## What are Laws of Cooling of bodies?

If a material body is heated up to a very high temperature and then let to cool down, it will take a certain time to reach at a lower temperature. Amount of heat lost per unit time is called cooling rate or rate of cooling. For small temperature difference between a body and its surrounding, the rate of cooling of the body is directly proportional to the temperature difference and the surface area exposed.

*The rate by which a body loses heat by radiation is called ***rate of cooling**.

The rate of cooling depends upon the following factors –

- Temperature of the cooling body.
- Temperature of the surrounding.
- Ambient conditions and wind speed.
- Nature of surface of the body.
- Surface area of the body.
- Nature of material of the body.

Many laws have been formulated stating the time taken for cooling of a hot body in a given atmosphere. These are called laws of cooling. Among these laws, two are most important –

- Newton’s laws of cooling.
- Stefan Boltzmann’s law.

__Newton’s Laws of Cooling__

According to Newton’s Laws of Cooling, rate of cooling depends upon following factors –

*Temperature*of the body.- Temperature of the surrounding medium.
- Nature and extent of exposed surface.

Let, a body of mass ( m ) at temperature ( T ) is kept in atmosphere at temperature ( T_0 ) .

Then, according to Newton’s law of cooling –

\text {Rate of loss of heat} \ \propto \ \text {Temperature difference}

So, \quad - \left ( \frac {dQ}{dt} \right ) \propto \left ( T - T_0 \right )

Thus, \quad - \left ( \frac {dQ}{dt} \right ) = k \left ( T - T_0 \right ) …….. (1)

From principle of *calorimetry*, expression for heat loss from a body is given by –

dQ = mc \ dT

Therefore, rate of heat loss or rate of cooling will be obtained by dividing both sides by ( dt ) . Thus –

\left ( \frac {dQ}{dt} \right ) = mc \left ( \frac {dT}{dt} \right ) ……… (2)

Comparing equations (1) and (2), we will get –

- mc \left ( \frac {dT}{dt} \right ) = k ( T - T_0 )

Or, \quad \left ( \frac {dT}{dt} \right ) = \left ( \frac {k}{mc} \right ) ( T - T_0 )

But \left ( \frac {k}{mc} \right ) is a constant. Let it is ( K ) . Then –

\left ( \frac {dT}{dt} \right ) = K ( T - T_0 )

Or, \quad \left [ \frac {dT}{( T - T_0 )} \right ] = - K \ dt

On integrating both sides, we will get –

\int \left [ \frac {1}{( T - T_0 )} \right ] dT = - K \int dt

Or, \quad \log \left ( T - T_0 \right ) = Kt + c ……. (3)

So, \quad ( T - T_0 ) = e^{- Kt + c}

Or, \quad T = T_0 + e^{c} e^{- Kt}

= T_0 + C e^{- Kt} ……. (4)

Equations (1), (2), (3) and (4) are different forms of mathematical representation of Newton’s law of cooling.

__Cooling curves__

Different curves obtained by following Newton’s laws of cooling are as follows –

__(A) Body Temperature and Time curve –__

Consider about the * “Body Temperature – Time curve”* as shown in figure (A).

- Time is plotted along X axis and body temperature is plotted along Y axis.
- Clearly the rate of fall of body temperature is initially higher and decreases exponentially.
- Hence, cooling rate also decreases exponentially.

__(B) Cooling Rate and Temperature difference curve –__

Consider about the * “Cooling Rate – Temperature difference”* curve as shown in figure (B).

- Temperature difference ( \theta - \theta_0 ) is plotted along X axis and cooling rate ( R ) is plotted along Y axis.
- Clearly the relation is a straight line passing through origin.
- Rate of cooling is directly proportional to the temperature difference i.e. ( R \ \propto \ \Delta \theta )

__(C) ____Cooling Rate – Body Temperature curve –__

Consider about the * “Cooling Rate – Body Temperature”* curve as shown in figure (C).

- Cooling rate ( R ) is plotted along X axis and body temperature ( KT ) is plotted along Y axis.
- Clearly the relation is a straight line which intercepts X axis.
- It indicates that rate of cooling is zero when ( T = T_0 )
- It also indicates that, cooling rate is negative when ( T < T_0 ) and for that portion of curve \left ( R = - KT_0 \right )

__(D) Cooling Rate – Time curve –__

Consider about the * “Cooling Rate – Time”* curve as shown in figure (D).

- Time is plotted along X axis and cooling rate in terms of \left [ \log_e \left ( T - T_0 \right ) \right ] is plotted along Y axis.
- Clearly the relation is a straight line with negative slope.
- It indicates that rate of cooling decreases with pass of time.

__Stefan Boltzmann Law of Cooling__

Stefan Boltzmann law states that –

*The total radiant heat power emitted from a hot surface is proportional to the fourth power of its absolute temperature.*

Consider about a *black body* which is kept in an enclosure at temperature ( T_0 ) . Then –

- Rate at which the black body will absorb
*radiation*from the enclosure is ( \sigma T^{4}_0 ) - Rate at which the black body emits radiation to enclosure is ( \sigma T^{4} )
- Therefore, the net loss of energy by the black body per unit time per unit area is \left [ E = \sigma \left ( T^4 - T^{4}_0 \right ) \right ]

If the body is not a black body, then the equation becomes –

\left ( \frac {dQ}{dt} \right ) = \epsilon \sigma \left ( T^4 - T^{4}_0 \right )

= e A \sigma \left ( T^4 - T^{4}_0 \right ) ……. (1)

From principle of *calorimetry*, rate of loss of energy by the body is –

\left ( \frac {dQ}{dt} \right ) = m c \left ( \frac {dT}{dt} \right ) …….. (2)

From equation (1) and (2) we have –

mc \left ( \frac {dT}{dt} \right ) = e A \sigma \left ( T^4 - T^{4}_0 \right )

Or, \quad \left ( \frac {dT}{dt} \right ) = \left ( \frac {eA \sigma}{mc} \right ) \left ( T^4 - T^{4}_0 \right ) ……… (3)

This equation gives cooling rate of a body.