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
Truong-Son N.
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")#

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