Question

Question 3 of 15, Step 1 of 11/15CorrectNewton's law of cooling is T = Ae + C, where is the temperature of the object at time t, and C 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 = 0.06889197 Round your answer to two decimal places.

Answer 1

Given,

`T=Ae^(-kt)+C`

C = 73°

A = 174°

T = 131°

k = 0.0688919

find t,

Steps,

#1 Replace

`131=174*e^(-0.0688919t)+73`

#2. Isolate e^-kt

`(131-73)/(174)=e^(-0.0688919t)`

#3. Simplify the left hand side

`(1)/(3)=e^(-0.0688919t)`

#4. Apply exponent rules

`\begin{gathered} ln((1)/(3))=ln(e^(-0.0688919t)) \\ ln((1)/(3))=-0.0688919t \end{gathered}`

#5. Solve for t

`t=(ln((1)/(3)))/(-0.0688919)=(ln(3))/(0.0688919)\approx15.94690...`

#6. Round to two decimal places

`t\approx15.95`

Answer: 15.95