Menu Top




Newton's Law of Cooling



Newton’S Law Of Cooling

When a hot object is placed in a cooler environment, it loses heat to the surroundings and cools down. Newton's Law of Cooling describes the rate at which this cooling occurs under specific conditions.


Statement of Newton's Law of Cooling

Newton's Law of Cooling states that the rate of heat loss of a body is directly proportional to the temperature difference between the body and its surroundings, provided the temperature difference is small and the nature of the exposed surface remains the same.

Mathematically, the rate of cooling ($ \frac{dT}{dt} $, the rate of change of temperature) or the rate of heat loss ($ \frac{dQ}{dt} $) is proportional to the temperature difference:

$ \frac{dQ}{dt} \propto (T - T_s) $

where $T$ is the temperature of the body and $T_s$ is the temperature of the surroundings (assuming $T > T_s$). The rate of heat loss is negative if $T < T_s$ (body gains heat and warms up). It is often stated as the rate of change of temperature:

$ \frac{dT}{dt} \propto -(T - T_s) $

Introducing a constant of proportionality $K$ (a positive constant that depends on the surface area, nature of the surface, and properties of the surroundings):

$ \frac{dT}{dt} = -K (T - T_s) $

The rate of heat loss is also related to the rate of change of temperature by $dQ = mc \, dT$, where $m$ is the mass and $c$ is the specific heat capacity of the body. So, $\frac{dQ}{dt} = mc \frac{dT}{dt}$.

$ mc \frac{dT}{dt} = -A' (T - T_s) $

Here, $A'$ is another constant that incorporates the surface properties and surrounding conditions. This is often written as:

$ \frac{dT}{dt} = -\frac{A'}{mc} (T - T_s) = -k (T - T_s) $

where $k = \frac{A'}{mc}$ is the cooling constant, which depends on the properties of the body (mass, specific heat capacity) and the properties of the surroundings and surface area.


Conditions for Validity

Newton's Law of Cooling is an approximation and is valid under certain conditions:

For larger temperature differences, the rate of heat loss by radiation is proportional to $(T^4 - T_s^4)$, which is not linear in $(T - T_s)$, so Newton's law becomes less accurate. However, for small temperature differences, $T^4 - T_s^4 \approx 4 T_s^3 (T - T_s)$, which is linear in $(T - T_s)$, thus validating the linear relationship in that limit.


Solving the Differential Equation

The equation $\frac{dT}{dt} = -k (T - T_s)$ is a first-order linear differential equation. Its solution gives the temperature of the body as a function of time:

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

where $T_0$ is the initial temperature of the body at $t=0$, and $T_s$ is the constant temperature of the surroundings. This equation shows that the temperature difference $(T - T_s)$ decreases exponentially with time, and the body asymptotically approaches the temperature of the surroundings.

Graph showing exponential cooling curve according to Newton's Law.

(Image Placeholder: A graph with Time on the x-axis and Temperature on the y-axis. Show a decreasing exponential curve starting from an initial temperature T0 above the surrounding temperature Ts. The curve approaches the horizontal line at Ts asymptotically.)


Experimental Determination of Cooling Constant (k)

The cooling constant $k$ (or the rate of cooling) can be determined experimentally by recording the temperature of a hot object placed in a constant-temperature environment at regular time intervals. A graph of temperature versus time can be plotted, and the slope $\frac{dT}{dt}$ at various temperatures can be found. Alternatively, a graph of $\frac{dT}{dt}$ versus $(T - T_s)$ should yield a straight line passing through the origin with a slope of $-k$ (or $k$ if plotting $|dT/dt|$ vs $T-Ts$).

The rate of cooling over a finite temperature interval from $T_1$ to $T_2$ in time $\Delta t$ can be approximated as the average rate of change, $\frac{T_1 - T_2}{\Delta t}$, being proportional to the average temperature difference, $\frac{T_1 + T_2}{2} - T_s$.

$ \frac{T_1 - T_2}{\Delta t} = K' \left(\frac{T_1 + T_2}{2} - T_s\right) $

This approximate form is often used in laboratory experiments to verify Newton's Law of Cooling or determine the cooling constant $K'$.


Applications

Example 1. A hot liquid cools from 80°C to 60°C in 5 minutes. If the surrounding temperature is 30°C, find the time it takes for the liquid to cool from 60°C to 50°C. (Assume Newton's Law of Cooling is applicable).

Answer:

According to Newton's Law of Cooling (approximate form for an interval):

$ \frac{T_{initial} - T_{final}}{\Delta t} = K \left(\frac{T_{initial} + T_{final}}{2} - T_s\right) $

where $K$ is a constant related to the body and surroundings.

Case 1: Cooling from 80°C to 60°C in 5 minutes.

$ T_{initial} = 80^\circ\text{C}, T_{final} = 60^\circ\text{C}, \Delta t = 5 \text{ minutes}, T_s = 30^\circ\text{C} $.

Average temperature of the liquid during this interval = $\frac{80 + 60}{2} = \frac{140}{2} = 70^\circ\text{C}$.

Average temperature difference = $70^\circ\text{C} - 30^\circ\text{C} = 40^\circ\text{C}$.

Rate of cooling = $\frac{80 - 60}{5} = \frac{20}{5} = 4 \, ^\circ\text{C/minute}$.

So, $ 4 = K (40) \implies K = \frac{4}{40} = 0.1 \, \text{minute}^{-1} $.

Case 2: Cooling from 60°C to 50°C.

$ T_{initial} = 60^\circ\text{C}, T_{final} = 50^\circ\text{C}, \Delta t = ?, T_s = 30^\circ\text{C} $.

Average temperature of the liquid during this interval = $\frac{60 + 50}{2} = \frac{110}{2} = 55^\circ\text{C}$.

Average temperature difference = $55^\circ\text{C} - 30^\circ\text{C} = 25^\circ\text{C}$.

Rate of cooling = $\frac{60 - 50}{\Delta t} = \frac{10}{\Delta t} \, ^\circ\text{C/minute}$.

Using Newton's Law of Cooling with the constant $K = 0.1 \, \text{minute}^{-1}$:

$ \frac{10}{\Delta t} = K (25) $

$ \frac{10}{\Delta t} = 0.1 \times 25 $

$ \frac{10}{\Delta t} = 2.5 $

$ \Delta t = \frac{10}{2.5} = \frac{100}{25} = 4 $ minutes.

The time it takes for the liquid to cool from 60°C to 50°C is 4 minutes.