NEWTON’S LAW OF COOLING

We all know that a hot cup of tea when left on the table cools down slowly depending on the external temperature. The greater the difference in temperature the faster is the rate of cooling. 

Body: We can explain Newton’s Law of Cooling by performing a simple experiment.

Let us take about 500ml of water in a calorimeter and cover it with a lid with two holes. Let us insert the stirrer through one hole and a thermometer through the other hole. The bulb of the thermometer must be immersed in the water. 

Let us take the initial temperature of the thermometer. Let the reading be marked as T1. Note that T1 is the temperature of the surroundings ( say 30°C)

Now let us heat the calorimeter till the thermometer attains a temperature of 40°C above room temperature (i.e., the temperature of the surroundings).

Now let us stop heating by removing the heat source

Using a stopwatch lets us measure the drop in temperature at fixed intervals (say at an interval of 5 minutes) till it comes back to the temperature of the surroundings.

If we plot a graph taking the above values with ∆T = T2 – T1 in the y-axis and time t in the x-axis then the graph will look like

                Fig 1 shows a graph depicting cooling hot water with time.

From the graph, we understand that cooling of hot water is dependent on the temperature of its surroundings i.e that is how hot the water is when compared to the surrounding temperature. From the graph, we also find out that the rate of cooling is faster in the beginning but slows down as the temperature of the water approaches the surrounding temperature.

The loss of heat to the surroundings is due to heat radiation. 

Derivation for Newton’s Laws of Cooling

The rate of loss of heat from a hot body is equal to (-DQ/DT) ∝ ∆T = (T2 –T1 ) of the hot body and the surroundings( applicable only for small differences of temperature).

We need to remember that any loss of heat by radiation depends on the type of the body and surface area exposed. 

Therefore,

Let us suppose k is a positive constant depending upon the type of the body surface and area exposed

Then (-dQ/dt) = k(T2 –T1 )……………………………(1)

Let us consider a body with mass m, specific heat capacity s and temperature T2.

T1 is the temperature of the surroundings

Let them fall in temperature denoted by dT₂ in time dt

∴ Rate of loss of Heat 

dQ/dt=ms(dT₂ /dt)………………(2)

From equations (1) and (2) we get

= – m s (dT₂ /dt) = k(T₂-T₁))

                          = dT₂ /(T₂ -T₁)=-kdt/ms=-Kdt…………(3)

Where K=k/ms

Integrating

logₑ( T₂ -T₁)=-Kt+c …………..(4)  

Or T₂ = T₁)+C ́e–Kt ; C ́=ec……….. (5)

Equation (5) is the representation of Newton’s Law of Cooling and helps us to calculate the time to cool a body through a particular range of temperature.

Therefore Equation for Newton’s Law of Cooling can be written as

 T₂ = T₁ + C′ e –Kt; where C′ = ec

Limitations of Newton’s Laws of Cooling

• The temperature difference between surroundings and the body must be small

• The Source of heat loss from the body to the surroundings must be through radiation only

• During the cooling process of the hot body, the temperature of the surroundings must be constant

For minute temperature differences, the cooling rate because of conduction, convection and radiation are directly proportional to the temperature differences. It results after approximation of heat transfer from a radiator to a room, the heat loss through the walls of a room or cooling of a hot teacup on the table.

Verification of Newton’s Law of cooling: 

T1( temperature of hot water) and T2 ( temperature of water in the calorimeter

Verification of Newton’s law of cooling can be done with the help of an experimental setup shown in Fig. 2(a). The set-up includes a double-walled vessel (V) with water-filled between the two walls. A calorimeter (C) made of copper-containing hot water is placed inside the double-walled vessel. With the help of corks, two thermometers are placed to detect the temperatures: T2 of water in the calorimeter and T1 of hot water in between the double walls respectively. After equal time intervals, the temperature of hot water in the calorimeter is noted. A graph is plotted between loge (T2 –T1 ) [or ln(T2 –T1 )]versus time (t). The graph is observed to be a straight line having a negative slope as shown in Fig. 2(b).

 This is in support of Eq. (5).

  Fig 2(a) and fig 2(b) show the experimental setup for Verification of Newton’s Law of cooling.

Let us understand the concept with an example:

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

Ans. The average temperature of 94 °C and 86 °C is 90 °C, which is 70 °C above the room temperature. Under these conditions, the pan cools 8 °C in 2 minutes.

Using Eq. (3), we have

Change in temperature / Time = K ∆T 

8°C /2 min = K(70°C)

The average of 69 °C and 71 °C is 70 °C, which is 50 °C above room temperature. K is the same for this situation as for the original.

2°C / Time = K (50 °C)

When we divide above two equations, we have

(8°C/2min) / 2°C/time = K(70°C)/K (50°C)

          Time = 0.7 min

                   = 42 s

Conclusion 

Heat is a form of energy that gets transferred from a body to another/the surrounding medium. This takes place due to the temperature differences between them. The degree of hotness of a body is known as its temperature. Sir Isaac Newton derived the fact that the rate of cooling of anybody/object is directly proportional to the excess temperature of the body/object over its surroundings. This law was subsequently known as Newton’s Law of Cooling. This is depicted by the formula 

  dQ/dT = -k( T2-T1)

Where k is a constant,

T1 is the temperature of the medium surrounding the body/object and

T2 is the temperature of the body/object.