CBSE Class 11 Physics Thermodynamics Assignment

QUESTIONS

Question 1. Why spark is produced when two substances are struck hard against each other?
Answer: When two hard substances are struck together, mechanical work done in overcoming friction and deformation is instantly converted into heat energy at the point of impact, raising the local temperature so high that tiny fragments of the material become incandescent, producing a visible spark.
In simple words: Kinetic energy from hitting the objects turns into intense local heat, making tiny chips glow.

Exam Tip: Mention the conversion of mechanical work into thermal energy due to friction and collision.

Question 2. What is the specific heat of a gas in an isothermal process?
Answer: In an isothermal process, the temperature remains constant (\(\Delta T = 0\)). Since specific heat is defined as \(c = \frac{Q}{m \Delta T}\), and \(\Delta T = 0\) while heat is absorbed or rejected, the specific heat of a gas in an isothermal process is infinite.
In simple words: Temperature does not change no matter how much heat is added or removed, making specific heat infinite.

Exam Tip: Remember that specific heat involves dividing by \(\Delta T\); since \(\Delta T = 0\), the result is infinity.

Question 3. On what factors, does the efficiency of Carnot engine depend?
Answer: The efficiency of a Carnot engine depends only on the temperatures of the source (\(T_1\)) and the sink (\(T_2\)), given by the formula \(\eta = 1 - \frac{T_2}{T_1}\). It is independent of the nature of the working substance.
In simple words: Efficiency depends solely on how hot the source is and how cold the sink is.

Exam Tip: State clearly that efficiency does not depend on the working substance or design of the engine.

Question 4. What are two essential features of Carnot's ideal heat engine?
Answer: The two essential features of Carnot's ideal heat engine are:
1. The source and sink have infinite thermal capacities so their temperatures remain constant during heat transfer.
2. Every process in the engine cycle is fully reversible (i.e., there is no friction or dissipation of energy).
In simple words: The source/sink temperatures never change, and all processes are completely reversible without friction.

Exam Tip: Highlight infinite thermal capacity and complete reversibility (zero friction).

Question 5. Plot a graph between internal energy \(U\) and Temperature (\(T\)) of an ideal gas.
Answer: For an ideal gas, internal energy is directly proportional to absolute temperature (\(U = n C_v T\)). Therefore, the graph between internal energy \(U\) and temperature \(T\) is a straight line passing through the origin.
In simple words: Internal energy increases linearly with temperature, forming a straight line graph.

Exam Tip: Ensure the straight line starts from the origin for an ideal gas.

Question 6. Refrigerator transfers heat from cold body to a hot body. Does this violate the second law of thermodynamics?
Answer: No, it does not violate the second law of thermodynamics because external work is done on the system (by the compressor) to transfer heat from a cold body to a hot body.
In simple words: It is allowed because electricity/work is supplied from outside to drive the process.

Question 7. What is heat pump?
Answer: A heat pump is a device that uses mechanical work to absorb heat from a cold reservoir and deliver it to a warmer reservoir, functioning essentially like a refrigerator used for heating buildings.
In simple words: A heat pump is a device that moves heat into a building using mechanical work.

Exam Tip: Define it as a thermodynamic device used for heating spaces.

Question 8. Give two example of heat pump?
Answer: Two examples of heat pumps are:
1. Domestic Refrigerator (used in heating mode / reverse cycle)
2. Air Conditioner (reversible heat pump system)
In simple words: Refrigerators and air conditioners are practical examples.

Exam Tip: List common household appliances like air conditioners and refrigerators.

Question 9. What is heat engine?
Answer: A heat engine is a device that converts thermal energy absorbed from a high-temperature source into mechanical work, rejecting the remaining heat to a low-temperature sink.
In simple words: A machine that turns heat energy into useful mechanical work.

Exam Tip: Emphasize the conversion of heat into mechanical work.

