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 .
The question is asking for
1 Answer
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Well, first, let's define what we have.
T is the current water temperature.T_0 = 46^@ "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
So, what we really have is:
T - T_s = (46 - T_s)e^(-kt)
From the two data points, which are
39 - T_s = (46 - T_s)e^(-10k)
33 - T_s = (46 - T_s)e^(-20k)
Dividing these equations gives:
(39 - T_s)/(33 - T_s) = e^(-10k + 20k) = e^(10k)
From this, the rate constant in terms of
k = 1/10ln((39 - T_s)/(33 - T_s))
Plugging this back into the original equation:
T - T_s = (46 - T_s)e^(-t/10ln((39 - T_s)/(33 - T_s)))
Now, suppose
39 - T_s = (46 - T_s)e^(-10/10ln((39 - T_s)/(33 - T_s)))
(39 - T_s)/(46 - T_s) = e^(-ln((39 - T_s)/(33 - T_s)))
= e^(ln((33 - T_s)/(39 - T_s)))
= (33 - T_s)/(39 - T_s)
Solving this now, we get:
(39 - T_s)^2 = (33 - T_s)(46 - T_s)
1521 - 78T_s + T_s^2 = 1518 - 79T_s + T_s^2
1521 + T_s = 1518
color(blue)(T_s = -3^@ "C")