Answer to Question #124043 in Differential Equations for Nii Laryea

Question

A bowl of water has a temperature of 50⁰C. It is put into a refrigerator where the temperature is 5⁰C. After 0.5 hours, the water is stirred and its temperature is measured to be
20⁰C. It is then left to cool further, take t in Newton's law of cooling to be measured in
minutes. Use Newton's law of cooling to predict when the temperature will be 10⁰C.

Answer 1

$T_s = 50^\circ C$ - starting temperature;

$T_0 = 5^\circ C$ - refrigerator's temperature;

$T_1 = 20^\circ C$ - after $t_1 = 0.5$ hours;

$T_2 = 10^\circ C$ - after $t_2 = ?$

$\dfrac{dT}{dt} = \alpha(T-T_0)$

$\dfrac{dT}{T-T_0} = \alpha dt$

$\int\limits_{T_s}^{T_1} \dfrac {dT}{T-T_0} = \int\limits_{0}^{t_1}\alpha dt \Rightarrow ln \dfrac{T_1 - T_0}{T_s - T_0} = \alpha t_1$

$\int\limits_{T_s}^{T_2} \dfrac {dT}{T-T_0} = \int\limits_{0}^{t_2}\alpha dt \Rightarrow ln \dfrac{T_2 - T_0}{T_s - T_0} = \alpha t_2$

Then $ln \dfrac{20 - 5}{50 - 5} = \alpha t_1= ln \dfrac{1}{3}$ ; $ln \dfrac{10 - 5}{50 - 5} = \alpha t_2= ln \dfrac{1}{9} = ln(\dfrac{1}{3})^2 = 2ln\dfrac{1}{3} = 2 \alpha t_1$;

Therefore, $\alpha t_2 = 2 \alpha t_1 \Rightarrow t_2 = 2t_1 = 2 \cdot0,5 hours = 1 hour$ (or 0.5 hours after the first measuring)