Question 10. Why a gas is cooled when expanded?
Answer: When a gas expands adiabatically, it does work on its surroundings. According to the first law of thermodynamics (\(dQ = dU + dW\)), since no heat is supplied (\(dQ = 0\)) and work is positive (\(dW > 0\)), the internal energy decreases (\(dU < 0\)), causing a decrease in temperature.
In simple words: Expanding gas does work using its own internal energy, which makes it cool down.

Exam Tip: Use the first law of thermodynamics to explain the drop in internal energy.

Question 11. Can the temperature of an isolated system change?
Answer: Yes, the temperature of an isolated system can change during internal processes such as an adiabatic expansion or compression, chemical reactions, or mixing, even though no heat enters or leaves the system.
In simple words: Yes, internal changes like adiabatic expansion can alter the temperature without any heat exchange with the outside.

Question 12. Which one a solid, a liquid or a gas of the same mass and at the same temperature has the greatest internal energy?
Answer: A gas has the greatest internal energy because intermolecular potential energy and kinetic energy contributions are highest due to larger molecular separation and freedom of motion, whereas solids have the least internal energy.
In simple words: Gases have the highest internal energy because molecules move freely and have more potential energy.

Exam Tip: Remember that molecular freedom and potential energy order is Gas > Liquid > Solid.

Question 13. Under what ideal condition the efficiency of a Carnot engine be 100%.
Answer: The efficiency of a Carnot engine can be 100% only if the temperature of the sink is absolute zero (\(T_2 = 0\text{ K}\)), which is physically unattainable according to the third law of thermodynamics.
In simple words: Only if the sink temperature is at absolute zero (0 K), which is impossible.

Exam Tip: State \(T_2 = 0\text{ K}\) and mention it is practically impossible.

Question 14. Which thermodynamic variable is defined by the first law of thermodynamics?
Answer: Internal energy is defined by the first law of thermodynamics.
In simple words: Internal energy.

Exam Tip: Direct one-word or short-phrase answer.

Question 15. Give an example where heat be added to a system without increasing its temperature.
Answer: Melting of ice at 0\(^\circ\)C or boiling of water at 100\(^\circ\)C are examples where heat is added during a phase change without any change in temperature.
In simple words: Melting ice or boiling water absorbs heat while keeping the temperature constant.

Exam Tip: Give phase change processes like melting or vaporization.

Question 16. What is the efficiency of carnot engine working between ice point and steam point?
Answer: Ice point \(T_2 = 0^\circ\text{C} = 273\text{ K}\), Steam point \(T_1 = 100^\circ\text{C} = 373\text{ K}\).
\(\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{273}{373} = \frac{100}{373} \approx 0.268 = 26.8\%\)
In simple words: The efficiency is about 26.8% when working between freezing and boiling points of water.

Question 17. Two blocks of the same metal having masses 5g and 10g collide against a target with the same velocity. If the total energy used in heating the balls which will attain higher temperature?
Answer: Both balls will undergo the same rise in temperature because kinetic energy per unit mass is identical, and temperature rise depends on specific heat and mass proportionately.
In simple words: Both attain the same temperature rise since kinetic energy per unit mass is the same.

Exam Tip: Focus on kinetic energy per unit mass and specific heat equality.

Question 18. What is the specific heat of a gas in an adiabatic process?
Answer: Zero, because no heat is exchanged with the surroundings (\(dQ = 0\)) during an adiabatic process.
In simple words: Zero, because no heat enters or leaves the system.

Question 1. What is the effect on the temperature of tea when it is stirred vigorously?
Answer: The temperature of the tea will rise slightly because mechanical work done in stirring is converted into heat energy within the liquid.
In simple words: Stirring does work on the tea, turning mechanical energy into heat and raising its temperature slightly.

Exam Tip: Relate mechanical work of stirring to frictional heat generation.

Question 2. Write two limitation of the first law of thermodynamics.
Answer: The two limitations are:
1. It does not indicate the direction of heat transfer (e.g., why heat cannot spontaneously flow from a colder body to a hotter body).
2. It gives no information about whether a proposed thermodynamic process is feasible or under what conditions conversion of heat into work is possible.
In simple words: It doesn't explain the direction of heat flow or whether a process can actually happen spontaneously.

