jbmow
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The temperature of a cup of coffee varies according to Newton's Law of Cooling: , where T is the water temperature, A is the room temperature, and k is a positive constant.

If the coffee cools from 180°F to 100°F in 10 minutes at a room temperature of 75°F, how long (to the nearest minute) will it take the water to cool to 80°F?

Respuesta :

wizard123
Starting with:
[tex]T = (T_0 - A) e^{-kt} + A[/tex]
T0 is initial temperature of coffee ---> 180
A = 75

Step 1: Find value for k
plug in T = 100, t = 10 min
[tex]100 = 105e^{-10k} + 75 \\ \\ \frac{25}{105} =\frac{5}{21}= e^{-10k} \\ \\ (\frac{5}{21})^{1/10} = e^{-k}[/tex]

This is just as good as finding an actual value for 'k' because it will be subbed into Cooling equation anyway. Subbing in e^(-k) is easier.

Step 2: Find value for 't' when T = 80
[tex]80 = 105 (\frac{5}{21})^{t/10} + 75 \\ \\ (\frac{5}{21})^{t/10} = \frac{5}{105} = \frac{1}{21} \\ \\ (\frac{5}{21})^t = (\frac{1}{21})^{10} \\ \\ ln(\frac{5}{21})^t = ln(\frac{1}{21})^{10} \\ \\ t = \frac{10 ln(\frac{1}{21})}{ln(\frac{5}{21})} = 21.21[/tex]

Therefore it takes about 21 minutes for coffee to cool from 180 to 80 degrees.

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