The second law of thermodynamics
State the second law precisely using Kelvin-Planck and Clausius forms, and show why the two forms are equivalent.
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ContinueTwo proposed devices can conserve energy and still be impossible. An engine that converts heat from one reservoir entirely into work, or a device whose sole effect is moving heat from cold to hot, violates the second law. The decisive words are cycle and sole effect: inspect what changes inside and outside the device before judging a claim.
Build on cycle energy accounting. A complete cycle restores the working substance’s state, excluding lasting internal storage as an explanation for the output. Use positive heat and work magnitudes when following transfer arrows, and kelvin for entropy or temperature-ratio calculations.
State the forbidden effects precisely
The Kelvin–Planck statement forbids a cyclic device whose sole effect is withdrawing heat from a single thermal reservoir and converting it entirely into work. For the specified two-reservoir heat engine, positive output therefore requires heat rejection; Wₒᵤₜ = Qₕ with Qₕ > 0 is forbidden when there is no other effect.
The Clausius statement forbids a cyclic device whose sole effect is transferring heat from a colder body to a hotter body. An ordinary two-reservoir refrigerator can do that with work supplied from outside, which is an additional effect. A heat-driven refrigerator can instead involve a third heat source; its heat input and other reservoir changes must be included before applying the statement.
Neither statement forbids every instance of cold-to-hot heat transfer or every one-off complete conversion of absorbed heat into work. A noncyclic ideal-gas isothermal expansion, for example, can have Δ U = 0 and Q = Wₒᵤₜ while changing volume. That change of state is another effect. Returning the gas to its starting state requires accounting for the return process too. The MIT statements of the second law make the sole-effect condition explicit.
Short worked examples
Testing a claimed perfect engine
Worked example 1
Perfect engine claim
Problem
A cyclic device claims to take 500 J from one reservoir and deliver 500 J of work, with no other effect. Test the claim against the first and second laws.
Show full solution
Separate energy balance from possibility
Method
Check the cycle balance and identify the device’s sole effect.
Reason
Conserving energy is necessary, but Kelvin–Planck also restricts complete conversion of single-reservoir heat.
Working
Answer
Δ U = 500-500 = 0 J satisfies the first law. The sole effect is to turn heat from one reservoir entirely into work.
Judge the claim
Working
Answer
The claim violates Kelvin–Planck and is impossible despite satisfying energy conservation. Its efficiency would be 500/500 = 1.
Testing a claimed work-free refrigerator
Worked example 2
Work-free refrigerator claim
Problem
Show full solution
Identify the direction and other effects
Method
Compare the two temperatures and check for supplied work.Reason
Clausius forbids cold-to-hot transfer as the sole effect of a cycle, rather than forbidding all cold-to-hot transfer.Working
Answer
Heat moves from 5°C to 25°C, and Wᵢₙ = 0. The cycle stores no net energy.Judge the claim
Working
Answer
The claim violates Clausius. A refrigerator can move this heat uphill only with another effect, such as work input.
Show how a Clausius violation would break Kelvin–Planck
Suppose a hypothetical work-free device transfers heat Qc from a cold reservoir to a hot reservoir. Pair it with an ordinary engine rejecting that same Qc to the cold reservoir. The cold-side transfers cancel. The engine supplies W = Qₕ-Qc, while the hot reservoir loses that same net heat Qₕ-Qc. The combined cyclic device would convert heat from one reservoir entirely into work: a Kelvin–Planck violation.
The matching heat transfers are essential. Merely naming an engine and a refrigerator does not establish the net effect; cancel their actual quantities across a combined boundary.
Guided practice 3
Expose the sole effect after coupling two devices
Problem
An ordinary cyclic engine absorbs 1000 J from a hot reservoir, rejects 600 J to a cold reservoir and delivers 400 J of work. A hypothetical Clausius-violating cyclic device transfers exactly 600 J from that cold reservoir back to the hot reservoir with zero work and no other effect. Consider the devices together. Find net heat gained by the cold reservoir, net heat withdrawn from the hot reservoir and exported work. Which second-law statement would the combined effect violate?
Work through the model, then compare your reasoning
Hints
Hint 1
Check the model
Compare changes in the reservoirs after both devices complete their cycles; the hypothetical device is not a physically allowed refrigerator.Show solution step by step
Cancel the cold-side transfers
Method
Add the two effects on each reservoir.Reason
Only net transfers across the combined boundary determine its sole effect.Working
Answer
The cold reservoir gains 600-600 = 0 J. The hot reservoir loses 1000-600 = 400 J. No lasting change remains inside either cyclic device.
Identify the resulting forbidden device
Method
Compare net reservoir heat withdrawal with exported work.Reason
The first law alone permits the energy balance; Kelvin–Planck forbids the sole effect.Working
Answer
Exported work is 400 J, exactly the heat withdrawn from the single remaining affected reservoir. Thus a Clausius violation would construct a Kelvin–Planck violation. The calculation is a contradiction argument, not a design for a work-free pump.
Prove the converse by matching work
Now suppose a hypothetical Kelvin–Planck-violating engine withdraws 600 J from one hot reservoir and supplies 600 J of work with no other effect. Use all that work to drive an ordinary refrigerator extracting 300 J from the cold reservoir and delivering 900 J to the same hot reservoir.
