NEB Class 12 • Physics • Thermodynamics
Second Law of Thermodynamics: NEB Class 12 Physics Guide
Add direction and feasibility to energy conservation: real processes are irreversible, engines reject heat and refrigerators require work.
Kelvin and Clausius statements · Heat engines, refrigerators and COP · Carnot limits, reversibility and entropy
Quick answer: The second law limits process direction and energy quality. No cyclic engine converts all heat from one reservoir into work. Heat does not flow spontaneously from cold to hot. A reversible Carnot engine sets an ideal upper efficiency η=1−Tc/Th using kelvin temperatures.
QuickScopeDirectionEnginesRefrigeratorCarnotEntropyMethodFAQ
Curriculum scope
Connect the second law to the first
The CDC Grade 12 page and curriculum control scope. The unit covers directional statements, heat engines/refrigerators, efficiency/COP, Carnot ideas, reversibility and entropy at the assigned depth.
Prerequisite: First Law guide. The first law balances energy; it would not forbid a hypothetical cyclic device converting one-reservoir heat wholly to work. The second law supplies the missing feasibility restriction.
Direction and irreversibility
Natural processes have a preferred direction
Heat flows spontaneously from hotter to colder bodies; reverse transfer needs work or another compensating effect. Friction converts organised mechanical energy into dispersed internal energy. Free expansion and mixing occur spontaneously but do not reverse by themselves.
Kelvin–Planck: impossible for a cyclic device to take heat from a single reservoir and convert it entirely into work without another effect. Clausius: impossible for a cyclic device’s sole effect to transfer heat from cold to hot. These are equivalent expressions of the second law, discussed by OpenStax.
Heat engines
An engine must reject some heat
A cyclic engine absorbs Qh from hot reservoir, produces work W and rejects Qc to cold reservoir. First law over cycle gives W=Qh−Qc. Efficiency η=W/Qh=1−Qc/Qh. Since Qc cannot be zero for a cyclic engine between finite reservoirs, η<1.
Engine example
An engine absorbs 1200 J and rejects 720 J. W=480 J and η=480/1200=0.40 or 40%. Energy balances; the second law asks whether claimed performance is feasible for reservoir temperatures.
Refrigerators and heat pumps
Work moves heat against its spontaneous direction
A refrigerator removes Qc from cold region, receives work W and delivers Qh=Qc+W to hot surroundings. Refrigerator COP=Qc/W. A heat pump’s heating COP=Qh/W=COP_R+1. COP may exceed one because it measures moved heat per work input, not energy-conversion efficiency.
Refrigerator example
A refrigerator removes 600 J using 200 J work. It rejects 800 J and COP=3.0. It does not create 800 J; 600 J came from cold space and 200 J from electrical work.
Reversible limit
Carnot efficiency depends only on reservoir temperatures
For a reversible engine between absolute temperatures Th and Tc, ηC=1−Tc/Th. Use kelvin. It is an upper bound; real engines are less efficient because of finite temperature differences, friction, turbulence and other irreversibilities.
Carnot limit
Between 600 K and 300 K, ηC=1−300/600=0.50. A claimed 60% cyclic engine between these reservoirs violates the Carnot bound.
The Carnot-cycle resource explains reversible comparisons. Reaching 100% would require Tc=0 K in the formula and is not physically attainable.
Entropy
Entropy tracks dispersal and irreversibility
For a reversible transfer, dS=dQrev/T. Entropy is a state function; for an isolated system, total entropy does not decrease. A reversible ideal process has zero total entropy production, while irreversible processes produce positive total entropy.
“Entropy is disorder” can be a rough analogy but is not enough for calculations. Use energy dispersal and state-function reasoning. The OpenStax entropy section shows how an irreversible path can have state-change entropy calculated through an imagined reversible path.
Solution method
Device diagram → energy balance → performance → feasibility
- Draw hot/cold reservoirs and arrows.
- Label magnitudes Qh, Qc and W.
- Apply Qh=W+Qc.
- Choose efficiency or COP definition.
- Convert reservoir temperatures to kelvin.
- Compare with Carnot or second-law direction.
Common mistakes: treating efficiency like COP, using Celsius in ratios, setting rejected heat to zero, reversing Qh/Qc arrows and assuming energy balance proves feasibility. For tuition call 9846662070; the MKS panel supports post-Grade-12 planning.
Questions students ask about the Second Law
Why is the first law not enough?
Energy conservation does not determine direction or maximum heat-to-work conversion; the second law adds those restrictions.
Can a heat engine be 100% efficient?
No cyclic engine can convert all heat from one reservoir into work without another effect.
What is Carnot efficiency?
For reversible operation between absolute temperatures Th and Tc, η=1−Tc/Th.
Why must temperatures be kelvin?
Thermodynamic temperature ratios in Carnot relations require the absolute scale.
Can heat flow from cold to hot?
Yes with work input, as in a refrigerator; it does not happen spontaneously as the sole effect.
Where can I get Thermodynamics tuition?
Call 9846662070 for current KTM Tuition options.
Frequently asked questions
Checked sources
References and related learning
CDC Nepal: Secondary Curriculum
OpenStax: Second-Law Statements
Continue with the Second Law Study Guide. Sources were checked on 2 August 2026; follow current CDC, NEB and school notices if requirements change.
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