Question

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Given Newton's law of cooling: dT = k (T-Z) ; where T = temperature of the cooling object, . dt T, - temperatue of the surrounding. A pie baked at 1&F is removed from an oven and put into the kitchen with a constant temperature of 70 °F. The pie cooled off to 160 °F after 5 minutes. Find a particular solution for the temperature of the pie. 6 marks]

Answer 1

Here, by Newton's cooling law dT/dt = k(T-T_s), Here T_s = 70 degree F integral dT/T-T_s = k integral dt Rightarrow log(T-T_s) = kt + c at t = 0, T = 180 degree F, T_s = 70 \r\n c = log(180-70) = log(110) Hence log(T-T_s) = kt + log 110 Rightarrow log T-T_s/110 = kt Rightarrow T = T_s + 110 e^kt Now, at t = 5 min T = 160 degree F K = 1/5 log 160-70/110 = 1/5 log 9/11 k = -0.0401341 Hence T = 70 + 11 e^-0.0401341t