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

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)=(58)/(101) \\ e^(-0.0688919t)=(58)/(101) \\ -0.0688919t=ln((58)/(101)) \\ t=-(1)/(0.0688919)ln((58)/(101)) \\ t=8.051418328 \\ t=8.05minutes\text{ \lparen to two decimal places\rparen} \end{gathered}`

Therefore, the answer is

`8.05minutes`