When a cold drink is taken from a refrigerator, its temperature is 5 degrees C. After 25 minutes in a 20 degrees C room its temperature has increased to 10 degrees C.(a) What is the temperature of the drink after 50 minutes?(b) When will its temperature by 15 degrees C?

Using Newton's Law of Cooling, [tex]\dfrac{dT}{dt} = k(T-T_s)[/tex], we have [tex]\dfrac{dT}{dt} = k(T - 20)[/tex] ([tex]T_s[/tex] is 20 degrees). Letting [tex]y = T - 20[/tex], we get [tex]\dfrac{dy}{dt} = ky[/tex], so [tex]y(t) = y(0)e^{kt}[/tex]. [tex]y(0) = T(0) - 20 = 5-20 = -15[/tex], so [tex]y(25) = y(0) e^{25k} = -15e^{25k}[/tex], and [tex]y(25) = T(25) - 20 = 10 - 20 = -10[/tex] so [tex]-15e^{25k} = -10 \ \ \Rightarrow \ \ e^{25k} = \frac{2}{3}[/tex]. Thus, [tex]25k = \ln\left(\frac{2}{3}\right)[/tex] and [tex]k = \frac{1}{25} \ln\left(\frac{2}{3}\right)[/tex], so [tex]y(t) = y(0)e^{kt} = -15e^{(1/25)\ln(2/3)t}[/tex]. More simply,
[tex]e^{25k} = \frac{2}{3} \ \ \Rightarrow \ \ e^k = \left( \frac{2}{3} \right)^{1/25}\\ \Rightarrow \ \ e^{kt} = \left( \frac{2}{3} \right)^{t/25}\\ \Rightarrow y(t) = -15 \cdot \left( \frac{2}{3}\right)^{t/25}[/tex]

(a) [tex]T(50) = 20 + y(50) = 20-15 \cdot \left( \frac{2}{3}\right)^{50/25}[/tex] =
[tex]20 - 15 \cdot\left(\frac{2}{3}\right)^2 = 20 - \frac{20}{3} = 13.\overline{3}{}^{\circ}C[/tex]

13.33333 °C

(b) [tex]15 = T(t) = 20 + y(t) = 20 - 15\cdot\left( \frac{2}{3} \right)^{t/15}[/tex]
[tex]\Rightarrow 15 \cdot \left(\frac{2}{3}\right)^{t/25} = 5 \Rightarrow \left( \frac{2}{3} \right)^{t/25} = \frac{1}{3} \Rightarrow[/tex]
[tex](t/25) \ln\left( \frac{2}{3}\right) = \ln\left( \frac{1}{3}\right) \rightarrow t = 25 \ln\left( \frac{1}{3}\right) / \ln\left( \frac{2}{3}\right) \approx 67.74\text{ min}[/tex]

67.74 min