JEE Important Chapter - Thermal Physics

The chapter Thermal Physics covers the most important concepts for JEE Advanced and other competitive exams. In the Thermal Physics chapter, students will study the basic heat transfer process and some laws related to the chapter.

The chapter on thermal physics begins with the very basic definition of Newton’s laws of cooling, the First law of thermodynamics and what are different other laws that are helpful in studying the chapter ahead. These concepts will lead us to a very important part of the chapter i.e. Reversible and Irreversible processes, Carnot engine, and efficiency. In this chapter, we will get to study many interesting and most important concepts such as thermal expansion, Specific heat capacity and Mean free path.

The concepts mentioned above will help us in studying the chapter  deeply along with the hands-on practice of many important numerical problems that will help us in acing the exams.

In this article, we will cover the important concepts and topics that will help students to revise and boost their preparations for JEE.

Important Topics of Thermal Physics

• Heat Transfer

• First Law of Thermodynamics and Heat Capacity

• Second Law of Thermodynamics

• Newton’s Law of Cooling

• Reversible and Irreversible Processes

• Carnot Engine

• Temperature and Thermal Expansion

• Specific Heat Capacity and Mean Free Path

• Thermodynamic Variables

Important Concepts of Thermal Physics

Different Thermodynamic Processes

List of Important Formulae of Thermal Physics

Solved Examples of Thermal Physics

1. A steam engine delivers 5.4 x 108 J of work per minute and services 3.6 x 109 J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?

Sol:

Given,

Work done by the boiler per minute (output work) = 5.4 x 108 J

Input work heat absorbed by boiler = 3.6 x 109 J

Hence efficiency = $\eta = \dfrac{\text{output}}{\text{input}}$

= $\dfrac{5.4\times 108} {3.6\times 109}$

=0.15

So the heat wasted per minute is equal to the difference between heat absorbed per minute and output work per minute.

= $\left(3.6 \times {10^9}\right)- \left (5.4 \times {10^8}\right) J$

=$3.06 \times {10^9} J$

Therefore, the efficiency of the engine is 0.15 and heat is wasted per minute.

Key Point: The relationship between the output work and input work is given by relation $\eta = \dfrac{\text{output}}{\text{input}}$

2. A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?

Sol:

Given,

The cylinder is undergoing adiabatic compression hence for a diatomic gas(hydrogen)  the value of $\gamma$ = 1.4 and as mentioned in the ques final volume = half of the initial one

$V_{2} = \frac{V_{1}}{2}$

$P_{2}$ = 1atm

Therefore for adiabatic process 

$P_{1}V_1^\gamma$ = $P_{2}V_2^\gamma$

$V_1^\gamma$ = $P_{2}V_2^\gamma$

$V_1^\gamma$ = $P_{2}(\frac{V_{1}}{2})^{\gamma}$

Hence $P_{2}= 2^{1.4}$  

=2.64atm

Therefore, 2.64 atm pressure of the gas increase if the gas is compressed to half its original volume.

Key point: In the case of adiabatic processes, the system is always an insulated one so that no heat can be absorbed or released by the system.

Previous Year Questions of Thermal Physics

1. Under isothermal conditions, a gas at 300 K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar . The work done by the gas is( Given that 1L bar=100J)(NEET 2019)

1. 5 kJ

2. 25 J

3. 30 J

4. -30 J 

Sol: 

Given,

Pressure of the gas = 2bar 

Change in volume = $\triangle V$ = 0.25-0.1 = 0.15 L

Since the process is isothermal conditions hence temperature of gas = 0K

From 1st law of thermodynamics 

$\triangle Q $ = $\triangle U$ + $\triangle W$

$\triangle Q $  = 0 which means 

 $\triangle U$ = - $\triangle W$

So the work done is calculated as $\triangle W$ = - $P\triangle V$

=-2$\times$ 0.15 = -0.3L-bar 

And as given in ques 1L-bar = 100 J 

Hence net work done in Joule is -30J Which is option (d)

Trick: As the given process is isothermal and by the formula of thermodynamics one can see that work done will be negative. So without calculations just by looking at the options, a student can go for an option with a negative sign.

2. A Carnot freezer takes heat from water at $0^{\circ} C$ inside it and rejects it to the room at a temperature of $27^{\circ}C$. The latent heat of ice is $336\times10^{3}$ J/kg. If 5 kg of water at $0^{\circ}C$ is converted into ice at $0^{\circ}C$ by the freezer, then the energy consumed by the freezer is close to:  (JEE 2016)

1. $1.67\times10^{5}$ J

2. $1.68 \times10^{6}$ J

3. $1.51\times10^{5}$ J

4. $1.71 \times10^{7}$ J

Sol: 

Given,

Temperature of sink = $T_{sink}$ = $0^{\circ} C$ = 273K

Temperature of source = $T_{source}$ = $27^{\circ} C$ = 300K

Now the latent energy required by freezer to convert ice to water will be the heat rejected by the freezer i.e. $Q_{sink}$ = mL

= 5 $\times 336 \times 10^{3}$ = $1680 \times 10^{3}$ J

From the formula of carnot engine

$\dfrac{T_{sink}}{T_{source}}$ = $\dfrac{Q_{sink}}{Q_{source}}$

Hence $\dfrac{273}{300}$ =$\dfrac{1680 \times 10^{3}}{Q_{source}}$

Or ${Q_{source}}$= $1846.1 \times 10^{3}$ J

So the energy consumed by the freezer = ${Q_{source}}$ - ${Q_{sink}}$

=$1846.1 \times 10^{3}$- $1680 \times 10^{3}$J

= $1.67\times10^{5}$ J which is option (a)

Trick: Here both the conversion of ice into water which is used as latent heat, formula of Carnot engine play an important role in solving the ques.

Practice Questions

1. A quantity of air at normal temperature is compressed (a) slowly (b) suddenly to one-third of its volume. Find the rise in temperature, if any in each case $\gamma$ = 1.4 (Ans-(a) zero (b) 150.6$^{\circ}C$)

2. The water of one kg mass at 373 K is converted into steam at the same temperature and at atmospheric pressure. On boiling, 1 cc of water takes a volume of 1671 cc. Calculate the change in internal energy of the system taking the heat of vaporization to be 540 cal/g.(Ans- 499.84k cal)

Conclusion

We conclude that the chapter ‘Thermal Physics’ has many important concepts that will help us in scoring better marks in NEET and JEE. In this article, we covered thermal expansion, Heat engines, Efficiency and all other important concepts from the chapter along with solved and previous year questions. Students can test their knowledge with the help of practice questions.