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JEE Main Previous Year Questions (2025): Thermodynamics
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Page 1 JEE Main Previous Year Questions (2025): Thermodynamics Q1:Three conductors of same length having thermal conductivity ?? ?? , ?? ?? and ?? ?? are connected as shown in figure. Area of cross sections of 1 st and 2 nd 3 rd conductor it is double of the 1 st conductor. The temperatures are given in the figure. In steady state condition, the value of ?? is ____ ° C. (Given : k 1 = 60Js -1 m -1 K -1 , k 2 = 120 Js -1 m -1 K -1 , k 3 = 135 Js -1 m -1 K -1 ) conductor are same and for Ans: 40 ?? 1 = 2?? ?? 1 ?? (As ?? 1 = ?? 2 = ?? 3 = ?? ?? 1 = ?? 2 = ?? /2 ?? 3 = ?? ) ?? 2 = 2?? ?? 2 ?? Page 2 JEE Main Previous Year Questions (2025): Thermodynamics Q1:Three conductors of same length having thermal conductivity ?? ?? , ?? ?? and ?? ?? are connected as shown in figure. Area of cross sections of 1 st and 2 nd 3 rd conductor it is double of the 1 st conductor. The temperatures are given in the figure. In steady state condition, the value of ?? is ____ ° C. (Given : k 1 = 60Js -1 m -1 K -1 , k 2 = 120 Js -1 m -1 K -1 , k 3 = 135 Js -1 m -1 K -1 ) conductor are same and for Ans: 40 ?? 1 = 2?? ?? 1 ?? (As ?? 1 = ?? 2 = ?? 3 = ?? ?? 1 = ?? 2 = ?? /2 ?? 3 = ?? ) ?? 2 = 2?? ?? 2 ?? ?? 3 = ?? ?? 3 ?? ?? 1 + ?? 2 = ?? 3 ? 100- ?? 2?? ?? 1 ?? + 100- ?? 2?? ?? 2 ?? = ?? - 0 ?? ?? 3 ?? ? 100- ?? 2 ( ?? 1 + ?? 2 ) = ?? 3 ?? ? 50( ?? 1 + ?? 2 )= ?? [?? 3 + ?? 1 + ?? 2 2 ] ? 50( 60 + 120 ) = ?? [135+ 60 + 120 2 ] ? 50 × 80 = ?? ( 135+ 90) ? ?? = 50 × 180 225 ? ?? = 40 ° ?? Q2: An ideal gas initially at ?? ° ?? temperature, is compressed suddenly to one fourth of its volume. If the ratio of specific heat at constant pressure to that at constant volume is ?? /?? , the change in temperature due to the thermodynamic process is ____ K. Ans: 273 Solution: We're given an ideal gas initially at ?? ?? = 0 ° C = 273 K. The gas is suddenly compressed to one- fourth of its initial volume, and we are told that ?? = ?? ?? ?? ?? = 3 2 . For an adiabatic process (one in which there is no heat exchange), if the process were reversible, the relation between the temperature and volume would be ?? ?? ?? -1 = constant. Even though the compression is "sudden" (and hence irreversible), the final equilibrium state of the gas is uniquely determined by its internal energy (which is a state function). That allows us to use the adiabatic relation between the initial and final states. Here are the steps: Write the adiabatic relation between the initial and final states: ?? ?? ?? ?? ?? -1 = ?? ?? ?? ?? ?? -1 . Since the gas is compressed to one-fourth of its volume, we have ?? ?? = 1 4 ?? ?? . Substitute this into the adiabatic relation: 273 ?? ?? ?? -1 = ?? ?? ( 1 4 ?? ?? ) ?? -1 . With ?? = 3 2 , ?? - 1 = 3 2 - 1 = 1 2 . So the relation becomes: Page 3 JEE Main Previous Year Questions (2025): Thermodynamics Q1:Three conductors of same length having thermal conductivity ?? ?? , ?? ?? and ?? ?? are connected as shown in figure. Area of cross sections of 1 st and 2 nd 3 rd conductor it is double of the 1 st conductor. The temperatures are given in the figure. In steady state condition, the value of ?? is ____ ° C. (Given : k 1 = 60Js -1 m -1 K -1 , k 2 = 120 Js -1 m -1 K -1 , k 3 = 135 Js -1 m -1 K -1 ) conductor are same and for Ans: 40 ?? 1 = 2?? ?? 1 ?? (As ?? 1 = ?? 2 = ?? 3 = ?? ?? 1 = ?? 2 = ?? /2 ?? 3 = ?? ) ?? 2 = 2?? ?? 2 ?? ?? 3 = ?? ?? 3 ?? ?? 1 + ?? 2 = ?? 