_{s}$.

Jun 15, 2018

I got $- {3}^{\circ} \text{C}$.

Well, first, let's define what we have.

• $T$ is the current water temperature.
• ${T}_{0} = {46}^{\circ} \text{C}$ is the starting water temperature.
• ${T}_{s}$ is the temperature of the fridge (the surroundings).
• $t$ is the time passed in minutes.
• $k$ is the rate constant of the cooling process, constant with respect to the surrounding temperature.

Also, there's a typo... should be ${e}^{- k t}$, since $T - {T}_{s}$ will decrease over time, but ${e}^{k t}$ will increase over time (${T}_{0} - {T}_{s}$ is constant).

So, what we really have is:

$T - {T}_{s} = \left(46 - {T}_{s}\right) {e}^{- k t}$

From the two data points, which are $\left({T}_{1} , {t}_{1}\right) = \left(39 , 10\right)$ and $\left({T}_{2} , {t}_{2}\right) = \left(33 , 20\right)$, we form two equations:

$39 - {T}_{s} = \left(46 - {T}_{s}\right) {e}^{- 10 k}$
$33 - {T}_{s} = \left(46 - {T}_{s}\right) {e}^{- 20 k}$

Dividing these equations gives:

$\frac{39 - {T}_{s}}{33 - {T}_{s}} = {e}^{- 10 k + 20 k} = {e}^{10 k}$

From this, the rate constant in terms of ${T}_{s}$ is:

$k = \frac{1}{10} \ln \left(\frac{39 - {T}_{s}}{33 - {T}_{s}}\right)$

Plugging this back into the original equation:

$T - {T}_{s} = \left(46 - {T}_{s}\right) {e}^{- \frac{t}{10} \ln \left(\frac{39 - {T}_{s}}{33 - {T}_{s}}\right)}$

Now, suppose $10$ minutes passed. If ${T}_{s} > 39$, it doesn't make sense.

$39 - {T}_{s} = \left(46 - {T}_{s}\right) {e}^{- \frac{10}{10} \ln \left(\frac{39 - {T}_{s}}{33 - {T}_{s}}\right)}$

$\frac{39 - {T}_{s}}{46 - {T}_{s}} = {e}^{- \ln \left(\frac{39 - {T}_{s}}{33 - {T}_{s}}\right)}$

$= {e}^{\ln \left(\frac{33 - {T}_{s}}{39 - {T}_{s}}\right)}$

$= \frac{33 - {T}_{s}}{39 - {T}_{s}}$

Solving this now, we get:

${\left(39 - {T}_{s}\right)}^{2} = \left(33 - {T}_{s}\right) \left(46 - {T}_{s}\right)$

$1521 - 78 {T}_{s} + {T}_{s}^{2} = 1518 - 79 {T}_{s} + {T}_{s}^{2}$

$1521 + {T}_{s} = 1518$

$\textcolor{b l u e}{{T}_{s} = - {3}^{\circ} \text{C}}$