Newton's Law of Cooling

Newton's Law of Cooling

The Phenomenon of Cooling

Every hot object left to itself eventually reaches the temperature of its surroundings. A cup of tea on a table, a hot pan taken off the stove, or a heated metal block in a room — all cool down over time. The question Newton asked was: how fast does this cooling happen, and what does the rate depend on?

The answer is not uniform. If you watch a hot object cool, you notice that it loses heat quickly at first, when it is much hotter than the room, and then more slowly as it approaches room temperature. The rate of cooling is not constant — it depends on how hot the object is relative to its surroundings.

The Experimental Observation

Take a calorimeter with 300 mL of water, fitted with a stirrer and a thermometer. Note the room temperature $T_1$. Heat the water until it is about 40 °C above room temperature, then remove the heat source. Stir gently and record the temperature every minute. Continue until the water is only about 5 °C above room temperature.

Plot a graph with $\Delta T = T_2 - T_1$ (the temperature difference between the water and the surroundings) on the y-axis and time $t$ on the x-axis. The curve you get looks like Fig. 10.19 — it falls steeply at first and then flattens out. This tells you that the rate of cooling is highest when the temperature difference is largest, and it decreases as the object cools.

Statement of Newton's Law of Cooling

Newton's law of cooling states that the rate of loss of heat by a body is directly proportional to the excess temperature of the body over its surroundings.

Mathematically, if $T_2$ is the temperature of the body and $T_1$ is the temperature of the surroundings (assumed constant), then

or

where $k$ is a positive constant that depends on the area and nature of the surface of the body. The negative sign indicates that heat is being lost — $dQ/dt$ itself is negative, so $-dQ/dt$ is positive.

The constant $k$ is not universal — it depends on:

• the surface area of the body (larger area → faster cooling)
• the nature of the surface (rough, dark surfaces radiate more efficiently than smooth, polished ones)
• the conditions of the surrounding medium (air currents, humidity, etc.)

Deriving the Temperature-Time Relation

Suppose a body of mass $m$ and specific heat capacity $s$ is at temperature $T_2$, and the surroundings are at constant temperature $T_1$. If the temperature falls by a small amount $dT_2$ in time $dt$, the heat lost is

The rate of loss of heat is therefore

Now apply Newton's law:

Substituting the expression for $dQ/dt$:

Rearrange to separate the variables:

Define a new constant $K = \frac{k}{m s}$, which has units of (time)$^{-1}$. Then

›Proof

Integrating the differential equation

Integrate both sides. The left side is integrated from the initial temperature $T_{2,0}$ at time $t=0$ to the temperature $T_2$ at time $t$:

$\int_{T_{2,0}}^{T_2} \frac{dT_2}{T_2 - T_1} = -K \int_0^t dt$

The left integral gives $\ln(T_2 - T_1)$ evaluated between the limits:

$\left[\ln(T_2 - T_1)\right]_{T_{2,0}}^{T_2} = -K t$

$\ln(T_2 - T_1) - \ln(T_{2,0} - T_1) = -K t$

$\ln\left(\frac{T_2 - T_1}{T_{2,0} - T_1}\right) = -K t$

Exponentiate both sides:

$\frac{T_2 - T_1}{T_{2,0} - T_1} = e^{-K t}$

$T_2 - T_1 = (T_{2,0} - T_1)\, e^{-K t}$

Or equivalently,

$T_2 = T_1 + (T_{2,0} - T_1)\, e^{-K t}$

The general solution can also be written as

where $c$ is the constant of integration. Or

where $C' = e^c$ is determined by the initial condition.

Verification of the Law

The experimental setup for verification is shown in Fig. 10.20(a). It consists of a double-walled vessel with water between the walls (this water acts as the surroundings at constant temperature $T_1$). A copper calorimeter containing hot water is placed inside this vessel. Two thermometers measure $T_2$ (water in the calorimeter) and $T_1$ (water between the walls).

Record $T_2$ at regular time intervals. Then plot $\ln(T_2 - T_1)$ against time $t$.

According to the derived equation $\ln(T_2 - T_1) = -K t + c$, this graph should be a straight line with a negative slope of magnitude $K$. Fig. 10.20(b) shows exactly this — a straight line, confirming Newton's law of cooling.

Worked Example

Example 10.8: A pan filled with hot food cools from 94 °C to 86 °C in 2 minutes when the room temperature is 20 °C. How long will it take to cool from 71 °C to 69 °C?

Solution:

For the first cooling interval (94 °C to 86 °C):

• Average temperature = $\frac{94 + 86}{2} = 90$ °C
• Excess above room temperature = $90 - 20 = 70$ °C
• Temperature drop = $94 - 86 = 8$ °C
• Time taken = 2 minutes

Using the rate equation $\frac{\text{Change in temperature}}{\text{Time}} = K \Delta T$:

For the second cooling interval (71 °C to 69 °C):

• Average temperature = $\frac{71 + 69}{2} = 70$ °C
• Excess above room temperature = $70 - 20 = 50$ °C
• Temperature drop = $71 - 69 = 2$ °C
• Let the time taken be $t$ minutes …