SOLUTION: 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

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Question 1096994: 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 181 degreesF in 3 ​minutes, how long will it take for the drink to cool to 100 degrees​F?
Answer by jorel1380(3719)   (Show Source): You can put this solution on YOUR website!
Using Newton's law of cooling, we get T(t)=T(s)+((T(0)-T(s))e^-kt, where T(s) is the surrounding temperature, T(t)is the temperature after time t, T(0) is the original temperature, and k is a constant. So, here we have:
T(3)=74+(202-74)e^-3k
181=74+(202-74)e^-3k
0.8359375=e^-3k
ln 0.8359375=ln e^-3k=-3k ln e =-3k
k=0.05973380981923699753874201106132
Then:
100=74+128*e^-0.05973380981923699753874201106132t
0.203125=e^-0.05973380981923699753874201106132t
ln 0.203125=ln e^-0.05973380981923699753874201106132t=-0.05973380981923699753874201106132t ln e= -0.05973380981923699753874201106132t
t=26.6839455 mins before the drink cools to 100 degrees
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