Certain quantity of water cools from 70$^{\circ}$C to 60$^{\circ}$C in the first 5 minutes and to 54$^{\circ}$C in the next 5 minutes. The temperature of the surroundings is

Question

A 200 g sample of water at 80°C is mixed with 100 g of water at 20°C. Assuming no heat loss to the surroundings, what is the final temperature of the mixture?

Answer 1

To solve this problem, we can apply Newton's Law of Cooling, which states that the rate of change of temperature of an object is proportional to the difference between the object's temperature and the ambient temperature (i.e., the surroundings). Mathematically, it is represented as:

\frac{d\theta}{dt} = -k(\theta - \theta_{\text{surroundings}})

Where:

\theta is the temperature of the object (water in this case).
\theta_{\text{surroundings}} is the ambient or surrounding temperature.
k is a constant.
t is time.

We are given two cooling intervals and need to find the surrounding temperature.

The water cools from 70^\circ \text{C} to 60^\circ \text{C} in the first 5 minutes.
It cools from 60^\circ \text{C} to 54^\circ \text{C} in the next 5 minutes.

According to Newton's Law of Cooling, the average temperature for each interval can be calculated as:

First interval average temperature: \theta_{\text{avg1}} = \frac{70 + 60}{2} = 65^\circ \text{C}
Second interval average temperature: \theta_{\text{avg2}} = \frac{60 + 54}{2} = 57^\circ \text{C}

According to the exponential cooling equation that follows Newton's Law, the cooling rate is also related to the average temperature difference with the surroundings. Thus, the difference in temperature decreases consistently:

For the first 5 minutes: \Delta\theta_1 = 10^\circ \text{C} (from 70 to 60)
For the next 5 minutes: \Delta\theta_2 = 6^\circ \text{C} (from 60 to 54)

Typically, for Newton's cooling law, the cooling pattern shows that the temperature drop halves with consecutive equal time intervals:

\frac{\Delta\theta_2}{\Delta\theta_1} = \frac{\theta_{\text{avg2}} - \theta_{\text{surroundings}}}{\theta_{\text{avg1}} - \theta_{\text{surroundings}}}

Substituting the known values:

\frac{6}{10} = \frac{57 - \theta_{\text{surroundings}}}{65 - \theta_{\text{surroundings}}}

Cross-multiply and solve for \theta_{\text{surroundings}}:

6(65 - \theta_{\text{surroundings}}) = 10(57 - \theta_{\text{surroundings}})

390 - 6\theta_{\text{surroundings}} = 570 - 10\theta_{\text{surroundings}}

4\theta_{\text{surroundings}} = 180

\theta_{\text{surroundings}} = 45^\circ \text{C}

Therefore, the temperature of the surroundings is 45^\circ \text{C}.

Answer 2