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
I got
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")#