Question

Newton’s Law of Cooling states that the rate at which the object's temperature decreases is directly proportional to the difference between the temperature of the object and the ambient temperature. A cup of tea (temperature = 90°C) is placed in a room whose temperature is 30°C. After five minutes, the temperature of the tea has dropped to 70°C. After how much time would the temperature of the tea be 50°C?

Answer 1

Question #75192 - Math - Differential Equations

Newton's Law of Cooling states that the rate at which the object's temperature decreases is directly proportional to the difference between the temperature of the object and the ambient temperature. A cup of tea (temperature = 90°C) is placed in a room whose temperature is 30°C. After five minutes, the temperature of the tea has dropped to 70°C. After how much time would the temperature of the tea be 50°C?

Solution: The rate of change of the temperature $\frac{dT(t)}{dt}$ , is (by Newton's Law of Cooling) proportional to the difference between the temperature of the object $T(t)$ and the ambient temperature $T_a$ . This means that:

$$\frac {d T (t)}{d t} = - r (T (t) - T a) = - r \cdot \Delta T (t)$$

$r$ - positive constant characteristic of the system

The solution of this differential equation:

$$\int \frac {1}{T - T a} d T = \int - r d t$$ $$\ln (T - T a) = - r t + C 1$$ $$T (t) = T a + e ^ {- r t + C 1} = T a + C \cdot e ^ {- r t}$$

For $t = 0$ :

$$T (0) = T a + C \cdot e ^ {- r \cdot 0} \rightarrow C = T (0) - T a = 9 0 - 3 0 = 6 0$$

For $t = 5$ :

$$T (5) = 7 0 = 3 0 + 6 0 \cdot e ^ {- r \cdot 5} \rightarrow r = \frac {\ln (3 / 2)}{5}$$

For $T(t) = 50$ :

$$5 0 = 3 0 + 6 0 \cdot e ^ {- \frac {\ln \left(\frac {3}{2}\right)}{5} t} = 3 0 + 6 0 \cdot \left(\frac {2}{3}\right) ^ {\frac {t}{5}} \rightarrow t = \frac {- 5 \cdot \ln (3)}{\ln (2 / 3)} \cong 1 3. 5 4 8$$

Answer: $t = \frac{-5 \cdot \ln(3)}{\ln(2/3)} \cong 13.548$ minutes

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