Question

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?

Answer 1

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

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

13.33333 °C

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

67.74 min