The combined external work is 600-600 = 0. The hot reservoir gains 900-600 = 300 J; the cold reservoir loses 300 J. Both devices complete cycles, leaving the sole effect of moving heat from cold to hot. That violates Clausius. Each forbidden device would therefore enable the other; the statements are equivalent.
Use entropy for a quantitative possibility test
For a cyclic device exchanging signed heat Qⱼ into the device from fixed-temperature reservoirs Tⱼ, the reservoirs’ total entropy change is -∑ⱼ Qⱼ/Tⱼ. The device’s cycle entropy change is zero, and the total must be nonnegative. Work transfers energy without carrying entropy in this account.
This condition can reject a heat-engine efficiency below one if it exceeds the reservoir-dependent Carnot limit. A consistent energy and entropy balance is necessary; it does not by itself provide a working mechanism or establish that measurements included every channel. In an isolated system, total entropy cannot decrease; a subsystem can lose entropy by exporting it.
Check a refrigeration claim with a third reservoir
Check your understanding 1
A cyclic refrigeration installation receives no external work. It absorbs 1000 J from a reservoir at 800 K and 250 J from a cold space at 250 K, then rejects 1250 J to a room at 300 K. These are its only transfers. Does the cold-to-hot transfer alone establish a Clausius violation? Check energy and total reservoir entropy, then state what those checks can and cannot establish.
Show answer
Energy balance closes: 1000 + 250-1250 = 0 J. The cold-to-hot transfer is not the sole effect, because the 800 K reservoir also supplies heat. Total reservoir entropy change is -1000/800-250/250 + 1250/300 = 1.9166… J/K > 0. The supplied account passes these necessary first- and second-law tests. It does not prove that a particular installation can produce the transfers; a mechanism, operating conditions and complete measurements would still be needed. Requiring work input for every possible refrigerator would omit the stated third-reservoir input.
Test the specified components as well as the external account
Apply the same cycle entropy test to each component with its own listed transfers. A proposal can have a balanced external account while requiring an impossible internal device. Another component’s positive entropy production does not remove the need for this test.
Check your understanding 2
A proposed installation has an engine between fixed reservoirs at 600 K and 300 K. In the stated interval, the engine absorbs 1200 J from the hot reservoir, rejects 800 J to the cold one and supplies 400 J of work. It sends 100 J of that work to a pump and exports the remaining 300 J. The pump claims to extract 500 J from the same cold reservoir and deliver 600 J to the same hot one. Both specifications claim complete cycles with no other transfers or effects. Calculate the reservoir entropy change associated with the engine alone, the pump alone and the combined account. Do the balanced combined energy and entropy accounts prove these specified components possible?
Show answer
The first-law balances both close: 1200-800-400 = 0 for the engine and 500 + 100-600 = 0 for the pump. The implied reservoir entropy sums are Δ S_(res,engine) = -1200/600 + 800/300 = +2/3 J/K, Δ S_(res,pump) = -500/300 + 600/600 = -2/3 J/K. Their aggregate is zero. Externally, the hot reservoir loses 1200-600 = 600 J, the cold one gains 800-500 = 300 J, and exported work is 400-100 = 300 J. That external account balances energy and has zero total reservoir entropy change.
Nevertheless, the specified pump is impossible: its state returns while its reservoirs lose net entropy, with no other effect to account for the decrease. Its heating COP would be 600/100 = 6, exceeding the reversible limit 600/(600-300) = 2 for those reservoirs. The engine’s positive entropy contribution cannot rescue that component.
The combined external figures could describe a different reversible engine, but they do not establish these proposed internal components. Check that your explanation applies the cycle and transfer conditions to the pump as well as the pair, rather than declaring every component valid from a total alone.
Practice set
Check your understanding 3
A cyclic device takes heat from only one reservoir and delivers net work, with no other effect. Which second-law statement does it violate?
Show hint
Determine what supplies the work when the device returns to its initial state.
Show answer
Kelvin–Planck: in a cycle the net energy stored is zero, so all the single-reservoir heat would become work as the sole effect.
Check your understanding 4
A cyclic refrigerator’s sole effect is to move heat from cold to hot without work input. Which statement does it violate?
Show hint
Identify the forbidden heat-transfer direction when there is no other effect.
Show answer
Clausius. Ordinary refrigerators avoid this violation by consuming work, so heat transfer is not their sole effect.
Check your understanding 5
Why does the phrase “operates in a cycle” matter in the second-law device statements?
Show hint
Could a device otherwise consume its initially stored energy or change state?
Show answer
A cycle restores the working substance’s state, excluding a lasting change inside the device as an additional effect. A one-off process may use stored energy or change state, so it is not the forbidden cyclic device.
Check your understanding 6
An engine has efficiency e = 0.62. Does this value alone establish a second-law violation?
Show hint
An efficiency below one must still be compared with the reservoir-dependent limit.
Show answer
No. It avoids the forbidden complete conversion, but may still exceed 1-Tc/Tₕ for its reservoirs. Their absolute temperatures are needed to test the Carnot bound.
What to keep
Keep the complete-cycle and sole-effect conditions. Prove equivalence by cancelling matched heat or work transfers across a combined boundary. Additional reservoirs or noncyclic state changes need their own account; energy conservation alone does not establish possibility.
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Syllabus and review details
No official syllabus alignment is listed for this lesson.