Newton's law of cooling states dT/dt = -k(T - T_env). If a pie cools from 100 deg C to 60 deg C in 10 minutes in a room at 20 deg C, what is the value of the cooling constant k?
Answer and explanation
Correct answer: A. k = 0.1 * ln(2) min^(-1)
- Step 1: T(t) - T_env = (T_0 - T_env) e^(-kt). Here T_0 = 100, T_env = 20, so T(t) - 20 = 80 e^(-kt). Step 2. At t = 10 min, T(10) = 60 => 60 - 20 = 80 e^(-10k) => 40 = 80 e^(-10k) => 1/2 = e^(-10k) => 10k = ln(2) => k = 0.1 ln(2). Answer: k = 0.1 ln(2) min^(-1). Why not the others:** Subtracting the room temperature (20 deg C) is essential.
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