WBJEE 2025 Heat and Thermodynamics Practice Questions with Solutions: WBJEE 2025 Heat and Thermodynamics practice questions with solutions are based on sub-topics, like the Laws of Thermodynamics, Heat, Work and Internal Energy, Isothermal & Adiabatic Processes etc. Most of the questions are numerical-based, and solving these can help you get a solid 4-8 marks in the WBJEE exam 2025. One way to score good marks in the WBJEE Physics paper (Paper II) is to practice WBJEE Thermodynamics questions from past years' question papers. Heat and Thermodynamics being a crucial chapter in the WBJEE 2025 Physics syllabus PDF carries an approximate weightage of 6%. In the upcoming WBJEE 2025 exam, 3-4 questions are expected from this topic. Start with the basic, easier questions and slowly move on to the more tricky, difficult ones as you regularly solve practice questions for WBJEE 2025 Heat and Thermodynamics. In this article, we have shared a list of probable WBJEE Thermodynamics sample questions so that you understand the question type and format better, and assess the approximate time spent in solving them.

Also Check - WBJEE 2025 Physics Topic-Wise Weightage & List of Important Topics

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List of WBJEE 2025 Heat and Thermodynamics Sub Topics for Exam Preparations

The chapter on WBJEE 2025 Heat and Thermodynamics is very important and you must prepare well for it to qualify the exam with flying colours. As discussed above, there are various numerical-based topics in this chapter, and to prepare well, you must be well-versed with all the topics. The table below outlines the WBJEE 2025 Heat and Thermodynamics key topics to be practiced regularly for an effective exam preparation. 

Also Check - 6-Month Preparation Strategy to Crack WBJEE 2025

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WBJEE 2025 Heat and Thermodynamics Expected Weightage

The Heat and Thermodynamics section from the WBJEE 2025 syllabus PDF holds at least 6% wightage and 3-4 expected number of questions in the exam. After understanding the expected questions and marks weightage, it will be easy for you to prepare accordingly. There's a good chance of scoring 4-8 marks in the Paper II if you diligently solve the WBJEE 2025 Heat and Thermodynamics practice questions. However, you should note that the data mentioned below is based on the previous year's paper assessment. 

WBJEE 2025 Heat and Thermodynamics Practice Questions with Solutions

The WBJEE Thermodynamics questions offer a huge advantage during the exam preparations. These sample questions not only give you a sneak peek into the probable questions but also leave you with the step-by-step solution method for in-depth knowledge. Here is a look at the highly expected WBJEE 2025 Heat and Thermodynamics practice questions with the solutions below. 

Q.1 When the room temperature becomes equal to the dew point, the relative humidity of the room is:

a) 100%

b) zero%

c) 70%

d) 85%

Correct Answer: a) 100%

Explanation: Dew point is the temperature at which the air can no longer hold any more water vapor. When the room temperature becomes equal to the dew point, the air is saturated with water vapor and the relative humidity is 100%. If the room temperature is below the dew point, the air cannot hold all of the water vapor that is present and some of the water vapor will condense out of the air. This is why you often see dew on surfaces in the morning when the temperature is cool and the air is humid. If the room temperature is above the dew point, the air can hold more water vapor than is currently present. This is why you often see fog in the morning when the temperature is cool and the air is humid. The fog is caused by the water vapor in the air condensing out of the air onto cold surfaces.

Q.2 Two substances A and B of same mass are heated at constant rate. The variation of temperature Θ of the substances with time t is shown in the figure. Choose the correct statement:-

a) Specific heat of A is greater than that of B

b) Specific heat of B is greater than that of A

c) Both have same specific heat

d) None of the above is true

Correct Answer: a) Specific heat of A is greater than that of B

Solution: To determine the correct statement, we analyze the relationship between specific heat and the slope of the temperature vs. time graph:

The heat supplied (QQQ) is related to the specific heat capacity (ccc) by Q=mcΔΘQ = mc\Delta \ThetaQ=mcΔΘ, where mmm is the mass and ΔΘ\Delta \ThetaΔΘ is the temperature change. Since the heat is supplied at a constant rate, the slope of the Θ\ThetaΘ-ttt graph (ΔΘ/Δt\Delta \Theta / \Delta tΔΘ/Δt) is inversely proportional to the specific heat capacity (ccc).

From the graph:

Substance A has a smaller slope (temperature rises more slowly).

Substance B has a steeper slope (temperature rises more quickly).

Thus, the specific heat of A is greater than that of B because a smaller slope indicates a higher heat capacity.

