Using Newton's Law of Cooling the proportionality constant for a thermos is 0.2231. A 130 degrees drink is left in an 82 degrees room temperature. How much time would it take for the drink to cool to 80 degrees.
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According to Newton's Law of Cooling, the rate of change of the temperature of a body at any time t is proportional to the difference between the temperature of the body at that moment of time and the ambient or environmental temperature. In the form of an equation this is: dT(t)/dt = -r*(T(t) - Tenv.)
The constant r in the system comprising the thermos, the drink in it and the environment is r = 0.2231. The environmental temperature is 82 degrees. The temperature fif the drink is originally 130 degrees. Solving the differential equation provides earlier gives: T(t) = Tenv + (T(0) - Tenv)*e^-rt
We want to determine the time when T(t) = 80 degrees.
80 = 82 + (130 - 82)*e^-0.2331*t
=> -2 = 48*e^-0.2331*t
=> e^-0.2331*t = -1/24
Now as e is a positive number, no exponent of e can give a negative result. The temperature of the drink can never fall to 80 degree.
This is also evident from the fact that the temperature of a liquid cannot fall below the temperature of its environment spontaneously; there is work required to be done for this to happen. If this were not the case, we would not need refrigerators. Just by waiting long enough all our cooling requirements would be satisfied.
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