A refrigerator makes its inside colder while warming the room around it. That is a useful reminder that moving heat and converting heat into work are different jobs. We will keep an energy account and then ask which directions are possible.
This chapter builds engine and refrigerator accounts with explicit signs, develops reversible temperature limits, and then derives those limits from entropy. Examples include finite bodies and free expansion so that entropy is learned as a state quantity rather than a slogan.
What you will learn to do
- Calculate engine work, efficiency and power without confusing net heat with heat input.
- Explain the cyclic scope of the second-law direction statements.
- Distinguish reversible motion from merely slow motion.
- Apply Kelvin-temperature Carnot bounds.
- Choose the correct useful output for refrigerator and heat-pump coefficients.
- Calculate entropy through a reversible comparison path.
- Include system and surroundings when testing the second law.
- Derive thermal-device limits from energy and entropy balances.
Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Heat engines, refrigerators and entropy Learning goals: - Calculate engine work, efficiency and power without confusing net heat with heat input. - Explain the cyclic scope of the second-law direction statements. - Distinguish reversible motion from merely slow motion. - Apply Kelvin-temperature Carnot bounds. - Choose the correct useful output for refrigerator and heat-pump coefficients. - Calculate entropy through a reversible comparison path. - Include system and surroundings when testing the second law. - Derive thermal-device limits from energy and entropy balances. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
BEFORE YOU READ
Bring back one idea
Try a sentence or a sketch before opening the comparison. If this is new, use the explanation to get started.
If an engine returns to its initial state after a cycle, must its heat intake equal its work output?
Compare your reasoning
Not generally. Net heat equals net work over the cycle, but part of the heat intake is usually rejected to a colder reservoir. Internal-energy change over a complete cycle is zero.
Revisit the explanation →01 / Heat engines, refrigerators and entropy
A cyclic engine transfers some heat into work and rejects the rest
A heat engine repeatedly returns its working substance to its initial state while receiving heat from a hotter source, delivering work and rejecting heat to a colder sink. Because the working substance completes a cycle, its net internal-energy change is zero.
In this chapter Q_h and Q_c are positive magnitudes: heat absorbed from the hot reservoir and heat rejected to the cold reservoir. W_out is positive delivered work. The first law then gives W_out = Q_h − Q_c. These magnitude symbols differ from signed heat entries used for individual process legs.
Efficiency is useful work output divided by heat input: η = W_out/Q_h = 1 − Q_c/Q_h. The denominator is the absorbed heat, not the net heat. Dividing work by net heat would give one for every closed cycle and would not measure engine performance.
A large efficiency does not by itself imply a large power output. Power is work per time and also depends on cycle size and repetition rate. A slow idealized cycle can have a high efficiency while delivering little power.
Q_h and Q_c are positive heat-transfer magnitudes per cycle. Use consistent per-cycle quantities or consistent rates.
WORKED EXAMPLE
Keep input heat distinct from net heat
An engine absorbs 1500 J and rejects 1050 J during each cycle. It completes 4 cycles per second. Find its efficiency and mechanical power output.
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Heat engines, refrigerators and entropy
Focus: A cyclic engine transfers some heat into work and rejects the rest
Learning goals:
- Calculate engine work, efficiency and power without confusing net heat with heat input.
- Explain the cyclic scope of the second-law direction statements.
- Distinguish reversible motion from merely slow motion.
- Apply Kelvin-temperature Carnot bounds.
- Choose the correct useful output for refrigerator and heat-pump coefficients.
- Calculate entropy through a reversible comparison path.
- Include system and surroundings when testing the second law.
- Derive thermal-device limits from energy and entropy balances.
Equations in this section (LaTeX):
W_{\mathrm{out}}=Q_h-Q_c,\qquad\eta=\frac{W_{\mathrm{out}}}{Q_h}=1-\frac{Q_c}{Q_h}
Meaning: Q_h and Q_c are positive heat-transfer magnitudes per cycle. Use consistent per-cycle quantities or consistent rates.
Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.
Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.02 / Heat engines, refrigerators and entropy
Energy conservation alone does not determine which processes can occur
The first law would allow a hypothetical cyclic device to absorb heat from one reservoir and convert all of it into work while returning to its initial state. The second law rules out that sole-result process. A two-reservoir heat engine must reject some heat under ordinary finite-temperature operation.
Another equivalent direction statement is that heat cannot be transferred cyclically from a colder reservoir to a hotter one as the sole result, without a compensating work input or other change. Refrigerators can move heat that way precisely because work is supplied.
These statements refer to complete devices and their surroundings. An individual gas expansion can convert absorbed heat entirely into work while changing the gas state. That does not contradict the cyclic-engine statement because restoring the initial state must also be included.
A reservoir is idealized as large enough that its temperature remains effectively constant during the transfers considered. If a finite hot body cools substantially, it is not one constant-temperature reservoir throughout the process, and simple two-temperature bounds need careful application.
STOP & THINK
An ideal gas expands isothermally and reversibly, with Q = W. Does this establish a cyclic engine of 100% efficiency?
- Yes, the one expansion is a full cycle
- No; the gas has changed state and the return process must be included
- Yes, if the gas is monatomic
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Heat engines, refrigerators and entropy Focus: Energy conservation alone does not determine which processes can occur Learning goals: - Calculate engine work, efficiency and power without confusing net heat with heat input. - Explain the cyclic scope of the second-law direction statements. - Distinguish reversible motion from merely slow motion. - Apply Kelvin-temperature Carnot bounds. - Choose the correct useful output for refrigerator and heat-pump coefficients. - Calculate entropy through a reversible comparison path. - Include system and surroundings when testing the second law. - Derive thermal-device limits from energy and entropy balances. Equations in this section (LaTeX): Build the physical reasoning before choosing an equation. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
03 / Heat engines, refrigerators and entropy
A reversible process leaves no net change when exactly undone
A reversible idealization can be reversed so that both system and surroundings return to their initial states. This requires more than slow motion. Friction, turbulence, free expansion and heat transfer across a finite temperature difference generate irreversibility even when part of a process is slow.
For reversible heat transfer, the interacting temperatures differ only infinitesimally. For reversible piston motion, the mechanical driving imbalance is also infinitesimal and dissipative effects are absent. Such conditions define a limiting benchmark for real processes.
A Carnot cycle consists of two reversible isothermal processes at reservoir temperatures T_h and T_c, joined by two reversible adiabatic processes. In the engine direction it absorbs heat at T_h, expands adiabatically to T_c, rejects heat isothermally there, and returns adiabatically to its initial temperature.
The ideal efficiency depends only on the two reservoir temperatures, not on selecting a special working substance. Real losses reduce the achievable efficiency. The reversible benchmark is a thermodynamic bound, not a promise of a practical finite-power device.
Both reservoir temperatures must be absolute. The subscript C denotes the reversible Carnot benchmark between these two reservoirs.
Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Heat engines, refrigerators and entropy
Focus: A reversible process leaves no net change when exactly undone
Learning goals:
- Calculate engine work, efficiency and power without confusing net heat with heat input.
- Explain the cyclic scope of the second-law direction statements.
- Distinguish reversible motion from merely slow motion.
- Apply Kelvin-temperature Carnot bounds.
- Choose the correct useful output for refrigerator and heat-pump coefficients.
- Calculate entropy through a reversible comparison path.
- Include system and surroundings when testing the second law.
- Derive thermal-device limits from energy and entropy balances.
Equations in this section (LaTeX):
\eta_{\mathrm C}=1-\frac{T_c}{T_h},\qquad 0<T_c<T_h
Meaning: Both reservoir temperatures must be absolute. The subscript C denotes the reversible Carnot benchmark between these two reservoirs.
Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.
Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.04 / Heat engines, refrigerators and entropy
Check a proposed engine against both energy and temperature limits
For any cyclic engine exchanging heat only with two fixed-temperature reservoirs, efficiency cannot exceed 1 − T_c/T_h. Equality requires reversible operation of the complete heat-transfer cycle. Passing the energy balance is necessary but not sufficient.
The same rejected-heat fraction that appears in efficiency has a lower bound: Q_c/Q_h ≥ T_c/T_h. Raising the hot-reservoir temperature or lowering the cold-reservoir temperature can increase the ideal efficiency, but these are changes in absolute temperatures.
A Celsius ratio can give a nonsensical bound because its zero is not an absolute thermal reference. Convert each reservoir temperature to kelvins before taking a ratio. Temperature differences may use Celsius intervals, but the individual values in the ratio may not.
The entropy balance later in this chapter will derive the bound directly. It also provides a way to diagnose a proposed device that conserves energy yet would require a negative total entropy change.
WORKED EXAMPLE
Reject an energy-consistent but thermodynamically impossible claim
A claimed cyclic engine works only between reservoirs at 600 K and 300 K. It absorbs 1200 J per cycle and is said to deliver 720 J of work. Check the claim.
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STOP & THINK
What is the maximum efficiency, as a decimal fraction, of a two-reservoir engine operating between 750 K and 300 K?
MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →
Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Heat engines, refrigerators and entropy Focus: Check a proposed engine against both energy and temperature limits Learning goals: - Calculate engine work, efficiency and power without confusing net heat with heat input. - Explain the cyclic scope of the second-law direction statements. - Distinguish reversible motion from merely slow motion. - Apply Kelvin-temperature Carnot bounds. - Choose the correct useful output for refrigerator and heat-pump coefficients. - Calculate entropy through a reversible comparison path. - Include system and surroundings when testing the second law. - Derive thermal-device limits from energy and entropy balances. Equations in this section (LaTeX): Build the physical reasoning before choosing an equation. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
05 / Heat engines, refrigerators and entropy
A refrigerator uses work to move heat against the temperature gradient
For a cyclic refrigerator, let Q_c be heat removed from the cold reservoir, Q_h heat delivered to the hot reservoir and W_in work supplied. All are positive magnitudes. Energy conservation gives Q_h = Q_c + W_in.
The useful effect of refrigeration is Q_c, so the coefficient of performance is COP_R = Q_c/W_in. It can exceed one because the device moves heat rather than converting all its work input into newly created cooling energy. There is no violation of energy conservation.
For a reversible refrigerator between constant temperatures, COP_R,C = T_c/(T_h − T_c). A smaller temperature lift allows a larger ideal coefficient. A real device has a lower coefficient because irreversibilities require more work for the same cooling duty.
Use rates consistently for continuously running equipment: cooling power divided by electrical or mechanical input power, with the device boundary specified. A per-cycle energy divided by a power is a time, not a coefficient of performance.
The Carnot expression is an upper bound for a refrigerator exchanging heat only with the two fixed-temperature reservoirs.
WORKED EXAMPLE
Compute rejected heat as well as cooling duty
A refrigerator removes 600 J from a cold compartment while receiving 200 J of work per cycle. Find its coefficient of performance and the heat delivered to the room.
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Heat engines, refrigerators and entropy
Focus: A refrigerator uses work to move heat against the temperature gradient
Learning goals:
- Calculate engine work, efficiency and power without confusing net heat with heat input.
- Explain the cyclic scope of the second-law direction statements.
- Distinguish reversible motion from merely slow motion.
- Apply Kelvin-temperature Carnot bounds.
- Choose the correct useful output for refrigerator and heat-pump coefficients.
- Calculate entropy through a reversible comparison path.
- Include system and surroundings when testing the second law.
- Derive thermal-device limits from energy and entropy balances.
Equations in this section (LaTeX):
Q_h=Q_c+W_{\mathrm{in}},\qquad\mathrm{COP}_{R}=\frac{Q_c}{W_{\mathrm{in}}},\qquad\mathrm{COP}_{R,C}=\frac{T_c}{T_h-T_c}
Meaning: The Carnot expression is an upper bound for a refrigerator exchanging heat only with the two fixed-temperature reservoirs.
Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.
Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.06 / Heat engines, refrigerators and entropy
The same transfer device has a different useful output when used for heating
A heat pump uses the same basic energy transfers as a refrigerator, but the desired effect is heat delivered to the warmer space. Its heating coefficient is COP_HP = Q_h/W_in. Since Q_h = Q_c + W_in, COP_HP = COP_R + 1 for the same operating cycle.
The reversible heating coefficient is T_h/(T_h − T_c). It can exceed one because some of the delivered heat was extracted from the colder surroundings. A resistance heater, by contrast, converts its input work into heat at roughly a one-to-one energy ratio at its local boundary.
The refrigerator and heat-pump coefficients are not competing efficiencies for one task. Their numerators describe different useful effects. Before choosing a formula, decide which reservoir is being cooled or heated and what the question counts as the input cost.
Opening a refrigerator door inside an otherwise closed room does not cool the room overall. The cooling and rejection occur within that room, while external work enters. The room gains the work input in the overall energy balance, apart from other environmental transfers.
WORKED EXAMPLE
Use a temperature lift to find a reversible benchmark
A reversible refrigerator operates between 270 K and 300 K and receives 120 J of work per cycle. Find its maximum cooling amount and the corresponding heat-pump coefficient.
MEMBERSThe approach, step-by-step working and check are for members.Unlock →
Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Heat engines, refrigerators and entropy Focus: The same transfer device has a different useful output when used for heating Learning goals: - Calculate engine work, efficiency and power without confusing net heat with heat input. - Explain the cyclic scope of the second-law direction statements. - Distinguish reversible motion from merely slow motion. - Apply Kelvin-temperature Carnot bounds. - Choose the correct useful output for refrigerator and heat-pump coefficients. - Calculate entropy through a reversible comparison path. - Include system and surroundings when testing the second law. - Derive thermal-device limits from energy and entropy balances. Equations in this section (LaTeX): Build the physical reasoning before choosing an equation. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
07 / Heat engines, refrigerators and entropy
Entropy is a state function calculated through a reversible path
Entropy S is a thermodynamic state property with units J/K. For a reversible infinitesimal heat transfer, dS = δQ_rev/T. Between two equilibrium states, integrate along any suitable reversible path connecting them. The result depends on the endpoints, even if the actual process was irreversible.
The subscript rev is essential. Substituting the actual heat of an irreversible process into Q/T is not generally valid. An insulated free expansion has actual Q = 0 but can still increase entropy. The hypothetical reversible path used for calculation need not have the same heat or work as the actual path.
For a reservoir at fixed absolute temperature T, losing heat magnitude Q changes its entropy by −Q/T, and gaining it changes entropy by +Q/T. Its constant temperature makes this particularly simple, even when the transfer to another body is irreversible.
Entropy is not merely a synonym for visible disorder. For calculations, use its thermodynamic definition and the system boundary. A verbal impression of order cannot replace the heat, temperature and state information needed to determine ΔS.
The integral is taken along a reversible path between the states. It does not assert that the actual process was reversible.
STOP & THINK
An insulated ideal gas freely expands and receives no heat. Must its entropy change be zero?
- Yes, because actual Q is zero
- No; entropy can increase through irreversibility
- Yes, because its final temperature is unchanged
MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →
Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Heat engines, refrigerators and entropy
Focus: Entropy is a state function calculated through a reversible path
Learning goals:
- Calculate engine work, efficiency and power without confusing net heat with heat input.
- Explain the cyclic scope of the second-law direction statements.
- Distinguish reversible motion from merely slow motion.
- Apply Kelvin-temperature Carnot bounds.
- Choose the correct useful output for refrigerator and heat-pump coefficients.
- Calculate entropy through a reversible comparison path.
- Include system and surroundings when testing the second law.