Question 3. Write the expressions for \(C_v\) and \(C_p\) of a gas in terms of gas constant R and \(\gamma\) where \(\gamma = C_p / C_v\).
Answer: We know that \(C_p - C_v = R\) and \(\gamma = \frac{C_p}{C_v}\), so \(C_p = \gamma C_v\).
Substituting this: \(\gamma C_v - C_v = R\)
\(C_v(\gamma - 1) = R \implies C_v = \frac{R}{\gamma - 1}\)
Consequently, \(C_p = \frac{\gamma R}{\gamma - 1}\)
In simple words: Derive using Mayer's relation and the ratio of specific heats.

Exam Tip: Show the derivation clearly starting from \(C_p - C_v = R\).

Question 4. No real engine can have an efficiency greater that of a carnot engine working between the same two temperatures. Why?
Answer: A Carnot engine is an ideal engine operating on reversible cycles with zero frictional losses and no thermal dissipation. Real engines suffer from unavoidable friction, heat losses to surroundings, and non-quasi-static processes, which lower their efficiency.
In simple words: Real engines have friction and heat losses, making them less efficient than the ideal Carnot engine.

Question 5. Why water at the base of a waterfall is slightly warmer than at the top?
Answer: Potential energy possessed by water at the top is converted into kinetic energy during its fall, and upon impact at the base, this mechanical energy is converted into heat energy, raising the water temperature.
In simple words: Falling water converts potential energy into heat upon impact at the bottom.

Question 6. When ice melts, the change in internal energy is greater than the heat supplied. Why?
Answer: According to the first law, \(dQ = dU + dW\). When ice melts into water, its volume decreases (\(dW\) is negative). Therefore, \(dU = dQ - dW = dQ + |dW|\), making the change in internal energy greater than the heat supplied.
In simple words: Melting ice contracts, so work is done on the system, making internal energy increase by more than the heat added.

Question 7. Explain why two isothermal curves never intersect.
Answer: If two isothermal curves intersected, it would mean that at the point of intersection, the gas would have two different temperatures simultaneously at the same pressure and volume, which is physically impossible.
In simple words: Intersection would mean two different temperatures at the exact same pressure and volume, which is impossible.

Question 8. An ideal monatomic gas is taken round the cycle ABCDA as shown. Calculate the work done during the cycle.
Answer: Work done during the cycle is equal to the area enclosed by the PV loop ABCDA.
Area = \((\text{Length of AB}) \times (\text{Length of BC})\)
\(\text{Length of AB} = 2V - V = V\)
\(\text{Length of BC} = 2P - P = P\)
Total Work Done = \(P \times V = PV\)
In simple words: Calculate the area inside the rectangle on the PV diagram.

Question 9. Can a room be cooled by opening the door of refrigerator in a closed room?
Answer: No, the room will actually heat up. A refrigerator extracts heat from its inside and ejects more heat (interior heat plus electrical work converted to heat) into the room, raising the net temperature.
In simple words: No, the refrigerator releases more heat into the room than it removes, making the room warmer.

Question 10. Explain what is meant by isothermal and adiabatic operations.
Answer: - Isothermal process: A process in which pressure and volume change while the temperature remains constant (\(\Delta T = 0\)).
- Adiabatic process: A process in which pressure, volume, and temperature change, but there is no heat exchange between the system and surroundings (\(dQ = 0\)).
In simple words: Isothermal keeps temperature constant; adiabatic allows no heat exchange.

Question 11. Two bodies at different temperatures \(T_1\) and \(T_2\), if brought in thermal contact do not necessarily settle to the mean temperature \((T_1 + T_2)/2\). Explain?
Answer: The final equilibrium temperature depends on the heat capacities (mass \(\times\) specific heat) of both bodies. They settle to the mean temperature only if their thermal capacities are identical.
In simple words: Final temperature depends on heat capacities, not just a simple average.

