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# Specific Heat

Top
 Sub Topics Heat is something, which produces the sensation of hotness or coldness to us. When we rub our hands, they become warm. Warm means hot. In fact, the kinetic energy of the moving hands is converted in the form of heat energy due to the friction between our hands. According to the kinetic theory of matter, the molecules of the substance have kinetic energy due to their motion. When a substance is heated, its molecule absorbs heat energy and move with higher kinetic energy. Substance having equal masses requires different amount of heat. This specific property of a substance is called specific heat capacity. In this section we will learn more about specific heat.

## Specific Heat Capacity

Specific heat capacity ($c$) of a substance is the heat required to produce unit temperature rise in unit mass of the substance.

## Specific Heat Formula

If someone increase the temperature of the substance, they will be an increase in the kinetic energy of its molecules - that is increase in its internal energy. So one need to supply energy. The energy needed to raise the temperature of an object is proportional to the increase in temperature and mass of the object

$Energy \propto mass \times Temperature\ rise$

Here the constant of proportionality which is depends on the substance, is called the Specific Heat Capacity ($c$).

So,
$Energy = mass \times specific\ heat\ capacity \times Temperature\ rise$

$q$ = $m \times c \times \Delta T$

The energy gained or lost as heat when a given mass of a substance is warmed or cooled can be calculated using above equation.

So the specific heat capacity,
$c$ = $\frac{q}{ m \times \Delta T}$

Here,
$c$ = The specific heat capacity
$q$ = It is the energy gained or lost.
$m$ = Mass of the substance
$\Delta T$ = $T_{final}- T_{initial}$ = The change in temperature.
• The units of the specific heat capacity are $\frac{J}{kgK}$ = $Jkg^{-1}K^{-1}$.

a) The specific heat capacity is also known as specific heat enthalpy.
b) If an substance is cooled or heated, there will a transfer of energy. which depends on following factors:
1. The magnitude of the temperature change
2. The identity of the material gaining or loosing energy.
3. The quantity of the materials

## Specific Heat Capacity Table

Specific Heat of the substances at $25^{0}C$ and Standard Atmospheric pressure.

## Specific Heat of Air

The Specific heat capacitance of the Air at different temperature are given in the below table

## Specific Heat of Gas

A gas can have the specific heat conductivity depending upon the conditions under which it is heated. Specific heat conductivity of a gas is defines in two ways.

Specific Heat conductivity of a gas at constant volume: The amount of heat required for raising the temperature of 1 g of gas through 1° C at the constant volume. It is denoted by $C_{v}$.

Specific Heat conductivity of a gas at constant pressure: The amount of heat required for raising the temperature of 1 g of gas through 1° C at the constant pressure. It is denoted by $C_{p}$.

## Specific Heat Ratio

The ratio of these two specific heats is called adiabatic constant ($\gamma$ ).

$\gamma$ = $\frac{C_{p}}{C_{v}}$

Here,

$C_{p}$ = Specific Heat conductivity of a gas at constant pressure
$C_{v}$ = Specific Heat conductivity of a gas at constant volume

The value of $\gamma$ is 1.67 for mono atomic gases, 1.4 for diatomic gases and 1.33 for tri or poly atomic gases.

## How To Calculate Specific Heat Capacity

The following are the problems of Specific heat capacity.

### Solved Examples

Question 1: How much energy must be transferred to raise the temperature of a cup of tea (250 ml) from 293.7 K to 368.8 K. Assume that the tea and water have the same density(1 g/ml), and specific heat capacity (4.184 J/gK).
Solution:

Given,
$c$  =   $4.184$ J/gK

Mass ($m$) = $\frac{250 ml}{1 g/ml}$ = $250$g

$\Delta T$ = $T_{final}-T_{initial}$  = $368.8 K$ - $293.7 K$ = $75.1$ K

We have,

$q$ = $m \times c \times \Delta T$

= $250 \times 4.184 \times 75.1$

= $78.554$kJ

Question 2: The specific heat capacity of water is $4200 Jkg^{-1}K^{-1}$. Calculate the energy needed to raise a bucket full of water, volume 1.4 litres from tap temperature, $10^{0}C$ to boiling point.

Solution:

Given,

$c$  =   $4200$ J/kgK

Mass ($m$) = $\frac{1.4 litres}{1 kg/l}$ = $1.4$kg

$\Delta T$ = $T_{final}-T_{initial}$  = $100 K$ - $10 K$ = $90$ K

We have,

$q$ = $m \times c \times \Delta T$

= $1.4 \times 4200 \times 90$

= $0.53$MJ