3 ? 100- ?? 2?? ?? 1 ?? + 100- ?? 2?? ?? 2 ?? = ?? - 0 ?? ?? 3 ?? ? 100- ?? 2 ( ?? 1 + ?? 2 ) = ?? 3 ?? ? 50( ?? 1 + ?? 2 )= ?? [?? 3 + ?? 1 + ?? 2 2 ] ? 50( 60 + 120 ) = ?? [135+ 60 + 120 2 ] ? 50 × 80 = ?? ( 135+ 90) ? ?? = 50 × 180 225 ? ?? = 40 ° ?? Q2: An ideal gas initially at ?? ° ?? temperature, is compressed suddenly to one fourth of its volume. If the ratio of specific heat at constant pressure to that at constant volume is ?? /?? , the change in temperature due to the thermodynamic process is ____ K. Ans: 273 Solution: We're given an ideal gas initially at ?? ?? = 0 ° C = 273 K. The gas is suddenly compressed to one- fourth of its initial volume, and we are told that ?? = ?? ?? ?? ?? = 3 2 . For an adiabatic process (one in which there is no heat exchange), if the process were reversible, the relation between the temperature and volume would be ?? ?? ?? -1 = constant. Even though the compression is "sudden" (and hence irreversible), the final equilibrium state of the gas is uniquely determined by its internal energy (which is a state function). That allows us to use the adiabatic relation between the initial and final states. Here are the steps: Write the adiabatic relation between the initial and final states: ?? ?? ?? ?? ?? -1 = ?? ?? ?? ?? ?? -1 . Since the gas is compressed to one-fourth of its volume, we have ?? ?? = 1 4 ?? ?? . Substitute this into the adiabatic relation: 273 ?? ?? ?? -1 = ?? ?? ( 1 4 ?? ?? ) ?? -1 . With ?? = 3 2 , ?? - 1 = 3 2 - 1 = 1 2 . So the relation becomes: 273 ?? ?? 1/2 = ?? ?? ( 1 4 ) 1/2 ?? ?? 1/2 . Cancel the common factor ?? ?? 1/2 (provided it is nonzero) and simplify: 273 = ?? ?? ( 1 2 ). Now solve for the final temperature ?? ?? : ?? ?? = 273× 2 = 546 K. The change in temperature is then: ??? = ?? ?? - ?? ?? = 546 - 273 = 273 K. So, the change in temperature due to the thermodynamic process is 273 K . Q3: The temperature of 1 mole of an ideal monoatomic gas is increased by ???? ° ?? at constant pressure. The total heat added and change in internal energy are ?? ?? and ?? ?? , respectively. If ?? ?? ?? ?? = ?? ?? then the value of ?? is ____ . JEE Main 2025 (Online) 24th January Morning Shift Ans: 15 Solution: ?? = ?? ?? ?? ??? and ??? = ?? ?? ?? ??? For a monoatomic ideal gas, the molar specific heats are given by: ?? ?? = 3 2 ?? and ?? ?? = ?? ?? + ?? = 5 2 ?? Thus, the ratio of the heat added to the change in internal energy is: ?? ??? = ?? ?? ?? ?? = 5 2 ?? 3 2 ?? = 5 3 Given that: ?? 1 ?? 2 = ?? 9 We equate the two ratios: 5 3 = ?? 9 Multiplying both sides by 9 : ?? = 5 3 × 9 = 15 Thus, the value of ?? is ???? . Q4: A container of fixed volume contains a gas at ???? ° ?? . To double the pressure of the gas, the temperature of gas should be raised to ____ ° ?? . JEE Main 2025 (Online) 29th January Morning Shift Ans: 327 Page 4 JEE Main Previous Year Questions (2025): Thermodynamics Q1:Three conductors of same length having thermal conductivity ?? ?? , ?? ?? and ?? ?? are connected as shown in figure. Area of cross sections of 1 st and 2 nd 3 rd conductor it is double of the 1 st conductor. The temperatures are given in the figure. In steady state condition, the value of ?? is ____ ° C. (Given : k 1 = 60Js -1 m -1 K -1 , k 2 = 120 Js -1 m -1 K -1 , k 3 = 135 Js -1 m -1 K -1 ) conductor are same and for Ans: 40 ?? 1 = 2?? ?? 1 ?? (As ?? 1 = ?? 2 = ?? 3 = ?? ?? 1 = ?? 2 = ?? /2 ?? 3 = ?? ) ?? 2 = 2?? ?? 2 ?? ?? 3 = ?? ?? 3 ?? ?? 1 + ?? 2 = ?? 3 ? 100- ?? 2?? ?? 1 ?? + 100- ?? 2?? ?? 2 ?? = ?? - 0 ?? ?? 3 ?? ? 100- ?? 2 ( ?? 1 + ?? 2 ) = ?? 