Question 1. Obtain an expression for work done in an isothermal process.
Answer: Consider \(n\) moles of an ideal gas expanding isothermally from volume \(V_1\) to \(V_2\).
Work done \(W = \int_{V_1}^{V_2} P \, dV\)
Since \(PV = nRT\), \(P = \frac{nRT}{V}\)
\(W = \int_{V_1}^{V_2} \frac{nRT}{V} \, dV = nRT \ln\left(\frac{V_2}{V_1}\right)\)
In base-10 logarithm: \(W = 2.303 nRT \log_{10}\left(\frac{V_2}{V_1}\right)\)
In simple words: Integrate pressure with respect to volume using the ideal gas law at constant temperature.

Question 2. Identify and name the Thermodynamic processes 1, 2, 3 as shown in figure.
Answer: Based on the slope on the PV diagram:
1. Adiabatic process (steepest slope)
2. Isothermal process (intermediate slope)
3. Isobaric process (horizontal line, constant pressure)
In simple words: Process 1 is adiabatic, 2 is isothermal, and 3 is isobaric.

Question 3. Two samples of gas initially at the same temperature and pressure are compressed from volume V to V/2 one sample is compressed isothermally and the other adiabatically in which case the pressure will be higher? Explain?
Answer: The pressure will be higher in the adiabatic compression because temperature increases during adiabatic compression due to work done on the gas, whereas temperature remains constant in isothermal compression.
In simple words: Pressure is higher in adiabatic compression because the gas gets heated up during the process.

Question 4. Explain briefly the principle of a heat pump. What is meant by coefficient of performance?
Answer: Principle: A heat pump absorbs heat from a low-temperature cold reservoir using external mechanical work and delivers heat to a high-temperature warm reservoir.
Coefficient of Performance (\(\beta\)): It is defined as the ratio of heat extracted from the cold reservoir (\(Q_2\)) to the work done (\(W\)):
\(\beta = \frac{Q_2}{W} = \frac{Q_2}{Q_1 - Q_2}\)
In simple words: It pumps heat using work, and its performance is the heat delivered/extracted divided by work input.

Question 5. When you blow on the back of your hand with your mouth wide open your breath feels warm about if you partially close your mouth form an "O" and then blow on your hand breath fells cool. Why?
Answer: When the mouth is wide open, air escapes slowly and undergoes little expansion, remaining warm. When blowing through a small "O" shape, air expands rapidly (throttling/adiabatic expansion), causing a drop in temperature.
In simple words: Fast expansion through a small opening cools the air down, making it feel cool.

Question 6. Is it a violation of the second law of thermodynamics to convert (a) Work completely in to heat? (b) Heat completely in to work? Why or why not?
Answer: (a) Converting work completely into heat does not violate the second law (e.g., friction turns work into heat directly).
(b) Converting heat completely into work without any other change violates the second law (Kelvin-Planck statement), as a cyclic engine cannot convert 100% of heat into work.
In simple words: Turning work into heat is fine, but turning 100% heat into work in a cycle violates the second law.

Question 7. State first law of thermodynamics on its basis establish the relation between two molar specific heat for a gas.
Answer: First law: \(dQ = dU + dW\).
For 1 mole of gas heated at constant volume (\(dV = 0\)): \(dQ_v = dU = C_v dT\).
For heating at constant pressure: \(dQ_p = dU + P \, dV = C_p dT\).
Since \(dU = C_v dT\) and \(P \, dV = R \, dT\) for 1 mole:
\(C_p dT = C_v dT + R \, dT\)
Dividing by \(dT\): \(C_p - C_v = R\) (Mayer's relation).
In simple words: Use the first law at constant volume and pressure to prove \(C_p - C_v = R\).

Question 8. Explain briefly the working principle of a refrigerator and obtain an expression for its coefficient of performance.
Answer: Working Principle: A refrigerator uses a volatile liquid refrigerant that absorbs heat from the cold interior (evaporator) and rejects heat to the surroundings outside (condenser) using electrical work supplied to a compressor.
Expression for COP (\(\beta\)):
\(\beta = \frac{\text{Heat extracted from cold reservoir}}{\text{Work done}} = \frac{Q_2}{W} = \frac{Q_2}{Q_1 - Q_2}\)
In simple words: A refrigerator removes heat from inside using work. COP measures efficiency as heat removed per unit work.

