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.

1 Answer
Jun 15, 2018

I got -3^@ "C".


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 e^(-kt), since T - T_s will decrease over time, but e^(kt) will increase over time (T_0 - T_s is constant).

So, what we really have is:

T - T_s = (46 - T_s)e^(-kt)

From the two data points, which are (T_1,t_1) = (39, 10) and (T_2,t_2) = (33, 20), we form two equations:

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 T_s is:

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 10 minutes passed. If T_s > 39, it doesn't make sense.

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")