We all know that hot water or milk when left on a table begins to cool, gradually. Ultimately it attains the temperature of the surroundings. To study how slow or fast a given body can cool on exchanging heat with its surroundings, let us perform the following activity. Fig 11.19 shows the graph of the cooling of hot water with respect to time. 

From the graph, you can infer how the cooling of hot water depends on the difference of its temperature from that of the surroundings. You will also notice that initially the rate of cooling is higher and decreases as the temperature of the body falls. The above activity shows that a hot body loses heat to its surroundings in the form of heat radiation. The rate of loss of heat depends on the difference in temperature between the body and its surroundings. Newton was the first to study, in a systematic manner, the relation between the heat lost by a body in a given enclosure and its temperature.

According to Newton’s law of cooling, the rate of loss of heat, \(-{d} Q / \mathrm{d} t\) of the body is directly proportional to the difference of temperature \(\Delta T=\left(T_2-T_1\right)\) of the body and the surroundings. The law holds good only for the small difference of temperature. Also, the loss of heat by radiation depends upon the nature of the surface of the body and the area of the exposed surface. We can write
\(
-\frac{d Q}{d t}=k\left(T_2-T_1\right) \dots(11.19)
\)
where \(k\) is a positive constant depending upon the area and nature of the surface of the body. Suppose a body of mass \(m\) and specific heat capacity \(s\) is at temperature \(T_2\). Let \(T_1\) be the temperature of the surroundings. If the temperature falls by a small amount \(\mathrm{d} T_2\) in time \({d} t\), then the amount of heat lost is
\(
\begin{aligned}
& {d} Q=m s {~d} T_2 \\
& \therefore \text { Rate of loss of heat is given by } \\
& \frac{d Q}{d t}=m s \frac{d T_2}{d t} \dots(11.20)\\
& \text { From Eqs. (11.15) and (11.16) we have } \\
& -m s \frac{d T_2}{d t}=k\left(T_2-T_1\right) \\
& \frac{d T_2}{T_2-T_1}=-\frac{k}{m s} d t=-K d t \dots(11.21)\\
& \text { where } K=k / m s \\
& \text { On integrating, } \\
& \log _{\mathrm{e}}\left(T_2-T_1\right)=-K t+c \dots(11.22)\\
& \text { or } T_2=T_1+C^{\prime} \mathrm{e}^{-K t} \text {; where } C^{\prime}=\mathrm{e}^{\mathrm{c}} \dots(11.23)
\end{aligned}
\)
Equation \(11.23\) enables you to calculate the time of cooling of a body through a particular range of temperatures.

For small temperature differences, the rate of cooling, due to conduction, convection, and radiation combined, is proportional to the difference in temperature. It is a valid approximation in the transfer of heat from a radiator to a room, the loss of heat through the wall of a room, or the cooling of a cup of tea on the table.

The cooling curve is a graph that shows the relationship between body temperature and time. The rate of temperature fall is determined by the slope of the tangent to the curve at any point.

In general we can write, \(T(t)=T_a+\left(T_h-T_a\right) e^{-k t}\)
where \(T(t)\) is the Temperature at time \(t, T_a\) is the Ambient temperature or temp of surroundings, \(T_h\) is the temperature of the hot object, \(k\) is the positive constant and \(t\) is the time.

Verification of Newton’s Law of cooling

Newton’s law of cooling can be verified with the help of the experimental setup shown in Fig. 11.20(a). The set-up consists of a double-walled vessel \((\mathrm{V})\) containing water between the two walls. A copper calorimeter (C) containing hot water is placed inside the double-walled vessel. Two thermometers through the corks are used to note the temperatures \(T_2\) of water in the calorimeter and \(T_1\) of hot water in between the double walls, respectively. The temperature of hot water in the calorimeter is noted after equal intervals of time. A graph is plotted between \(\log _e\left(T_2-T_1\right)\) [or \(\ln \left(T_2-T_1\right)\) ] and time \((t)\). The nature of the graph is observed to be a straight line having a negative slope as shown in Fig. 11.20(b). This is in support of Eq. \(11.22\).

Example 1: \(11.8\) A pan filled with hot food cools from \(94^{\circ} \mathrm{C}\) to \(86^{\circ} \mathrm{C}\) in 2 minutes when the room temperature is at \(20^{\circ} \mathrm{C}\). How long will it take to cool from \(71^{\circ} \mathrm{C}\) to \(69^{\circ} \mathrm{C}\)?

