Chapter 7.4, Problem 16E

To determine

To calculate : at what time the temperature of coffee occur

Answer to Problem 16E

At $20.27\text{ minutes}$

Explanation of Solution

Given information : The question is

A freshly brewed cup of coffee has temperature ${\text{95}}^{\circ }C$ in a ${20}^{\circ }C$ room. When its temperature is ${70}^{\circ }C$ , its cooling at a rate of $1^{\circ }C$ per minute

Calculation :

  $\begin{array}{l}\frac{dT}{dt}=k\left(T-T_s\right)\\ \frac{dT}{dt}=k\left(T_f-T_s\right)=-1^{\circ }/\min \\ k\left({70}^{\circ }-{20}^{\circ }\right)=-1^{\circ }/\min \\ k=-\frac{1^{\circ }}{{50}^{\circ }}/\min =0.02/\min \\ \frac{d{T}_f}{dt}=-1^{\circ }/\min \\ y\left(t\right)=T\left(t\right)-T_s\\ \frac{dy}{dt}=\frac{dT}{dt}=k\left(T-T_s\right)=ky\\ k\cdot y\left(t\right)=k\cdot y\left(0\right)e^{kt}\\ y\left(t_f\right)=\left(T_f-T_s\right)\\ y\left(t_f\right)=y\left(0\right)e^{k{t}_f}\\ 50=75e^{-0.02t_f}\\ e^{-0.02t_f}=\frac{50}{75}\\ \ln \left(e^{-0.02t_f}\right)=\ln \left(\frac{50}{75}\right)\\ -0.02t_f=\ln \left(\frac{50}{75}\right)\\ t_f=-\frac{\ln \left(\frac{50}{75}\right)}{0.02}=20.27\end{array}$