- Derive thermal-device limits from energy and entropy balances.
Equations in this section (LaTeX):
\Delta S=\int_i^f\frac{\delta Q_{\mathrm{rev}}}{T}
Meaning: The integral is taken along a reversible path between the states. It does not assert that the actual process was reversible.
Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.
Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.08 / Heat engines, refrigerators and entropy
The second law constrains the combined system and surroundings
The entropy of a particular system may decrease when it transfers heat out. The second law does not forbid that local decrease. It requires the total entropy change of an isolated combined system, including the relevant surroundings, to be nonnegative.
For a reversible process, total entropy generation is zero. For an irreversible process it is positive. Energy remains conserved in both cases; entropy generation measures a different restriction on how energy transfers and state changes can occur.
Heat flowing from a hotter reservoir to a colder one illustrates the sign. If an amount Q leaves T_h and enters T_c, the combined change is Q(1/T_c − 1/T_h), positive when T_h > T_c. Reversing that transfer as the sole result would make the total negative.
Do not add entropy changes using Celsius temperatures in denominators. The reciprocal-temperature dependence needs kelvins. Also ensure that every relevant body is included: omitting a reservoir can make a permissible system entropy decrease look like a violation.
Equality describes a reversible ideal process for the complete isolated combination; strict positivity indicates irreversibility.
WORKED EXAMPLE
Check the direction of a heat transfer with entropy
A heat amount of 600 J passes from a reservoir at 500 K to one at 300 K. No other changes occur. Find the entropy change of each reservoir and their total.
MEMBERSThe approach, step-by-step working and check are for members.Unlock →
Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Heat engines, refrigerators and entropy
Focus: The second law constrains the combined system and surroundings
Learning goals:
- Calculate engine work, efficiency and power without confusing net heat with heat input.
- Explain the cyclic scope of the second-law direction statements.
- Distinguish reversible motion from merely slow motion.
- Apply Kelvin-temperature Carnot bounds.
- Choose the correct useful output for refrigerator and heat-pump coefficients.
- Calculate entropy through a reversible comparison path.
- Include system and surroundings when testing the second law.
- Derive thermal-device limits from energy and entropy balances.
Equations in this section (LaTeX):
\Delta S_{\mathrm{system}}+\Delta S_{\mathrm{surroundings}}\ge0
Meaning: Equality describes a reversible ideal process for the complete isolated combination; strict positivity indicates irreversibility.
Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.
Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.09 / Heat engines, refrigerators and entropy
Use a convenient reversible comparison path for gas states
For a fixed amount of ideal gas with constant C_V, a reversible heat increment can be written δQ_rev = nC_V dT + p dV. Dividing by T and using p/T = nR/V gives dS = nC_V dT/T + nR dV/V.
Integration produces ΔS = nC_V ln(T_f/T_i) + nR ln(V_f/V_i). This is a state-change relation, so it can be applied to the equilibrium endpoints of an irreversible process as well. The actual heat and work need not equal those along the reversible path used to derive it.
For an isothermal ideal-gas expansion, the temperature term vanishes and entropy increases by nR ln(V_f/V_i). The same endpoint change occurs in ideal free expansion into vacuum, even though actual heat input is zero there.
For a reversible adiabatic ideal-gas process, the two logarithmic terms cancel and ΔS = 0. An adiabatic irreversible process need not have zero entropy change. Thus adiabatic and isentropic are not interchangeable without additional conditions.
Fixed amount and composition of ideal gas with constant C_V. The endpoint state relation does not require the actual path to be reversible.
WORKED EXAMPLE
Calculate entropy for a process with zero actual heat
One mole of ideal gas freely expands in an insulated rigid outer vessel from V to 2V, ending at the same temperature. Use R = 8.31 J/(mol·K). Find the gas entropy change.
MEMBERSThe approach, step-by-step working and check are for members.Unlock →
Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Heat engines, refrigerators and entropy
Focus: Use a convenient reversible comparison path for gas states
Learning goals:
- Calculate engine work, efficiency and power without confusing net heat with heat input.
