Newton's Law Cooling Explained

Newton's Law of Cooling describes the rate at which an object cools down. It states that the rate of heat loss is proportional to the temperature difference between the object and its surroundings. The mathematical form is:

$T(t) = T_{room} + (T_0 - T_{room})e^{-kt}$

Where:

• $T(t)$ is the temperature at time $t$.
• $T_0$ is the initial temperature.
• $T_{room}$ is the ambient room temperature.
• $k$ is the cooling constant.
• $t$ is the time.

For problems involving successive equal time intervals, a useful relationship derived from the law is:

$ \frac{T_1 - T_{room}}{T_0 - T_{room}} = \frac{T_2 - T_{room}}{T_1 - T_{room}} $

Cooling Problem Setup

We are given the following information:

• Initial temperature ($T_0$): $75^{\circ}C$
• Temperature after 5 minutes ($T_1$): $65^{\circ}C$
• Room temperature ($T_{room}$): $25^{\circ}C$

We need to find the temperature ($T_2$) after the next 5 minutes (at $t = 10$ minutes).

Calculating Temperature Ratios

Calculate the temperature differences between the body and the room at the specified times:

• Initial difference: $\Delta T_0 = T_0 - T_{room} = 75^{\circ}C - 25^{\circ}C = 50^{\circ}C$
• Difference after 5 mins: $\Delta T_1 = T_1 - T_{room} = 65^{\circ}C - 25^{\circ}C = 40^{\circ}C$

Let $\Delta T_2 = T_2 - T_{room}$ be the temperature difference after the next 5 minutes. Applying the ratio form for equal time intervals (5 minutes):

$ \frac{\Delta T_2}{\Delta T_1} = \frac{\Delta T_1}{\Delta T_0} $

Final Temperature Calculation

Solve for $\Delta T_2$:

$ \Delta T_2 = \frac{(\Delta T_1)^2}{\Delta T_0} $

Substitute the values:

$ \Delta T_2 = \frac{(40^{\circ}C)^2}{50^{\circ}C} = \frac{1600}{50} = 32^{\circ}C $

Finally, calculate the temperature $T_2$ after 10 minutes:

$ T_2 = T_{room} + \Delta T_2 = 25^{\circ}C + 32^{\circ}C = 57^{\circ}C $

The temperature of the body at the end of the next 5 minutes is $57^{\circ}C$.