Question

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 surroundings. A cup of coffee at 200° in a room of temperature 70^° is stirred continuously and reaches 100° after 10 min. At what time was it at 120°?

Short answer

The time at which the temperature of the coffee cup is${120}^{°}$is $6.52\min$.

Step-by-step solution

Definition

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.

Given parameters

The rate at which the temperature of an object changes is proportional to the difference between its temperature and that of its surroundings.

Find time at  120°

The differential equation according to the given condition is

$\frac{dT}{dt}=\alpha (T-70)$

Where $\alpha$ is the proportionality constant, notice that chosen temperature of the coffee is a function of time, but the temperature of the surrounding is to be constant because the surrounding is too large and won't be affected significantly by the heat that will gain from the coffee cup.

Now solve the differential equation

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

Now when the initial condition is at t=0 and the temperature of the coffee cup is$T=200°$, the differential equation becomes$\ln \left(130\right)=C$

Now when the temperature of the cup after 10 minutes is$T=100°$, the differential equation becomes

$$\begin{gathered} \ln \left(30\right)=10\alpha +\ln \left(130\right) \\ \alpha =\frac{\ln \left(\frac{3}{13}\right)}{10} \end{gathered}$$

Therefore, the time at which the temperature of the coffee cup is ${120}^{°}$is$$\begin{gathered} t=\frac{\ln \left(T-70\right)-C}{\alpha } \\ =10\frac{\ln \left(50\right)-\ln \left(130\right)}{\ln \left(\frac{3}{13}\right)} \\ =10\frac{\ln \left(\frac{5}{13}\right)}{\ln \left(\frac{3}{13}\right)} \\ =6.52\min \end{gathered}$$

Hence, the time at which the temperature of the coffee cup is${120}^{°}$is$6.52\min$.