Question

: You place a cup of 210°F coffee on a table in a room that is 68°F, and 10 minutes later, it is 200°F . Approximately how long will it be before the coffee is 180°F? Use Newton's law of cooling: 180°F

A: 1 hour
B: 15 minutes
C: 35 minutes
D: 45 minutes

: A body was found at 6 a.m. Outdoors on a day when the temperature was 50°F . The medical examiner found the temperature of the body to be 66°F . What was the approximate time of death? Use newtons law of cooling, with k= 0.1947

T(t) = TA + (To – TA)e^-kt

A: 5 a.m.
B: 3 a.m.
C: 2 a.m.
D: Midnight (12 a.m.)

Answer 1

1)

$$\begin{array}{l} T_{0} = 210 \\ T_{A} = 68 \\ t_{1} = 10 \\ \end{array}$$ $$\begin{array}{l} T_{1} = T(t_{1}) = 200 \\ T_{2} = T(t_{2}) = 180 \\ t_{2} = ? \\ \end{array}$$ $$\begin{array}{l} T(t) = T_{A} + (T_{0} - T_{A}) e^{-kt} \\ T_{1} = T_{A} + (T_{0} - \overline{T_{A}}) e^{-kt_{1}} \\ e^{-kt_{1}} = \frac{T_{1} - T_{A}}{T_{0} - T_{A}} \\ -kt_{1} = \ln \frac{T_{1} - T_{A}}{T_{0} - T_{A}} \quad k = \frac{1}{t_{1}} \ln \frac{T_{0} - T_{A}}{T_{1} - T_{A}} \\ T_{2} = T_{A} + (T_{0} - T_{A}) e^{-kt_{2}} \\ e^{-kt_{2}} = \frac{T_{2} - T_{A}}{T_{0} - T_{A}} \\ -kt_{2} = \ln \frac{T_{2} - T_{A}}{T_{0} - T_{A}} \quad t_{2} = \frac{1}{k} \ln \frac{T_{0} - T_{A}}{T_{2} - T_{A}} \\ t_{2} = t_{1} \ln \frac{T_{0} - T_{A}}{T_{2} - T_{A}} / \ln \frac{T_{0} - T_{A}}{T_{1} - T_{A}} \approx 3,25 t_{1} = 32.5 \\ \end{array}$$

The answer closest is $C$: 35 minutes

2)

$$\begin{array}{l} t_{1} = 6 \\ T_{A} = 50 \\ T_{1} = 66 \\ k = 0,1947 \\ t_{0} = ? \\ \end{array}$$ $$\begin{array}{l} T_{0} = 96,8 \\ (\text{normal human's temperature}) \\ \end{array}$$ $$\begin{array}{l} T(t) = T_{A} + (T_{0} - T_{A}) e^{-k(t - t_{0})} \\ T_{1} = T_{A} + (T_{0} - \overline{T_{A}}) e^{-k(t_{1} - t_{0})} \\ e^{-k(t_{1} - t_{0})} = \frac{T_{1} - T_{A}}{T_{0} - T_{A}} \\ -kt_{1} = \ln \frac{T_{1} - T_{A}}{T_{0} - T_{A}} \\ t_{1} = t_{0} = \frac{1}{k} \ln \frac{T_{0} - T_{A}}{T_{1} - T_{A}} \\ t_{0} = t_{1} - \frac{1}{k} \ln \frac{T_{0} - T_{A}}{T_{1} - T_{A}} = 6 - 5.5 = 0.5 \text{ (hours)} \\ \end{array}$$

The closest answer is $D$ (12a.)