Knowing $T - T_{s} = (T_{0} - T_{s}) e^{k t}$ $T-T_s=(T_0 - T_s)e^(kt)$ , 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?

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Knowing $T - T_{s} = (T_{0} - T_{s}) e^{k t}$ $T-T_s=(T_0 - T_s)e^(kt)$ , 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 $T_{s}$ $T_s$ .

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Answer 1 · 2018-06-15T23:20:56

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

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

$T$ $T$ is the current water temperature.
$T_{0} = 46^{\circ} C$ $T_0 = 46^@ "C"$ is the starting water temperature.
$T_{s}$ $T_s$ is the temperature of the fridge (the surroundings).
$t$ $t$ is the time passed in minutes.
$k$ $k$ is the rate constant of the cooling process, constant with respect to the surrounding temperature.

Also, there's a typo... should be $e^{- k t}$ $e^(-kt)$ , since $T - T_{s}$ $T - T_s$ will decrease over time, but $e^{k t}$ $e^(kt)$ will increase over time ( $T_{0} - T_{s}$ $T_0 - T_s$ is constant).

So, what we really have is:

$T - T_{s} = (46 - T_{s}) e^{- k t}$ $T - T_s = (46 - T_s)e^(-kt)$

From the two data points, which are $(T_{1}, t_{1}) = (39, 10)$ $(T_1,t_1) = (39, 10)$ and $(T_{2}, t_{2}) = (33, 20)$ $(T_2,t_2) = (33, 20)$ , we form two equations:

$39 - T_{s} = (46 - T_{s}) e^{- 10 k}$ $39 - T_s = (46 - T_s)e^(-10k)$
$33 - T_{s} = (46 - T_{s}) e^{- 20 k}$ $33 - T_s = (46 - T_s)e^(-20k)$

Dividing these equations gives:

$\frac{39 - T_{s}}{33 - T_{s}} = e^{- 10 k + 20 k} = e^{10 k}$ $(39 - T_s)/(33 - T_s) = e^(-10k + 20k) = e^(10k)$

From this, the rate constant in terms of $T_{s}$ $T_s$ is:

$k = \frac{1}{10} ln (\frac{39 - T_{s}}{33 - T_{s}})$ $k = 1/10ln((39 - T_s)/(33 - T_s))$

Plugging this back into the original equation:

$T - T_{s} = (46 - T_{s}) e^{- \frac{t}{10} ln (\frac{39 - T_{s}}{33 - T_{s}})}$ $T - T_s = (46 - T_s)e^(-t/10ln((39 - T_s)/(33 - T_s)))$

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

$39 - T_{s} = (46 - T_{s}) e^{- \frac{10}{10} ln (\frac{39 - T_{s}}{33 - T_{s}})}$ $39 - T_s = (46 - T_s)e^(-10/10ln((39 - T_s)/(33 - T_s)))$

$\frac{39 - T_{s}}{46 - T_{s}} = e^{- ln (\frac{39 - T_{s}}{33 - T_{s}})}$ $(39 - T_s)/(46 - T_s) = e^(-ln((39 - T_s)/(33 - T_s)))$

$= e^{ln (\frac{33 - T_{s}}{39 - T_{s}})}$ $= e^(ln((33 - T_s)/(39 - T_s)))$

$= \frac{33 - T_{s}}{39 - T_{s}}$ $= (33 - T_s)/(39 - T_s)$

Solving this now, we get:

${(39 - T_{s})}_{s}^{2} = (33 - T_{s}) (46 - T_{s})$ $(39 - T_s)^2 = (33 - T_s)(46 - T_s)$

$1521 - 78 T_{s} + T_{s}^{2} = 1518 - 79 T_{s} + T_{s}^{2}$ $1521 - 78T_s + T_s^2 = 1518 - 79T_s + T_s^2$

$1521 + T_{s} = 1518$ $1521 + T_s = 1518$

$T_{s} = - 3^{\circ} C$ $color(blue)(T_s = -3^@ "C")$

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Knowing $T - T_{s} = (T_{0} - T_{s}) e^{k t}$ $T-T_s=(T_0 - T_s)e^(kt)$ , 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 $T_{s}$ $T_s$ .

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Knowing T-T_s=(T_0 - T_s)e^(kt)T−Ts=(T0−Ts)ektT-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_sTsT_s.

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Knowing $T - T_{s} = (T_{0} - T_{s}) e^{k t}$ $T-T_s=(T_0 - T_s)e^(kt)$ , 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 $T_{s}$ $T_s$ .