3 ?? ? 50( ?? 1 + ?? 2 )= ?? [?? 3 + ?? 1 + ?? 2 2 ] ? 50( 60 + 120 ) = ?? [135+ 60 + 120 2 ] ? 50 × 80 = ?? ( 135+ 90) ? ?? = 50 × 180 225 ? ?? = 40 ° ?? Q2: An ideal gas initially at ?? ° ?? temperature, is compressed suddenly to one fourth of its volume. If the ratio of specific heat at constant pressure to that at constant volume is ?? /?? , the change in temperature due to the thermodynamic process is ____ K. Ans: 273 Solution: We're given an ideal gas initially at ?? ?? = 0 ° C = 273 K. The gas is suddenly compressed to one- fourth of its initial volume, and we are told that ?? = ?? ?? ?? ?? = 3 2 . For an adiabatic process (one in which there is no heat exchange), if the process were reversible, the relation between the temperature and volume would be ?? ?? ?? -1 = constant. Even though the compression is "sudden" (and hence irreversible), the final equilibrium state of the gas is uniquely determined by its internal energy (which is a state function). That allows us to use the adiabatic relation between the initial and final states. Here are the steps: Write the adiabatic relation between the initial and final states: ?? ?? ?? ?? ?? -1 = ?? ?? ?? ?? ?? -1 . Since the gas is compressed to one-fourth of its volume, we have ?? ?? = 1 4 ?? ?? . Substitute this into the adiabatic relation: 273 ?? ?? ?? -1 = ?? ?? ( 1 4 ?? ?? ) ?? -1 . With ?? = 3 2 , ?? - 1 = 3 2 - 1 = 1 2 . So the relation becomes: 273 ?? ?? 1/2 = ?? ?? ( 1 4 ) 1/2 ?? ?? 1/2 . Cancel the common factor ?? ?? 1/2 (provided it is nonzero) and simplify: 273 = ?? ?? ( 1 2 ). Now solve for the final temperature ?? ?? : ?? ?? = 273× 2 = 546 K. The change in temperature is then: ??? = ?? ?? - ?? ?? = 546 - 273 = 273 K. So, the change in temperature due to the thermodynamic process is 273 K . Q3: The temperature of 1 mole of an ideal monoatomic gas is increased by ???? ° ?? at constant pressure. The total heat added and change in internal energy are ?? ?? and ?? ?? , respectively. If ?? ?? ?? ?? = ?? ?? then the value of ?? is ____ . JEE Main 2025 (Online) 24th January Morning Shift Ans: 15 Solution: ?? = ?? ?? ?? ??? and ??? = ?? ?? ?? ??? For a monoatomic ideal gas, the molar specific heats are given by: ?? ?? = 3 2 ?? and ?? ?? = ?? ?? + ?? = 5 2 ?? Thus, the ratio of the heat added to the change in internal energy is: ?? ??? = ?? ?? ?? ?? = 5 2 ?? 3 2 ?? = 5 3 Given that: ?? 1 ?? 2 = ?? 9 We equate the two ratios: 5 3 = ?? 9 Multiplying both sides by 9 : ?? = 5 3 × 9 = 15 Thus, the value of ?? is ???? . Q4: A container of fixed volume contains a gas at ???? ° ?? . To double the pressure of the gas, the temperature of gas should be raised to ____ ° ?? . JEE Main 2025 (Online) 29th January Morning Shift Ans: 327 Solution: To solve this problem, we can use the combined gas law in terms of pressure ( ?? ) and temperature ( ?? ). For a gas with a fixed volume, the relationship is given as: ?? 1 ?? 1 = ?? 2 ?? 2 where: ?? 1 is the initial pressure, ?? 1 is the initial temperature in Kelvin, ?? 2 is the final pressure, ?? 2 is the final temperature in Kelvin. Given that the initial temperature ?? 1 = 27 ° ?? , we need to convert it to Kelvin: ?? 1 = 27 ° ?? + 273 .15 = 300 .15 K The final pressure ?? 2 is twice the initial pressure ( ?? 2 = 2?? 1 ). Substituting these values into the gas law equation gives: ?? 1 300 .15 = 2?? 1 ?? 2 Solving for ?? 2 : Cancel ?? 1 from both sides: 1 300 .15 = 2 ?? 2 Rearrange to solve for ?? 2 : ?? 2 = 2 × 300 .15 = 600.3 K Convert the final temperature back to Celsius: ?? 