Introduction To Heat, Internal Energy And Work: Definition, Formula

We already know what heat is. It is a form of Energy, but it always comes into the picture when energy is being transferred from one system to another. Suppose that we are in some initial state 'a' and we want to get to some final state 'b'. We can do that through various processes, as shown in the figure, and the heat energy released or absorbed in all the processes will be different.

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This Story also Contains

1. Heat
2. Work
3. Different Types of Works and the Formulas
4. Internal Energy or Intrinsic Energy
5. Some Solved Examples
6. Summary

Heat

Heat is the energy transfer due to the difference in temperature. Heat is a form of energy which the system can exchange with the surroundings if they are at different temperatures. The heat flows from higher temperature to lower temperature.

Heat is expressed as 'q'

Heat absorbed by the system = +q

Heat evolved by the system = - q

Work

It is the energy transfer due to the difference in pressure that is, the mode of energy transfer.

Types of work

(i) Mechanical Work (Pressure volume work) = Force x Displacement

(ii) Electrical Work = Potential difference x charge flow , VQ = EnF

(iii) Expansion Work $=\mathrm{P} \times \Delta \mathrm{V}=-\mathrm{P}_{\text {ext. }}\left[\mathrm{V}_2-\mathrm{V}_1\right]$

$\mathrm{P}=$ external pressure And $\Delta \mathrm{V}=$ increase or decrease in volume.

(iv) Gravitational Work = mgh

Here m = mass of body,

g = acceleration due to gravity

h = height moved.

Units: dyne cm or erg (C.G.S.)

Newton meter (joule)

(i) If the gas expands, [V₂> V₁] and work is done by the system and W is negative.

(ii) If the gas [V₂ < V₁] and work is done on the system and W is positive.

Different Types of Works and the Formulas

(i) Work done in a reversible isothermal process

$\begin{aligned} & \mathrm{W}=-2.303 \mathrm{nRT} \log _{10} \frac{\mathrm{V}_2}{\mathrm{~V}_1} \\ & \mathrm{~W}=-2.303 \mathrm{nRT} \log _{10} \frac{\mathrm{P}_1}{\mathrm{P}_2}\end{aligned}$

(ii) Work done in an irreversible isothermal process

Work $=-\mathrm{P}_{\text {ext. }}\left(\mathrm{V}_2-\mathrm{V}_1\right)$
That is, Work $=-\mathrm{P} \times \Delta \mathrm{V}$

Internal Energy or Intrinsic Energy

The energy stored within a substance is called its internal energy. The absolute value of internal energy cannot be determined.

Or

It is the total energy of a substance depending upon its chemical nature, temperature, pressure, and volume, amount of substrate. It does not depend upon path in which the final state is achieved.

$\begin{aligned} & \mathrm{E}=\mathrm{E}_{\mathrm{t}}+\mathrm{E}_{\mathrm{r}}+\mathrm{E}_{\mathrm{V}}+\mathrm{E}_{\mathrm{e}}+\mathrm{E}_{\mathrm{n}}+\mathrm{E}_{\mathrm{PE}}+\mathrm{E}_{\mathrm{B}} \\ & \mathrm{E}_{\mathrm{t}}=\text { Transitional energy } \\ & \mathrm{E}_{\mathrm{r}}=\text { Rotational energy } \\ & \mathrm{E}_{\mathrm{PE}}=\text { Potential energy } \\ & \mathrm{E}_{\mathrm{B}}=\text { Bond energy }\end{aligned}$

The exact measurement of it is not possible so it is determined as $\Delta \mathrm{E}$ as follows:

$\begin{aligned} & \Delta \mathrm{E}=\Sigma \mathrm{E}_{\mathrm{p}}-\Sigma \mathrm{E}_{\mathrm{R}} \\ & \Delta \mathrm{E}=\mathrm{E}_{\mathrm{f}}-\mathrm{E}_{\mathrm{i}} \\ & \mathrm{Here} \mathrm{E}_{\mathrm{f}}=\text { final internal energy } \\ & \mathrm{Ei}=\text { Initial internal energy } \\ & \mathrm{Ep}=\text { Internal energy of products } \\ & \mathrm{Er}=\text { Internal energy of reactants }\end{aligned}$

Facts about Internal Energy

• It is an extensive property.
• Internal energy is a state property.
• The change in internal energy does not depend on the path by which the final state is reached.
• Internal energy for an ideal gas is a function of temperature only so when the temperature is kept constant $\Delta \mathrm{E}$ is zero for an ideal gas.
  $
  \mathrm{E} \propto \mathrm{T}
  $

$\Delta \mathrm{E}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}\left[\mathrm{C}_{\mathrm{v}}\right.$ is the heat capacity at constant volume

• For a cyclic process $\Delta \mathrm{E}$ is zero. (E = state function), $E \propto T$
• For an ideal gas, it is totally kinetic energy as there is no molecular interaction.
• Internal energy for an ideal gas is a function of temperature only hence, when the temperature is kept constant it is zero.
• At constant volume (Isochoric) $\mathrm{Q}_{\mathrm{v}}=\Delta \mathrm{E}$
• For exothermic process, is negative as $\mathrm{E}_{\mathrm{R}}>\mathrm{E}_{\mathrm{P}}$ but For endothermic process $\Delta \mathrm{E}$ is positive as $\mathrm{E}_{\mathrm{R}}>\mathrm{E}_{\mathrm{P}}$.
• It is determined by using a Bomb calorimeter of the system.

