Question

Newton's law of cooling is T = AeW! + C, where is the temperature of the object at time t, and is the constant temperature of the surrounding medium. Supposethat the room temperature is 73, and the temperature of a cup of coffee is 174' when it is placed on the table. How long will it take for the coffee to cool to 131" fork = 0.06889192 Round your answer to two decimal places.

Answer 1

Solution

We are given the equation

$T=Ae^{-kt}+C$

Room temperature, C = 73 degrees

Temperature at time t = 0, is T = 174 degrees

$\begin{gathered} T=Ae^{-kt}+C \\ 174=Ae^0+73 \\ 174=A+73 \\ A=174-73 \\ A=101 \end{gathered}$

Therefore, the equation becomes

$T=101e^{-kt}+73$

We want to find t = ?, when T = 131 degrees and k = 0.0688919

$\begin{gathered} T=101e^{-0.0688919t}+73 \\ 131=101e^{-0.0688919t}+73 \\ 131-73=101e^{-0.0688919t} \\ 58=101e^{-0.0688919t} \\ e^{-0.0688919t}=\frac{58}{101} \\ e^{-0.0688919t}=\frac{58}{101} \\ -0.0688919t=ln(\frac{58}{101}) \\ t=-\frac{1}{0.0688919}ln(\frac{58}{101}) \\ t=8.051418328 \\ t=8.05minutes\text{ \lparen to two decimal places\rparen} \end{gathered}$

Therefore, the answer is

$8.05minutes$