2 = 600.3 K - 273 .15 = 327 .15 ° C Therefore, to double the pressure of the gas, the temperature should be raised to approximately ?????? ° ?? . Q5: ?? ?? is the specific heat ratio of monoatomic gas A having 3 translational degrees of freedom. ?? ?? is the specific heat ratio of polyatomic gas ?? having 3 translational, 3 rotational degrees of freedom and ?? vibrational mode. If ?? ?? ?? ?? = (?? + ?? ?? ), then the value of ?? is ____ . JEE Main 2025 (Online) 2nd April Morning Shift Ans: 3 Solution: To determine the value of ?? , we start by considering the specific heat ratios for gases A and B . For gas A , which is monoatomic with 3 translational degrees of freedom: The degrees of freedom, ?? ?? , is 3 . Therefore, ?? ?? = ?? ?? +2 ?? ?? = 3+2 3 = 5 3 . For gas B , which is polyatomic with 3 translational, 3 rotational degrees of freedom, and 1 vibrational mode: The total degrees of freedom, ?? ?? , is 3 + 3 + 2 × 1 = 8 (since one vibrational mode contributes 2 degrees of freedom: 1 kinetic +1 potential). Page 5 JEE Main Previous Year Questions (2025): Thermodynamics Q1:Three conductors of same length having thermal conductivity ?? ?? , ?? ?? and ?? ?? are connected as shown in figure. Area of cross sections of 1 st and 2 nd 3 rd conductor it is double of the 1 st conductor. The temperatures are given in the figure. In steady state condition, the value of ?? is ____ ° C. (Given : k 1 = 60Js -1 m -1 K -1 , k 2 = 120 Js -1 m -1 K -1 , k 3 = 135 Js -1 m -1 K -1 ) conductor are same and for Ans: 40 ?? 1 = 2?? ?? 1 ?? (As ?? 1 = ?? 2 = ?? 3 = ?? ?? 1 = ?? 2 = ?? /2 ?? 3 = ?? ) ?? 2 = 2?? ?? 2 ?? ?? 3 = ?? ?? 3 ?? ?? 1 + ?? 2 = ?? 3 ? 100- ?? 2?? ?? 1 ?? + 100- ?? 2?? ?? 2 ?? = ?? - 0 ?? ?? 3 ?? ? 100- ?? 2 ( ?? 1 + ?? 2 ) = ?? 3 ?? ? 50( ?? 1 + ?? 2 )= ?? [?? 3 + ?? 1 + ?? 2 2 ] ? 50( 60 + 120 ) = ?? [135+ 60 + 120 2 ] ? 50 × 80 = ?? ( 135+ 90) ? ?? = 50 × 180 225 ? ?? = 40 ° ?? Q2: An ideal gas initially at ?? ° ?? temperature, is compressed suddenly to one fourth of its volume. If the ratio of specific heat at constant pressure to that at constant volume is ?? /?? , the change in temperature due to the thermodynamic process is ____ K. Ans: 273 Solution: We're given an ideal gas initially at ?? ?? = 0 ° C = 273 K. The gas is suddenly compressed to one- fourth of its initial volume, and we are told that ?? = ?? ?? ?? ?? = 3 2 . For an adiabatic process (one in which there is no heat exchange), if the process were reversible, the relation between the temperature and volume would be ?? ?? ?? -1 = constant. Even though the compression is "sudden" (and hence irreversible), the final equilibrium state of the gas is uniquely determined by its internal energy (which is a state function). That allows us to use the adiabatic relation between the initial and final states. Here are the steps: Write the adiabatic relation between the initial and final states: ?? ?? ?? ?? ?? -1 = ?? ?? ?? ?? ?? -1 . Since the gas is compressed to one-fourth of its volume, we have ?? ?? = 1 4 ?? ?? . Substitute this into the adiabatic relation: 273 ?? ?? ?? -1 = ?? ?? ( 1 4 ?? ?? ) ?? -1 . With ?? = 3 2 , ?? - 1 = 3 2 - 1 = 1 2 . So the relation becomes: 273 ?? ?? 1/2 = ?? ?? ( 1 4 ) 1/2 ?? ?? 1/2 . Cancel the common factor ?? ?? 