Question

Newton's law of cooling is T = A * e ^ (- d * i) + C_{i} where T is the temperature of the object at time and C is the constant temperature of the surrounding mediumSuppose that the room temperature is 71and the temperature of a cup of coffee is 172 degrees when it is placed on the table. How long will it take for the coffee to cool to 129 degrees for k = 0.0459279 Round your answer to two decimal places

Answer 1

Step-by-step explanation

`\begin{gathered} T=Ae^(-kt)+C \\ 129=172e^(-0.0459279t)+71 \\ switch\text{ sides} \\ 172e^{\left\{-0.0459279t\right\}}+71=129 \\ 172e^{\left\{-0.0459279t\right\}}=129-71 \\ 172e^{\left\{-0.0459279t\right\}}=58 \\ Divide\text{ both sides by 172} \\ (172e^(-0.0459279t))/(172)=(58)/(172) \\ e^(-0.0459279t)=(29)/(86) \\ Apply\text{ exponent rules} \\ -0.0459279t=\ln \left((29)/(86)\right) \\ t=-(\ln \left((29)/(86)\right))/(0.0459279) \\ t=23.67\text{ minutes} \end{gathered}`

Answer: 23.67 minutes