Answer :
We found the cooling constant and calculated that it will take approximately 31.2 minutes for the coffee to cool from 200°F to 180°F in a room at 68°F by using Newton's Law of Cooling.
We will use Newton's Law of Cooling. The law states the rate of cooling is proportional to the difference in temperature between the object and its surroundings.
[tex]\[T(t) = T_s + (T_0 - T_s) \cdot e^{-kt}\][/tex]
Where:
- T(t) is the temperature at time t
- T_s is the surrounding temperature (68°F)
- T_0 is the initial temperature (210°F)
- k is the cooling constant
- t is time
First, we will find the cooling constant k.
Setting up the equation:
200 = 68 + (210 - 68) * [tex]e^{-10k}[/tex]
132 = 142 * [tex]e^{-10k}[/tex]
0.9296 = [tex]e^{-10k}[/tex]
-10k = ln(0.9296)
k ≈ 0.00709
Now, we need to find the time t when the coffee reaches 180°F:
180 = 68 + (210 - 68) * [tex]e^{-0.00709t}[/tex]
112 = 142 * [tex]e^{-0.00709t}[/tex]
0.7887 = [tex]e^{-0.00709t}[/tex]
-0.00709t = ln(0.7887)
t ≈ 31.2 minutes
So, it will take approximately 31.2 minutes for the coffee to cool to 180°F.