1/2 (provided it is nonzero) and simplify: 273 = ?? ?? ( 1 2 ). Now solve for the final temperature ?? ?? : ?? ?? = 273× 2 = 546 K. The change in temperature is then: ??? = ?? ?? - ?? ?? = 546 - 273 = 273 K. So, the change in temperature due to the thermodynamic process is 273 K . Q3: The temperature of 1 mole of an ideal monoatomic gas is increased by ???? ° ?? at constant pressure. The total heat added and change in internal energy are ?? ?? and ?? ?? , respectively. If ?? ?? ?? ?? = ?? ?? then the value of ?? is ____ . JEE Main 2025 (Online) 24th January Morning Shift Ans: 15 Solution: ?? = ?? ?? ?? ??? and ??? = ?? ?? ?? ??? For a monoatomic ideal gas, the molar specific heats are given by: ?? ?? = 3 2 ?? and ?? ?? = ?? ?? + ?? = 5 2 ?? Thus, the ratio of the heat added to the change in internal energy is: ?? ??? = ?? ?? ?? ?? = 5 2 ?? 3 2 ?? = 5 3 Given that: ?? 1 ?? 2 = ?? 9 We equate the two ratios: 5 3 = ?? 9 Multiplying both sides by 9 : ?? = 5 3 × 9 = 15 Thus, the value of ?? is ???? . Q4: A container of fixed volume contains a gas at ???? ° ?? . To double the pressure of the gas, the temperature of gas should be raised to ____ ° ?? . JEE Main 2025 (Online) 29th January Morning Shift Ans: 327 Solution: To solve this problem, we can use the combined gas law in terms of pressure ( ?? ) and temperature ( ?? ). For a gas with a fixed volume, the relationship is given as: ?? 1 ?? 1 = ?? 2 ?? 2 where: ?? 1 is the initial pressure, ?? 1 is the initial temperature in Kelvin, ?? 2 is the final pressure, ?? 2 is the final temperature in Kelvin. Given that the initial temperature ?? 1 = 27 ° ?? , we need to convert it to Kelvin: ?? 1 = 27 ° ?? + 273 .15 = 300 .15 K The final pressure ?? 2 is twice the initial pressure ( ?? 2 = 2?? 1 ). Substituting these values into the gas law equation gives: ?? 1 300 .15 = 2?? 1 ?? 2 Solving for ?? 2 : Cancel ?? 1 from both sides: 1 300 .15 = 2 ?? 2 Rearrange to solve for ?? 2 : ?? 2 = 2 × 300 .15 = 600.3 K Convert the final temperature back to Celsius: ?? 2 = 600.3 K - 273 .15 = 327 .15 ° C Therefore, to double the pressure of the gas, the temperature should be raised to approximately ?????? ° ?? . Q5: ?? ?? is the specific heat ratio of monoatomic gas A having 3 translational degrees of freedom. ?? ?? is the specific heat ratio of polyatomic gas ?? having 3 translational, 3 rotational degrees of freedom and ?? vibrational mode. If ?? ?? ?? ?? = (?? + ?? ?? ), then the value of ?? is ____ . JEE Main 2025 (Online) 2nd April Morning Shift Ans: 3 Solution: To determine the value of ?? , we start by considering the specific heat ratios for gases A and B . For gas A , which is monoatomic with 3 translational degrees of freedom: The degrees of freedom, ?? ?? , is 3 . Therefore, ?? ?? = ?? ?? +2 ?? ?? = 3+2 3 = 5 3 . For gas B , which is polyatomic with 3 translational, 3 rotational degrees of freedom, and 1 vibrational mode: The total degrees of freedom, ?? ?? , is 3 + 3 + 2 × 1 = 8 (since one vibrational mode contributes 2 degrees of freedom: 1 kinetic +1 potential). Therefore, ?? ?? = ?? ?? +2 ?? ?? = 8+2 8 = 10 8 = 5 4 . Now, the relation given is: ?? ?? ?? ?? = (1 + 1 ?? ) Substitute the values of ?? ?? and ?? ?? : 5 3 5 4 = 5 3 × 4 5 = 20 15 = 4 3 Equating to the given form: 4 3 = 1 + 1 ?? Thus, we have: 4 3 - 1 = 1 ?? 1 3 = 1 ?? Solving for ?? : ?? = 3 Hence, the value of ?? is 3 . Q6: The internal energy of air in ?? ?? × ?? ?? × ?? ?? sized room at 1 atmospheric pressure will be ____ × ???? ?? ?? (Consider air as diatomic molecule) JEE Main 2025 (Online) 2nd April Evening Shift Ans: 12 Solution: To determine the internal energy of the air in a room that measures 4 m× 4 m× 3 m at 1 atmospheric pressure, we can follow these steps using the properties of a diatomic gas: The volume of the room, ?? , is calculated as: ?? = 4 m× 4 m× 3 m= 48 m 3 For a diatomic gas like air, the internal energy ?? is given by the formula: ?? = ?? ?? ?? ?? = ?? 5???? 2 Where: ?? is the number of moles, ?? ?? is the molar heat capacity at constant volume, ?? is the ideal gas constant, and ?? is the temperature in Kelvin. Using the ideal gas law, ???? = ?????? , we can substitute for ?????? : ?????? = ???? Therefore, the internal energy ?? becomes: ?? = 5 2 ???? Substituting the given values for atmospheric pressure ?? = 10 5 Pa ( 1 atm ) and volume ?? =Read More