- Explain the cyclic scope of the second-law direction statements.
- Distinguish reversible motion from merely slow motion.
- Apply Kelvin-temperature Carnot bounds.
- Choose the correct useful output for refrigerator and heat-pump coefficients.
- Calculate entropy through a reversible comparison path.
- Include system and surroundings when testing the second law.
- Derive thermal-device limits from energy and entropy balances.
Equations in this section (LaTeX):
\Delta S=nC_V\ln\!\left(\frac{T_f}{T_i}\right)+nR\ln\!\left(\frac{V_f}{V_i}\right)
Meaning: Fixed amount and composition of ideal gas with constant C_V. The endpoint state relation does not require the actual path to be reversible.
Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.
Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.10 / Heat engines, refrigerators and entropy
A changing-temperature body requires an integral
A finite body warming or cooling through a temperature range is not a constant-temperature reservoir. With constant heat capacity C and an appropriate negligible-work model, its entropy change is C ln(T_f/T_i), obtained by integrating C dT/T along a reversible comparison path.
If two finite bodies equilibrate in an insulated enclosure, first determine their final temperature from energy conservation. Then calculate each entropy change using absolute temperatures and add them. One entropy change can be negative while the total remains positive.
The energy balance uses a heat-capacity-weighted arithmetic mean for the final temperature. The entropy expression uses logarithms. These different mathematical forms encode why heat exchange across a finite temperature difference generates entropy even though no energy leaves the combined system.
A phase change at an equilibrium transition temperature can contribute mL/T_transition to entropy in the reversible comparison. Any warming or cooling before or after that transition needs its own integral. As in calorimetry, do not compress several physically different stages into one inappropriate formula.
WORKED EXAMPLE
Check an insulated equilibration of two equal-capacity bodies
Two bodies, each with constant heat capacity 200 J/K, start at 400 K and 300 K. They exchange heat in an insulated enclosure with negligible work until reaching a common temperature. Find the final temperature and total entropy change.
MEMBERSThe approach, step-by-step working and check are for members.Unlock →
Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Heat engines, refrigerators and entropy Focus: A changing-temperature body requires an integral Learning goals: - Calculate engine work, efficiency and power without confusing net heat with heat input. - Explain the cyclic scope of the second-law direction statements. - Distinguish reversible motion from merely slow motion. - Apply Kelvin-temperature Carnot bounds. - Choose the correct useful output for refrigerator and heat-pump coefficients. - Calculate entropy through a reversible comparison path. - Include system and surroundings when testing the second law. - Derive thermal-device limits from energy and entropy balances. Equations in this section (LaTeX): Build the physical reasoning before choosing an equation. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
11 / Heat engines, refrigerators and entropy
Derive the engine and refrigerator limits from one entropy balance
For a cyclic engine between two reservoirs, the working substance returns to its original entropy. The reservoir changes are −Q_h/T_h and +Q_c/T_c. The second law therefore requires Q_c/T_c − Q_h/T_h ≥ 0.
Rearranging gives Q_c/Q_h ≥ T_c/T_h, and inserting that into η = 1 − Q_c/Q_h gives η ≤ 1 − T_c/T_h. Equality requires zero total entropy generation. This derivation shows why a device can satisfy its energy balance yet violate the efficiency bound.
For a refrigerator, the reservoir changes reverse: the cold reservoir loses Q_c/T_c and the hot gains Q_h/T_h. Requiring Q_h/T_h − Q_c/T_c ≥ 0, together with W_in = Q_h − Q_c, gives COP_R ≤ T_c/(T_h − T_c).
Use both balances as independent checks. First verify energy conservation. Then compute the entropy change of the reservoirs and any other participating bodies. A negative total means some claimed transfer, omitted interaction or model assumption is inconsistent with the second law.
