A glass of milk at $\mathbf{38}\mathbf{°}$is removed from the refrigerator and left in a room at a temperature $\mathbf{70}\mathbf{°}$^°. If the temperature of the milk is$\mathbf{54}\mathbf{°}$ after 10min , what will its temperature be in half an hour? (See Problem 27.)

Newton's law of cooling states that the speed at that Associate in object cools is proportional to the distinction in temperature between the item and also the object's surroundings.

When a glass of milk is in the refrigerator its temperature is$38°$ and after 10mins the temperature of the milk is $54°$at room temperature${70}^{°}$

According to Newton's law of cooling, the rate at which the temperature of an object changes is proportional to the difference between its temperature and that of its surrounding. Therefore, the differential equation is$\frac{dT}{dt}=\alpha \left(T-70\right)$ where,$\alpha$is the proportionality constant. Notice that, the chosen temperature of the milk is a function of time, but the temperature of the surrounding to be constant, because the surrounding is too large and won't be affected significantly by the heat that is received from the milk glass.

Now solve the differential equation.

$$\begin{gathered} \int \frac{dT}{T-70}=\int \alpha dt \\ \ln \left|T-70\right|=\alpha t+C \end{gathered}$$

Apply the initial condition when t=0 and the temperature,$T=38°$ the integration constant C is

$$\begin{gathered} \ln \left|-32\right|=C \\ C=\ln \left(32\right) \end{gathered}$$

Now apply the second initial condition and the value of the constant C to find the proportionality constant.

role="math" localid="1664279356947" $$\begin{gathered} \ln \left|54-70\right|=10\alpha +\ln \left(32\right) \\ \alpha =\frac{\ln \left(\frac{1}{2}\right)}{10} \end{gathered}$$

Evaluate the temperature of the milk after 30 minutes.

$$\begin{gathered} \ln \left|T-70\right|=\frac{\ln \left(\frac{1}{2}\right)}{10}\times 30+\ln \left(32\right) \\ \ln \left|T-70\right|=\ln \left(\frac{1}{8}\right)+\ln \left(32\right) \\ \left|T-70\right|=\frac{32}{8} \\ T-70=\pm 4 \end{gathered}$$

Solve further.

$T=66°$

Hence, the temperature of the milk after 30 minutes is$66°$.