FAQs on JEE Main Previous Year Questions (2025): Thermodynamics
| 1. What is the first law of thermodynamics and how does it apply to closed systems? |  |
Ans. The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed in an isolated system. In the context of closed systems, it implies that the change in internal energy of the system is equal to the heat added to the system minus the work done by the system. Mathematically, it is expressed as ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added, and W is the work done. This principle is fundamental in understanding energy transfers and transformations in thermodynamic processes.
| 2. Can you explain the concept of entropy and its significance in thermodynamics? | |
Ans. Entropy is a measure of the disorder or randomness in a system. In thermodynamics, it reflects the number of ways a system can be arranged, with higher entropy indicating greater disorder. The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time; it can only increase or remain constant. This law implies that natural processes tend to move toward a state of maximum entropy, which is crucial for understanding the direction of spontaneous processes and the efficiency of heat engines.
| 3. What is the difference between isothermal and adiabatic processes? | |
Ans. Isothermal processes occur at a constant temperature, meaning that any heat added to the system is used to do work, keeping the internal energy constant. In contrast, adiabatic processes occur without any heat exchange with the surroundings, meaning that all the energy changes are due to work done on or by the system, which leads to a change in temperature. Understanding the differences between these processes is essential for analyzing various thermodynamic cycles and their efficiencies.
| 4. How does the Carnot theorem relate to the efficiency of heat engines? | |
Ans. The Carnot theorem states that no heat engine operating between two thermal reservoirs can be more efficient than a Carnot engine operating between the same reservoirs. The efficiency of a Carnot engine is determined by the temperatures of the hot (T₁) and cold (T₂) reservoirs, expressed as η = 1 - (T₂/T₁). This theorem sets an upper limit on the efficiency of all real heat engines and emphasizes the importance of temperature differences in improving engine performance.
| 5. What are the applications of the laws of thermodynamics in real-life systems? | |
Ans. The laws of thermodynamics have numerous applications in real-life systems, including engines, refrigerators, and heat pumps. For instance, the first law is applied in calculating energy inputs and outputs in engines, while the second law is crucial in understanding the efficiency of thermal machines and the limitations of energy conversion processes. Other applications include climate control systems, biological processes, and industrial chemical reactions, highlighting the pervasive role of thermodynamics in science and engineering.
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