WORKED EXAMPLE
Quantify irreversibility in a permissible engine
An engine between 600 K and 300 K absorbs 1200 J and delivers 400 J of work per cycle. Find rejected heat, efficiency and reservoir entropy change.
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Heat engines, refrigerators and entropy Focus: Derive the engine and refrigerator limits from one entropy balance Learning goals: - Calculate engine work, efficiency and power without confusing net heat with heat input. - Explain the cyclic scope of the second-law direction statements. - Distinguish reversible motion from merely slow motion. - Apply Kelvin-temperature Carnot bounds. - Choose the correct useful output for refrigerator and heat-pump coefficients. - Calculate entropy through a reversible comparison path. - Include system and surroundings when testing the second law. - Derive thermal-device limits from energy and entropy balances. Equations in this section (LaTeX): Build the physical reasoning before choosing an equation. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
BRING THE IDEAS TOGETHER
What should stay with you
- A cyclic working substance has no net stored-energy or entropy change.
- Engine efficiency uses absorbed heat as its denominator.
- Energy conservation alone does not establish thermodynamic feasibility.
- Reversible two-reservoir bounds depend on Kelvin temperatures.
- A refrigerator moves heat and can have COP greater than one.
- For the same cycle, heat-pump COP is refrigerator COP plus one.
- Entropy is a state function; Q_actual/T is not a universal entropy-change formula.
- An individual body can lose entropy while the isolated total increases.
- Adiabatic irreversible processes can generate entropy.
RETRIEVE, THEN EXPLAIN
Try it without the examples
Work these out before checking. If an answer surprises you, identify which assumption or step needs another look. A correct choice is most useful when you can explain why the alternatives fail.
SELF CHECK 1
A cyclic engine absorbs 800 J and rejects 560 J. What work does it deliver?
Answer in J.
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SELF CHECK 2
A refrigerator removes 450 J using 150 J of input work per cycle. What is its COP?
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SELF CHECK 3
A reservoir at 400 K absorbs 200 J. What is its entropy change?
Answer in J/K.
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SELF CHECK 4
A hot body loses entropy as it cools. Is that alone a second-law violation?
- Yes, no body may ever lose entropy
- No; include the entropy change of the surroundings
- Only if its temperature is in kelvins
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SELF CHECK 5
Why may a heat pump deliver more heat than its work input?
- It creates energy
- It also extracts heat from a colder region
- The first law does not apply to cycles
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SELF CHECK 6
Which condition is sufficient to call a process reversible?
- It is slow, regardless of friction
- It can be undone with no net change in system and surroundings
- It conserves energy
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APPLY THE CHAPTER
Take the ideas into a problem
Try a complete problem next. Commit to a method before revealing a hint, then compare your reasoning with the worked solution.
An engine efficiency needs the positive heat legsAdvanced · Hints and guided reasoningAn energy-balanced engine can still be impossibleAdvanced · Hints and guided reasoningAccount for the work as well as coolingIntermediate · Hints and guided reasoningBalance energy, then compare entropyIntermediate · Hints and guided reasoningHow much heat must an ideal engine absorb?Intermediate · Hints and guided reasoningCOME BACK TO THE IDEA
What can you explain later?
Try these from memory at your next study session, then again a few days later. Say your reasoning aloud or sketch it before looking. Adjust the gap to what you can explain.
Next study session
Why can a refrigerator’s coefficient of performance exceed one without creating energy?
Compare your reasoning
It measures heat removed from the cold space divided by work supplied. The work helps move existing thermal energy; it is not the sole source of the transferred heat.
Revisit the explanation →A few days later: change the situation
Two bodies at different temperatures exchange heat in an isolated enclosure. Can their total energy stay fixed while total entropy increases?
Compare your reasoning
Yes. Energy is redistributed between them, while spontaneous transfer across a finite temperature difference produces entropy. Conservation of energy does not require conservation of entropy.
Revisit the explanation →Compare the physical reason, the assumptions and the conclusion. Your wording can differ. If one part is missing, revisit that section and try explaining it again after a break.
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