# Newtons Law of cooling

Newtons law of cooling state that for small difference in temperature between a body and it surroundings, the rate of gain or loss of heat by the body is proportional to the difference in temperature between the body and the surroundings. The Newtons law of cooling only holds in small range of temperature.

The rate of cooling of liquid according to  Newtons law of cooling as it is defined above  depend on the followings: its temperature, the temperature of the enclosure, the area of the exposed surface and the nature of the container.

## How to determine the specific heat capacity using Newtons Law of cooling.

The equipment needed for this Nowtons law of cooling   experiment is: Calorimeter, Thermometer, stop watch and starrier.

Method: First measure the mass of the Copper caloriemeter, when it is empty and record it. Now fill the calorimeter with water but not to the brim, weigh the Calorimeter again as it contains water.  Heat it to about 200C  above room temperature.

The thermometer is used in measurement of the temperature of the liquid at regular time interval of about half a minute  and Stop watch is use to measure the time.

The liquid should be continuously stirred during the cooling period and reweigh the calorimeter and its contents at the end of the experiment.

Repeat the experiment with equal volume of water instead of the liquid and also with the same temperature range, used for the liquid

From the time interval taken by the two liquid to cool through the same temperature range, the specific heat capacity can be calculated from the following;

Let the mass of calorimeter =m

Let the mass of water =m1

Let the time taken by the water to cool from Ѳ12 =t1

Let the mass of the liquid =m2

Let the time taken by the liquid to cool from Ѳ12=t2

Let the specific heat capacity of water =Cw

Let the specific heat capacity of the liquid =C2

Let the specific heat capacity of the calorimeter =C

Heat lost by the water is given by

lost = m1Cw11)

Rate of cooling of water and calorimeter

$= \frac { { (m }_{ 1 }{ c }_{ w }+{ mc) }{ (\theta }_{ 1 }-{ \theta }_{ 2 }) }{ { t }_{ 1 } }$

the rate of cooling of the liquid and calorimeter

$=\frac { { (m }_{ 2 }{ c }_{ 2 }+{ mc) }{ (\theta }_{ 1 }-{ \theta }_{ 2 }) }{ { t }_{ 2 } }$

Since they cooled under the same conditions, we can say that

$= \frac { { (m }_{ 1 }{ c }_{ w }+{ mc) }{ (\theta }_{ 1 }-{ \theta }_{ 2 }) }{ { t }_{ 1 } } = \frac { { (m }_{ 2 }{ c }_{ 2 }+{ mc) }{ (\theta }_{ 1 }-{ \theta }_{ 2 }) }{ { t }_{ 2 } }$

$\frac { { (m }_{ 1 }{ c }_{ w }+mc) }{ { t }_{ 1 } } =\quad \frac { { (m }_{ 2 }{ c }_{ 2 }+mc) }{ { t }_{ 2 } }$

$\therefore { \quad c }_{ 2 }=[\frac { { (m }_{ 1 }{ c }_{ w }+mc) }{ { t }_{ 1 } } { t }_{ 2 }-mc]\frac { 1 }{ { m }_{ 2 } }$