Therefore, the answer is the specific heat of A is greater than that of B

Q.3 One mole of an ideal monoatomic gas expands along the polytrope PV³ = constant from V₁ to V₂ at a constant pressure P₁. The temperature during the process is such that molar specific heat C_V = 3R/2. The total heat absorbed during the process can be expressed as:

a) P₁V₁ (V²₁/V²₂+1)

b) P₁V₁ (V²₁/V₂²-1)

c) P₁V₁(V³₁/V₂²-1)

d) P₁V₁(V₁/V₂²-1)

Correct Answer: b) P₁V₁ (V²₁/V₂²-1)

Solution:

Given pV¹ = C, CV = 3E/2 , n=1 ,p;

For polytropic process,

pV³ = C

Therefore x = 3

C = C_V + R/1-x = 3R/2 + R/1-3 = R

Heat supplied (X) = nC​ΔT

= 1xR x (T₂-T₁) = R(T₂-T₁)

Now, as pV = nRT....(I)

Therefore, pV = RT (therefore n = 1)

Therefore, R = pV/T = p₁T₁/T₁

Also, pV³ = V

= (RT/V)V³ = C

= TV² = C = T₁V²₁ = T2V₂²

Hence proved T₂ = T₁V²₁/V₂²

Heat supplied (Q) = p₁V₁/T₁ [T₁V²₁/V₂² - T₁] = p₁V₁ [V²₁/V₂² - 1]

Q4. An experiment takes 10 minutes to raise temperature of water from 0°C to 100°C and another 55 minutes to convert it totally into steam by a stabilized heater. The latent heat of vaporization comes out to be:

a) 530 cal/g

b) 540 cal/g

c) 550 cal/g

d) 560 cal/g

Correct Answer: c) 550 cal/g

Solution:

The ratio of time taken for heating to vaporization is:

Time ratio = 10/55 = 2/11

The heat required to raise the temperature of water from 0^∘C to 100^∘C is:

Q_heating​=m⋅c⋅ΔT,

where m is the mass of water, c = 1cal/g^∘C, and ΔT = 100^∘C.

Simplifying: Q_heating = 100m cal.

The heat required to vaporise the water at 100^∘C:

Q_vaporisation = m.L,

where L is the latent heat of vaporisation.

The heat supplied is proportional to the time, so:

Q_heating/Q_vaporisation = Time for Heating/ Time for vaporisation = 10/55 = 2/11

Substitute Q_heating = 100m and Q_vaporisation = mL:

100m/ml = 2/11

Cancel m and solve for L:

100/L = 2/11

Rearrange:

L = 100x11/2 = 550 cal/g.

Q5. At what temperature will the rms speed of air molecules be double that of NTP?

a) 519°C

b) 619°C

c) 719°C

d) 819°C

Correct Answer: a) 519°C

Explanation: The root mean square speed of air molecules will be double that of normal temperature and pressure (NTP) at 519°C.

Q6. When 100 g of boiling water at 100° C is added into a calorimeter containing 300 g of cold water at 10°C, temperature of the mixture becomes 20°C. Then, a metallic block of mass 1 kg at 10° C is dipped into the mixture in the calorimeter. After reaching thermal equilibrium, the final temperature becomes 19° C. What is the specific heat of the metal in CGS unit?

a) 0.01

b) 0.3

c) 0.09

d) 0.1

Correct Answer: b) 0.3

Explanation: The specific heat of the metal block, after reaching thermal equilibrium in the calorimeter, is 0.3 J/g°C. This result is obtained by balancing the heat lost by 100 g of boiling water cooling from 100°C to 20°C (8000 cal) and the heat gained by 300 g of cold water warming from 10°C to 20°C (3000 cal), along with the heat absorbed by the 1 kg metal block warming from 10°C to 19°C. The total heat supplied equals the total heat lost, and solving this equation gives the specific heat of the metal as 0.3 J/g°C.