Question 9. State zeroth law of thermodynamics. How does it lead to the concept of temperature?
Answer: Zeroth Law: If two systems A and B are separately in thermal equilibrium with a third system C, then they are also in thermal equilibrium with each other.
Concept of Temperature: This law implies the existence of a property common to systems in thermal equilibrium, which is called temperature. Two systems in equilibrium share the same temperature.
In simple words: If A and B are at the same temperature as C, they are at the same temperature as each other. This property is temperature.

Question 10. What is a cyclic process? Show that the net work done during a cyclic process is equal to the area enclosed by the path of the cycle on a PV diagram.
Answer: Cyclic Process: A thermodynamic process in which a system returns to its initial state after undergoing a series of changes.
Proof: Net work \(W = \oint P \, dV\). On a PV diagram, \(\int P \, dV\) represents the area under the expansion curve minus the area under the compression curve, which equals the net enclosed area of the loop.
In simple words: A cycle returns to start. The work done is the enclosed area on the PV graph.

Question 11. State second law of thermodynamics.
Answer: - Kelvin-Planck statement: It is impossible for any device operating on a cyclic process to receive heat from a single thermal reservoir and produce a net amount of work.
- Clausius statement: It is impossible for any device to transfer heat from a colder body to a warmer body without the input of external work.
In simple words: Heat cannot spontaneously flow from cold to hot, and 100% heat cannot be turned into work in a cycle.

Question 12. An ideal engine works between temperatures \(T_1\) and \(T_2\). It drives an ideal refrigerator that works between temperatures \(T_3\) and \(T_4\). Find the ratio \(Q_3 / Q_1\) in terms of \(T_1, T_2, T_3\) and \(T_4\).
Answer: For the ideal engine operating between \(T_1\) and \(T_2\), efficiency is:
\(\eta = \frac{W}{Q_1} = 1 - \frac{T_2}{T_1} \implies W = Q_1 \left(1 - \frac{T_2}{T_1}\right)\)
For the refrigerator operating between \(T_3\) and \(T_4\), the coefficient of performance is:
\(\beta = \frac{Q_3}{W} = \frac{T_3}{T_4 - T_3} \implies W = Q_3 \left(\frac{T_4 - T_3}{T_3}\right)\)
Equating work \(W\) supplied by the engine to the refrigerator:
\(Q_1 \left(\frac{T_1 - T_2}{T_1}\right) = Q_3 \left(\frac{T_4 - T_3}{T_3}\right)\)
Therefore, \(\frac{Q_3}{Q_1} = \left(\frac{T_1 - T_2}{T_1}\right) \left(\frac{T_3}{T_4 - T_3}\right)\)
In simple words: Equate the work produced by the engine to the work consumed by the refrigerator.

Question 1. Describe briefly carnot engine and obtain an expression for its efficiency.
Answer: A Carnot engine is an idealized reversible heat engine operating between two temperatures \(T_1\) (source) and \(T_2\) (sink) via a 4-step cycle: (1) Isothermal expansion, (2) Adiabatic expansion, (3) Isothermal compression, and (4) Adiabatic compression.
Efficiency \(\eta\) is given by:
\(\eta = \frac{\text{Work done}}{\text{Heat absorbed}} = \frac{W}{Q_1} = \frac{Q_1 - Q_2}{Q_1} = 1 - \frac{Q_2}{Q_1}\)
For temperatures \(T_1\) and \(T_2\):
\(\eta = 1 - \frac{T_2}{T_1}\)
In simple words: Derive Carnot efficiency using the four steps of the cycle and temperatures of source and sink.

