Newton’s Law of Cooling

When hot water in a vessel is kept on a table, it begins to cool gradually. To study how a given body cools on exchanging heat with its surroundings, the following experiment is performed.

Setup:

• A calorimeter is filled up to two-thirds of its capacity with boiling water and is covered.
• A thermometer is fixed through a hole in the lid. Its position is adjusted so that the bulb of the thermometer is fully immersed in the water.
• The calorimeter vessel is kept in a constant temperature enclosure or in open air, since room temperature will not change much during the experiment.​

Procedure

1. Note the temperature on the thermometer at one-minute intervals.
2. Continue until the temperature of the water decreases by about 25°C.

Observation

• Initially, the rate of cooling is higher.
• The rate of cooling decreases as the temperature of the water falls — the water cools more slowly as it approaches room temperature.​

A graph of temperature T (along the y-axis) is plotted against time t (along the x-axis). This graph is called the Cooling Curve.

Fig 7.13 (a) — Cooling Curve: T vs t

• The curve shows that cooling is rapid initially and slows down over time.
• A tangent is drawn to the curve at suitable points.
• The slope of each tangent gives the rate of fall of temperature (dT/dt) at that temperature.​

Slope at point A = \[\lim_{\Delta t\to0}\frac{\Delta T}{\Delta t}=\frac{dT}{dt}\]

Fig. 7.13 (b) — Rate of Cooling vs Temperature Difference: dT/dt vs (T − T₀)

• Taking (0, 0) as the origin, a graph of dT/dt is plotted against the corresponding temperature difference (T−T₀).
• The curve is a straight line, confirming direct proportionality between the rate of cooling and the temperature difference.

"The rate of loss of heat (dT/dt) of a body is directly proportional to the difference in temperatures (T−T₀) of the body and the surroundings, provided the difference in temperatures is small."

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 of the body and its surroundings.​

Proportionality Form

\[\frac{dT}{dt}\propto(T-T_0)\]

Introducing the constant of proportionality C:

\[\frac{dT}{dt}=C\left(T-T_0\right)\]

T = Temperature of the body at time t
T₀ = Temperature of the surroundings (constant)
C = Constant of proportionality
\[\frac {dT}{dt}\] = Rate of fall of temperature (rate of cooling)

Problem: A metal sphere cools at the rate of 1.6°C/min when its temperature is 70°C. At what rate will it cool when its temperature is 40°C? The temperature of the surroundings is 30°C.

Given

Solution

Step 1 — Apply Newton's Law at temperature T₁ to find C:

\[\left(\frac{dT}{dt}\right)_1=C\left(T_1-T_0\right)\]

1.6 = C(70 − 30)

1.6 = 40 C ⇒ C = \[\frac {1.6}{40}\] = 0.04 min⁻¹ 

Step 2 — Apply Newton's Law at temperature T₂:

\[\left(\frac{dT}{dt}\right)_2=C\left(T_2-T_0\right)=0.04\times(40-30)=0.04\times10\]

\[{\left(\frac{dT}{dt}\right)_2=0.4°\mathrm{C/min}}\]

Result: The rate of cooling drops by a factor of four when the difference in temperature of the metal sphere and its surroundings drops by a factor of four – directly confirming Newton's Law of Cooling.

• A hot body loses heat to its surroundings in the form of heat radiation.​
• The rate of cooling is directly proportional to the temperature difference between the body and its surroundings.
• The cooling curve (T vs t) shows rapid initial cooling that gradually slows down.
• Plotting \[\frac {dT}{dt}\] vs (T−T₀) gives a straight line through the origin, confirming Newton's law.​
• Mathematically: dT/dt = C(T − T₀), where C is the constant of proportionality.​
• The rate of cooling is proportional to — not independent of — the temperature difference. A 4× drop in temperature difference produces a 4× drop in cooling rate.