Q.7 A metallic block of mass 20 kg is dragged with a uniform velocity of 0.5 ms-1 on a horizontal table for 2.1 s. The coefficient of static friction between the block and the table is 0.10. What will be the maximum possible rise in temperature of the metal block, if the specific heat of the block is 0.1 CGS unit? Assume g = 10 ms-2 and uniform rise in temperature throughout the whole block. [Ignore absorption of heat by the table]:

a) 0.0025°C

b) 0.025°C

c) 0.001°C

d) 0.05°C

Correct Answer: a) 0.0025°C

Solution:

The metallic block of mass 20 kg is dragged with a uniform velocity of 0.5 m/s on a horizontal table for 2.1 seconds. The coefficient of static friction between the block and the table is 0.10. The frictional force opposing the motion is given by F_riction = μ⋅m⋅g where μ is 0.10, , = 20kg, and g = 10m/s². This results in a frictional force of 20N. The work done by friction is W = F_riction ⋅distance, and since the block is dragged 1.05 metres, the work done is 20N⋅1.05m - 21J. This work done by friction is converted into heat energy (Q) which raises the temperature of the block.

The heat energy (Q) can be related to the rise in temperature (ΔT) using the formula Q = m⋅c⋅ΔT, where c = 0.1 cal/g °C (specific het of the block). The total energy absorbed is 21J (or 21 cal), and solving for ΔT, we get:

ΔT = 21/2000 x 0.1 = 21/200 = 0.105°C.

Thus, the maximum possible rise in the temperature of the metal block is 0.025°C.

Q.8 The water equivalent of a calorimeter is 10 g and it contains 50 g of water at 15°C. Some amount of ice, initially at  -10°C is dropped in it and half of the ice melts till equilibrium is reached. What was the initial amount of ice that was dropped (when specific heat of ice = 0.5 cal gm-1° C-1, specific heat of water = 10 cal gm-1°C-1 and latent heat of melting of ice = 80 cal gm-1)?

a) 10g

b) 18g

c) 20g

d) 30g

Correct Answer: b) 18g

Explanation: To determine the initial amount of ice dropped into the calorimeter, we need to consider the heat exchange between the water, the calorimeter, and the ice. The water in the calorimeter has a mass of 50 g and is initially at 15°C. The water equivalent of the calorimeter itself is 10 g. As heat is transferred, the water and the calorimeter cool down to 0°C, and half of the ice, initially at -10°C, melts. The heat lost by the water and the calorimeter is equal to the heat absorbed by the ice. The specific heat of water is 10 cal/g°C, the specific heat of ice is 0.5 cal/g°C, and the latent heat of melting is 80 cal/g. Setting up the equation based on heat lost and heat absorbed, we solve for the mass of ice. After calculations, it was found that the initial amount of ice dropped was 18 g.

Q.9 300 g of water at 25°C is added to 100 g of ice at 0°C. The final temperature of the mixture is:

a) 12.5°C

b) 0°C

c) 25°C

d) 50°C

Correct Answer: b) 0°C

Explanation: To determine the final temperature of the mixture, we need to consider the heat exchange between the 300 g of water at 25°C and the 100 g of ice at 0°C. The water will lose heat as it cools down, and this heat will be used to melt the ice. The heat lost by the water, which is calculated using the formula Q_water=m_water​⋅cwater​ ΔT_water is 7500 cal. The heat required to melt the ice and warm it to 0°C is calculated using Q_{melt ice} = m_ice⋅L_melt + m_ice⋅c_ice ⋅ ΔT_ice, which is 8000 cal. Since the heat lost by the water (7500 cal) is not sufficient to fully melt the ice (8000 cal), the final temperature of the mixture will be 0°C, where the ice has just melted.

Q.10 Two black bodies A and B have equal surface areas are maintained at temperatures 27°C and 177°C respectively. What will be the ratio of the thermal energy radiated per second by A to that by B?

a) 4:9

b) 2:3

c) 16:81

d) 27:177

Correct Answer: c) 16:81

Explanation: The thermal energy radiated per second by a black body is governed by the Stefan-Boltzmann law, which states that the energy radiated is proportional to the fourth power of its absolute temperature. In this case, two black bodies, A and B, have equal surface areas and are maintained at temperatures of 27°C and 177°C, respectively. To calculate the ratio of the thermal energy radiated by A to that by B, we first convert the temperatures to Kelvin: A at 300 K and B at 450 K. The ratio of radiated energy is then given by (T_A/T_B)⁴, which simplifies to (2/3)⁴ = 16/81. Therefore, the thermal energy radiated per second by A is 16 times less than that radiated by B. The correct ratio is 16:81.

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Boost your exam preparation with WBJEE 2025 mock test alongside practice questions. Take 2-3 mock exams every week to assess your overall preparation and performance. You may mark or highlight the questions that are overlapping in the mock paper and the WBJEE 2025 Heat and Thermodynamics practice paper, as it will help you identify and focus on better. Spend time to track progress after each attempt and work on your weaknesses before the final exam. 

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