Question 1096993: Estabon poured himself a hot beverage that had a temperature of 202 degreesF then set it on the kitchen table to cool. The temperature of the kitchen was a constant 74degreesF. If the drink cooled to 181degreesF in 3 minutes, how long will it take for the drink to cool to 100degreesF?
Answer by jorel1380(3719)   (Show Source):
You can put this solution on YOUR website! Newton's law of cooling states that T(t) = Ts + (T0 – Ts) e^-kt where, t = time, T(t) = temperature of the given body at time t, Ts = surrounding temperature, To = initial temperature of the body, and k is a constant. So, here we have:
181=74+(202-74)e^-3k
0.8359375=e^-3k
ln 0.8359375=ln e^-3k=-3k ln e=-3k
k=0.059733809819
So, in order to reach 100 degrees:
100=74+(202-74)e^-0.059733809819t
0.203125=e^-0.059733809819t
ln 0.203125=ln e^-0.059733809819t=-0.059733809819t ln e=-0.059733809819t
t=26.68 mins before the drink cools to 100 degrees
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