Solution:

The average temperature of \(94^{\circ} \mathrm{C}\) and \(86^{\circ} \mathrm{C}\) is \(90^{\circ} \mathrm{C}\), which is \(70^{\circ} \mathrm{C}\) above the room temperature. Under these conditions, the pan cools \(8^{\circ} \mathrm{C}\) in 2 minutes.
Using Eq. (11.21), we have
\(\frac{\text { Change in temperature }}{\text { Time }}=K \Delta T\)
\(\frac{8^{\circ} \mathrm{C}}{2 \min }=K\left(70^{\circ} \mathrm{C}\right)\)
The average of \(69^{\circ} \mathrm{C}\) and \(71^{\circ} \mathrm{C}\) is \(70^{\circ} \mathrm{C}\), which is \(50^{\circ} \mathrm{C}\) above room temperature. \(K\) is the same for this situation as for the original.
\(
\frac{2^{\circ} \mathrm{C}}{\text { Time }}=K\left(50^{\circ} \mathrm{C}\right)
\)
When we divide the above two equations, we have
\(
\begin{aligned}
& \frac{8^{\circ} \mathrm{C} / 2 \mathrm{~min}}{2^{\circ} \mathrm{C} / \text { time }}=\frac{K\left(70^{\circ} \mathrm{C}\right)}{K\left(50^{\circ} \mathrm{C}\right)} \\
& \begin{array}{l}
\text { Time }=0.7 \mathrm{~min} \\
=42 \mathrm{~s}
\end{array}
\end{aligned}
\)

Example 2: Water is heated to \(80^{\circ} \mathrm{C}\) for \(10 \mathrm{~min}\). How much would be the temperature in degree Celsius, if \(\mathbf{k}=\mathbf{0} .056\) per \(\mathrm{min}\) and the surrounding temperature is \(25^{\circ} \mathrm{C}\)?

Solution:

Given,
The ambient temperature \(T_s\) is \(25^{\circ} \mathrm{C}\),
The temperature of water \(T_0\) is \(80^{\circ} \mathrm{C}\).
The time that water is heated is \(t 10 \mathrm{~min}\).
The value of constant \(k\) is \(0.056\).
According to Newton’s law of cooling,
\(
T(t)=T_s+\left(T_0-T_s\right) e^{-k t}
\)
Substitute the above data in the above expression,
\(
\begin{aligned}
& T(t)=25+(80-25) e^{-(0.056 \times 10)} \\
& T(t)=25+55 e^{-(0.056 \times 10)} \\
& T(t)=25+31.42 \\
& T(t)=56.42
\end{aligned}
\)
After 10 min the temperature cools down from \(80^{\circ} \mathrm{C}\) to \(56.42^{\circ} \mathrm{C}\).

Example 3: The oil is heated to \(70^{\circ} \mathrm{C}\). It cools to \(50^{\circ} \mathrm{C}\) after 6 minutes. Calculate the time taken by the oil to cool from \(50^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C}\) given the surrounding temperature \(\mathrm{Ts}=25^{\circ} \mathrm{C}\).

Solution:

Given,
The temperature of oil after 6 min i.e. \(T(t)\) is equal to \(50^{\circ} \mathrm{C}\).
The ambient temperature \(T_s\) is \(25^{\circ} \mathrm{C}\).
The temperature of oil, \(T_0\) is \(70^{\circ} \mathrm{C}\).
The time to cool to \(50^{\circ} \mathrm{C}\) is \(6 \mathrm{~min}\).
According to Newton’s law of cooling,
\(
\begin{aligned}
& T(t)=T_s+\left(T_0-T_s\right) e^{-k t} \\
& \left(T(t)-T_s\right) /\left(T_o-T_s\right)=e^{-k t} \\
& -k t=\ln \left[\left(T(t)-T_s\right) /\left(T_o-T_s\right)\right] \dots(1)
\end{aligned}
\)
Substitute the above data in Newton’s law of cooling expression,
\(
\begin{aligned}
& -k t=\ln [(50-25) /(70-25)] \\
& -k=(\ln 0.55556) / 6 \\
& k=0.09796
\end{aligned}
\)
The average temperature is equal to \(45^{\circ} \mathrm{C}\)
Substitute the values in equation (1),
\(-(0.09796) t=\ln [(45-25) /(70-25)]\)
\(-0.09796 t=\ln (0.44444)\)
\(0.09796 t=0.81093\)
\(t=0.09796 / 0.58778=8.278 \mathrm{~min}\).

Example 4: A body at a temperature \(40^{\circ} \mathrm{C}\) is kept in a surrounding of constant temperature \(20^{\circ} \mathrm{C}\). It is observed that its temperature falls to \(35^{\circ} \mathrm{C}\) in 10 minutes. Find how much more time will it take for the body to attain a temperature of \(30^{\circ} \mathrm{C}\).

Solution:

According to Newton’s law of cooling
\(
q_f=q_i e^{-k t}
\)
Now, for the interval in which temperature falls from \(40^{\circ} \mathrm{C}\) to \(35^{\circ} \mathrm{C}\).
\(
\begin{aligned}
& (35-20)=(40-20) e^{-(10 k)} \\
& e^{-10 k}=3 / 4 \\
& -10 k=(\ln 4 / 3) \\
& k=0.2876 / 10 \\
& k=0.02876
\end{aligned}
\)
Now, for the next interval;
\(
\begin{aligned}
& (30-20)=(35-20) e^{-k t} \\
& 10=15 e^{-k t} \\
& e^{-k t}=2 / 3 \\
& -k t=\ln (2 / 3) \\
& t=0.40546 / k
\end{aligned}
\)
Substitute the value of \(k\) in the above equation,
\(
\begin{aligned}
& t=0.40546 / 0.02876 \\
& t=14.098 \mathrm{~min} .
\end{aligned}
\)