Question 2. Define adiabatic process. Derive an expression for work done during adiabatic process.
Answer: Definition: An adiabatic process is one in which no heat enters or leaves the system (\(dQ = 0\)).
Derivation: For an adiabatic process, \(PV^\gamma = K\) (constant).
\(W = \int_{V_1}^{V_2} P \, dV = \int_{V_1}^{V_2} \frac{K}{V^\gamma} \, dV\)
\(W = K \left[ \frac{V^{-\gamma + 1}}{-\gamma + 1} \right]_{V_1}^{V_2} = \frac{K}{1 - \gamma} (V_2^{1-\gamma} - V_1^{1-\gamma})\)
Since \(K = P_1V_1^\gamma = P_2V_2^\gamma\):
\(W = \frac{P_2V_2 - P_1V_1}{1 - \gamma} = \frac{nR(T_1 - T_2)}{\gamma - 1}\)
In simple words: Integrate PV to the power gamma between initial and final states.

Question 3. Why a gas has two principal specific heat capacities? What is the significance of \(C_p - C_v\) and \(C_p / C_v\), where symbols have usual meaning.
Answer: A gas has two specific heats because when heated, it can expand significantly, doing external work requiring extra heat if pressure is kept constant, unlike solids and liquids.
- Significance of \(C_p - C_v = R\): It represents the external work done by 1 mole of gas when heated through 1\(^\circ\)C at constant pressure.
- Significance of \(\gamma = C_p / C_v\): It is the adiabatic index which determines the atomicity (monoatomic, diatomic, etc.) and elasticity of gases.
In simple words: Gases expand when heated, needing different amounts of heat at constant pressure vs constant volume.

Question 1. When a system is taken from state A to state B along the path ACB, 80 k cal of heat flows into the system and 30 k cal of work is done.
(a) How much heat flows into the system along path ADB if the work done is 10 k cal?
(b) When the system is returned from B to A along the curved path the work done is 20 k cal. Does the system absorb or librate heat?
(c) If \(U_A = 0\) and \(U_D = 40\) k cal, find the heat absorbed in the process.
Answer:
(a) Given for path ACB: \(dQ_{ACB} = 80\text{ kcal}\), \(dW_{ACB} = 30\text{ kcal}\).
Change in internal energy \(dU = dQ - dW = 80 - 30 = 50\text{ kcal}\).
Since internal energy is a state function, \(dU_{ADB} = dU_{ACB} = 50\text{ kcal}\).
For path ADB: \(dW_{ADB} = 10\text{ kcal}\).
\(dQ_{ADB} = dU_{ADB} + dW_{ADB} = 50 + 10 = 60\text{ kcal}\).
(b) For return path BA: \(dW_{BA} = -20\text{ kcal}\), \(dU_{BA} = -50\text{ kcal}\).
\(dQ_{BA} = dU_{BA} + dW_{BA} = -50 + (-20) = -70\text{ kcal}\).
The system liberates 70 kcal of heat.
(c) Given \(U_A = 0\), \(U_D = 40\text{ kcal}\).
\(dU_{AD} = U_D - U_A = 40 - 0 = 40\text{ kcal}\).
Work done along AD (\(dW_{AD} = 10\text{ kcal}\)):
\(dQ_{AD} = dU_{AD} + dW_{AD} = 40 + 10 = 50\text{ kcal}\).
In simple words: Use the principle that internal energy change is path-independent.

Question 2. \(\frac{1}{2}\) mole of helium is contained in a container at S.T.P. How much heat energy is needed to double the pressure of the gas, keeping the volume constant? Heat capacity of gas is \(3\text{ J g}^{-1}\text{ K}^{-1}\).
Answer: Number of moles \(n = \frac{1}{2}\).
For helium (monatomic), molar mass \(M = 4\text{ g/mol}\).
Molar heat capacity at constant volume \(C_v = 3 \times 4 = 12\text{ J mol}^{-1}\text{ K}^{-1}\).
Since volume is constant (\(\Delta V = 0\)), pressure is proportional to temperature (\(P \propto T\)).
To double pressure, temperature must be doubled (\(T_2 = 2T_1\)), so \(\Delta T = T_1 = 273\text{ K}\).
Heat energy needed: \(Q = n C_v \Delta T = \frac{1}{2} \times 12 \times 273 = 1638\text{ J}\).
In simple words: Calculate heat using \(Q = n C_v \Delta T\) where doubling pressure doubles the absolute temperature.

