Chapter 6.4, Problem 31E

To determine

To find: (a) how much longer to take the soup to cool ${35}^{\circ }\text{C}$ .and

(b) how long to take cool ${90}^{\circ }\text{C to }{35}^{\circ }\text{C}$

Answer to Problem 31E

(a) $17.53$ minutes and

(b) $13.26$ minutes

Explanation of Solution

Given information:

The cup of soup cooled from ${190}^{\circ }\text{C to }{60}^{\circ }\text{C}$

  $10\min$ in a room where the temperature was ${20}^{\circ }\text{C}$

Calculation:

(a)To use the formula:

  $T-T_s=\left(T_0-T_s\right)e^{-kt}$ where $T_s$ is the surrounding temperature , $T_0$ is the initial temperature .

At time $t=0$

  $\begin{array}{c}T-T_s=\left(T_0-T_s\right)e^{-kt}\\ \Rightarrow 60-20=(90-20)e^{-10k}\\ \Rightarrow 40=70e^{-10k}\\ \Rightarrow \frac{4}{7}=e^{-10k}\\ \Rightarrow \ln \frac{4}{7}=\ln e^{-10k}\\ \Rightarrow \ln \frac{4}{7}=-10k\\ \Rightarrow k=-\frac{\ln \frac{4}{7}}{10}\\ \Rightarrow k\approx 0.056\end{array}$

Now wait for the soup to cool to ${35}^{\circ }\text{C}$ :

  $\begin{array}{c}T-T_s=\left(T_0-T_s\right)e^{-kt}\\ \Rightarrow 35-20=(60-20)e^{-0.056t}\\ \Rightarrow 15=40e^{-0.056t}\\ \Rightarrow \frac{3}{8}=e^{-0.056t}\\ \Rightarrow \ln \frac{3}{8}=\ln e^{-0.056t}\\ \Rightarrow \ln \frac{3}{8}=-0.056t\\ \Rightarrow t=-\frac{\ln \frac{3}{8}}{0.056}\\ \Rightarrow t\approx 17.53\end{array}$

(b)

  $\begin{array}{c}T-T_s=\left(T_0-T_s\right)e^{-kt}\\ \Rightarrow 35+15=(90+15)e^{-0.056t}\\ \Rightarrow 50=105e^{-0.056t}\\ \Rightarrow \frac{10}{21}=e^{-0.056t}\\ \Rightarrow \ln \frac{10}{21}=\ln e^{-0.056t}\\ \Rightarrow \ln \frac{10}{21}=-0.056t\\ \Rightarrow t=-\frac{\ln \frac{10}{21}}{0.056}\\ \Rightarrow t\approx 13.26\end{array}$

Therefore the required (a) $17.53$ minutes longer to take the soup to cool ${35}^{\circ }\text{C}$ .and

(b) $13.26$ minutes long to take cool ${90}^{\circ }\text{C to }{35}^{\circ }\text{C}$ .