We have the Newton's law of cooling:
[tex]T=Ae^{-kt}+C[/tex]
where:
A: initial difference of temperature between the object and the room temperature.
C: room temperature.
k: constant of cooling.
T: temperature of the object at time t.
We know that the initial temperature of the coffee is A+C = 174° and the room temperature is C = 73°.
Given a constant k = 0.0688919, we have to calculate the time t for which the coffe temperature T is 131°.
We can solve it as:
[tex]\begin{gathered} T=131 \\ Ae^{-kt}+C=131 \\ (174-73)e^{-0.0688919t}+79=131 \\ 101e^{-0.0688919t}=131-79 \\ \frac{101}{101}e^{-0.0688919t}=\frac{52}{101} \\ \ln(e^{-0.0688919t})=\ln(\frac{52}{101}) \\ -0.0688919t\approx-0.6638768 \\ t\approx\frac{-0.6638768}{-0.0688919} \\ t\approx9.64 \end{gathered}[/tex]
Answer: 9.64 minutes.