Question 3. The volume of steam produced by \(1\text{ g}\) of water at \(100^\circ\text{C}\) is \(1650\text{ cm}^3\). Calculate the change in internal energy during the change of state given \(J = 4.2 \times 10^7\text{ erg cal}^{-1}\text{ g} = 98\text{ J cm/s}^2\), latent heat of steam = \(540\text{ cal/g}\).
Answer: Mass of water \(m = 1\text{ g} = 10^{-3}\text{ kg}\).
Volume of \(1\text{ g}\) water = \(1\text{ cm}^3\).
Volume of steam \(V_2 = 1650\text{ cm}^3\).
Change in volume \(\Delta V = 1650 - 1 = 1649\text{ cm}^3 = 1649 \times 10^{-6}\text{ m}^3\).
Heat supplied \(Q = mL = 1 \times 540 = 540\text{ cal} = 540 \times 4.2 \times 10^7\text{ erg} = 22.68 \times 10^9\text{ erg}\).
Work done \(W = P \Delta V = (1.013 \times 10^6\text{ dynes/cm}^2) \times 1649\text{ cm}^3 = 1.67 \times 10^9\text{ erg}\).
Change in internal energy \(dU = Q - W = 22.68 \times 10^9 - 1.67 \times 10^9 = 21.01 \times 10^9\text{ erg}\).
In simple words: Subtract work done during vaporization from latent heat supplied to find internal energy change.

Question 4. What is the coefficient of performance \((\beta)\) of a carnot refrigerator working between \(30^\circ\text{C}\) and \(0^\circ\text{C}\)?
Answer: Given temperatures: \(T_2 = 0^\circ\text{C} = 273\text{ K}\), \(T_1 = 30^\circ\text{C} = 303\text{ K}\).
Coefficient of performance \(\beta = \frac{T_2}{T_1 - T_2}\)
\(\beta = \frac{273}{303 - 273} = \frac{273}{30} = 9.1\)
In simple words: Use COP formula with temperatures in Kelvin.

Question 5. Calculate the fall in temperature when a gas initially at \(72^\circ\text{C}\) is expanded suddenly to eight times its original volume. \((\gamma = 5/3)\).
Answer: Initial temperature \(T_1 = 72^\circ\text{C} = 72 + 273 = 345\text{ K}\).
Volume ratio \(V_2 = 8V_1\).
For adiabatic expansion: \(T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}\)
\(T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma-1} = 345 \times \left(\frac{1}{8}\right)^{\frac{5}{3} - 1} = 345 \times \left(\frac{1}{8}\right)^{2/3}\)
\(T_2 = 345 \times \frac{1}{4} = 86.25\text{ K}\)
Fall in temperature \(\Delta T = T_1 - T_2 = 345 - 86.25 = 258.75\text{ K}\).
In simple words: Use the adiabatic relation between temperature and volume to find final temperature, then subtract.

Question 6. A steam engine intake steam at \(200^\circ\text{C}\) and after doing work exhausts it directly in air at \(100^\circ\text{C}\) calculate the percentage of heat used for doing work. Assume the engine to be an ideal engine?
Answer: Intake temperature \(T_1 = 200^\circ\text{C} = 473\text{ K}\).
Exhaust temperature \(T_2 = 100^\circ\text{C} = 373\text{ K}\).
Efficiency \(\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{373}{473} = \frac{100}{473} \approx 0.211 = 21.1\%\)
In simple words: Calculate Carnot efficiency using the given Kelvin temperatures to find percentage of heat used.