$\begin{aligned} & \Delta \mathrm{E}=\frac{\mathrm{Z} \times \Delta \mathrm{T} \times \mathrm{m}}{\mathrm{W}} \\ & \mathrm{Z}=\text { Heat capacity of Bomb calorimeter } \\ & \Delta \mathrm{T}=\text { Rise in temperature } \\ & \mathrm{w}=\text { Weight of substrate (amount) } \\ & \mathrm{m}=\text { Molar mass of substrate }\end{aligned}$

Recommended topic video on (Introduction To Heat, Internal Energy And Work )

Some Solved Examples

Example 1: Which of the following is true about heat energy?

1)It can be stored

2)It is an ordered form of energy

3) It is a disordered form of energy

4)Heat energy and work

Solution

Any type of energy transfer between the system & surrounding due to temperature difference is known as Heat. It is considered a disordered form of energy.

Hence, the answer is the option (3).

Example 2: Work done is a

1)State function

2) Path function

3)Both (1) and (2)

4)None of the above

Solution

Between any two states 1 and 2, the work done is dependent on the path by which the process is carried out. Hence, work is a path function.

Hence, the answer is the Option (2).

Example 3: Five moles of an ideal gas at 1 bar and 298K are expanded into a vacuum to double the volume. The work done is:

1)C_V(T₂ – T₁)

2)$-R T\left(V_2-V_1\right)$

3)$-R T \ln \frac{V_2}{V_1}$

4) zero

Solution

The expansion is done in a vacuum that is in the absence of ext. pressure.

If ext. pressure = 0

So, Work Done = zero

Hence, the answer is the option (4).

Example 4: An ideal gas is allowed to expand from 1L to 10L against a constant external pressure of 1 bar. The work done in kJ is :

1)10

2)-9

3)-2

4) -0.9

Solution

We know that,

$\begin{aligned} \mathrm{W} & =-\mathrm{P}_{\mathrm{ext}}\left(\mathrm{V}_2-\mathrm{V}_1\right) \\ & =-1 \times(10-1) \\ & =-9 \text { bar }- \text { litre }\end{aligned}$

We know the conversion of bar-litre to Joules :

1bar - litre $=100 \mathrm{~J}$

$\therefore \mathrm{W}=-9 \times 100 \mathrm{~J}=-0.9 \mathrm{~kJ}$

Hence, the answer is (-0.9 kJ).

Example 5: One mol ideal gas is taken in a container of Volume 5L, it is then reversibly compressed to a volume of 3L. The temperature was kept constant during the process (310 K). Find out the work (in KJ) done to compress the gas.

1) 1.3

2)-1.5

3)-13

4)13

Solution

As we have learned,
The work done in the reversible isothermal process is $\mathrm{W}=-2.303 \mathrm{nRT} \log \left(\mathrm{V}_2 / \mathrm{V}_1\right)$
We have,

$\begin{aligned} & n=1 \\ & \mathrm{~T}=310 \mathrm{~K} \\ & \mathrm{~V}_1=5 \mathrm{~L} \\ & \mathrm{~V}_2=3 \mathrm{~L} \\ & \mathrm{R}=8.314 \mathrm{~J} / \mathrm{mol} . \mathrm{K}\end{aligned}$

Putting all the values we get,$\mathrm{W}=-2.303 \times 1 \times 8.314 \times 310 \times \log \left(\frac{3}{5}\right)=1316 \mathrm{~J}$

$\therefore \mathrm{W}=1.316 \mathrm{~kJ}$

Hence, the answer is the option(1).

Summary

Heat and work are two different modes of transmitting energy from one body to another. Understanding the difference between heat and work lies at the root of thermodynamics. Heat implies the transference of thermal energy from one system to another, while work means the flow of mechanical energy from one system to another. This difference between microscopic motion, that is, heat, and macroscopic motion, that is, work, forms a very important part of how thermodynamic processes work. Heat may be converted into work and vice versa, but they are not the same thing. In terms of the first law of thermodynamics, heat, and work are additive contributions to the total internal energy of a system. Because of the second law of thermodynamics, however, there is always some restriction over the amount of heat that may be converted into work.