Question 7. A perfect carnot engine utilizes an ideal gas the source temperature is 500K and sink temperature is 375K. If the engine takes 600k cal per cycle from the source, calculate
(i) The efficiency of engine
(ii) Work done per cycle
(iii) Heat rejected to sink per cycle.
Answer: Given \(T_1 = 500\text{ K}\), \(T_2 = 375\text{ K}\), \(Q_1 = 600\text{ kcal}\).
(i) Efficiency \(\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{375}{500} = 1 - 0.75 = 0.25 = 25\%\).
(ii) Work done per cycle \(W = \eta Q_1 = 0.25 \times 600\text{ kcal} = 150\text{ kcal}\).
(iii) Heat rejected to sink \(Q_2 = Q_1 - W = 600 - 150 = 450\text{ kcal}\).
In simple words: Use efficiency formula, then multiply by heat input to find work and heat rejected.

Question 8. Two carnot engines A and B are operated in series. The first one A receives heat at 900 K and reject to a reservoir at temperature T K. The second engine B receives the heat rejected by the first engine and in turn rejects to a heat reservoir at 400 K calculate the temperature T when
(i) The efficiencies of the two engines are equal
(ii) The work output of the two engines are equal
Answer:
(i) If efficiencies are equal: \(1 - \frac{T}{900} = 1 - \frac{400}{T} \implies \frac{T}{900} = \frac{400}{T} \implies T^2 = 360000 \implies T = 600\text{ K}\).
(ii) If work outputs are equal: \(W_A = W_B \implies Q_1(1 - T/900) = Q_2(1 - 400/T)\).
Solving with intermediate heat balance yields \(T = 650\text{ K}\).
In simple words: Equate either efficiencies or work outputs to solve for intermediate temperature T.

Question 9. Ten mole of hydrogen at NTP is compressed adiabatically so that its temperature become 400\(^\circ\)C How much work is done on the gas? what is the increase in the internal energy of the gas (\(R = 8.4\text{ J mol}^{-1}\text{ K}^{-1}\), \(\gamma = 1.4\)).
Answer: Initial temperature at NTP \(T_1 = 273\text{ K}\), final temperature \(T_2 = 400^\circ\text{C} = 673\text{ K}\).
Number of moles \(n = 10\).
Work done in adiabatic process: \(W = \frac{n R (T_1 - T_2)}{\gamma - 1} = \frac{10 \times 8.4 \times (273 - 673)}{1.4 - 1}\)
\(W = \frac{84 \times (-400)}{0.4} = -8.4 \times 10^4\text{ J}\)(Work done on the gas is \(+8.4 \times 10^4\text{ J}\)).
Increase in internal energy \(dU = -W = 8.4 \times 10^4\text{ J}\).
In simple words: Calculate work using temperature difference, noting that work done on the gas increases internal energy.

Question 10. The temperature \(T_1\) and \(T_2\) of the two heat reservoirs in an ideal carnot engine be \(1500^\circ\text{C}\) and \(500^\circ\text{C}\) respectively. Which of these increasing \(T_1\) by \(100^\circ\text{C}\) or decreasing \(T_2\) by \(100^\circ\text{C}\) would result in a greater improvement in the efficiency of the engine.
Answer: Initial temperatures: \(T_1 = 1500^\circ\text{C} = 1773\text{ K}\), \(T_2 = 500^\circ\text{C} = 773\text{ K}\).
Initial efficiency \(\eta_0 = 1 - \frac{773}{1773} = \frac{1000}{1773} = 56.4\%\).
Case 1: Increase \(T_1\) by \(100^\circ\text{C}\) (\(T_1' = 1873\text{ K}\)):
\(\eta_1 = 1 - \frac{773}{1873} = \frac{1100}{1873} = 58.7\%\) (Increase of \(2.3\%\)).
Case 2: Decrease \(T_2\) by \(100^\circ\text{C}\) (\(T_2' = 673\text{ K}\)):
\(\eta_2 = 1 - \frac{673}{1773} = \frac{1100}{1773} = 62.0\%\) (Increase of \(5.6\%\)).
Therefore, decreasing \(T_2\) results in a greater improvement in efficiency.
In simple words: Test both changes and compare which gives a higher boost to efficiency.