NEB Class 12 Physics Second Law of Thermodynamics Notes in PDF Complete Handwritten. Physics Notes 2082: All Chapters | New Curriculum | Class 12 Physics Notes download.
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NEB Class 12 Physics Second Law of Thermodynamics Note Handwritten in PDF
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Why the First Law is Insufficient
The first law says energy is conserved. But it cannot explain why:
- Heat flows from hot to cold spontaneously, never the reverse
- A scrambled egg never unscrambles itself
- A glass that falls and breaks never reassembles spontaneously
These observations require the Second Law of Thermodynamics.
Two Statements of the Second Law
Kelvin-Planck Statement: It is impossible to construct a heat engine that operates in a cycle and converts all heat absorbed from a single heat source into work, with no other effect.
β No engine can be 100% efficient. Some heat must always be rejected.
Clausius Statement: It is impossible for heat to flow spontaneously from a colder body to a hotter body without the expenditure of work.
β Refrigerators need external work input to move heat “uphill.”
Both statements are equivalent β each can be proved from the other.
Heat Engine and Efficiency
A heat engine operates between a hot source (Tβ) and cold sink (Tβ):
- Absorbs Qβ from hot source
- Does work W
- Rejects Qβ to cold sink
Efficiency: Ξ· = W/Qβ = (Qβ β Qβ)/Qβ = 1 β Qβ/Qβ
Since Qβ is always > 0, efficiency is always < 1 (less than 100%).
Carnot Cycle β The Ideal Engine
Carnot’s cycle is a theoretical ideal engine operating between two temperatures.
Four stages:
- Isothermal expansion at Tβ (absorbs Qβ, does work)
- Adiabatic expansion (temperature drops from Tβ to Tβ, does work)
- Isothermal compression at Tβ (rejects Qβ, work done on gas)
- Adiabatic compression (temperature rises from Tβ to Tβ)
Carnot efficiency: Ξ·_Carnot = 1 β Tβ/Tβ = (Tβ β Tβ)/Tβ
This is the maximum possible efficiency of any engine working between Tβ and Tβ. Real engines are always less efficient because of friction, heat loss, and non-ideal processes.
Example: If Tβ = 600K and Tβ = 300K β Ξ· = 1 β 300/600 = 50%
Otto Cycle (Petrol Engine) and Diesel Cycle
Otto cycle:
- Uses spark ignition
- Two isochoric (constant volume) and two adiabatic processes
- Compression ratio r = V_max/V_min (typically 8β10)
- Efficiency: Ξ· = 1 β 1/r^(Ξ³β1)
- Higher compression ratio β higher efficiency
Diesel cycle:
- Uses compression ignition (no spark plug)
- Higher compression ratio (14β22) β more efficient than petrol
- Has an isobaric (constant pressure) heat addition stage
- More fuel efficient but heavier engine
Refrigerator and COP
A refrigerator is a heat engine running in reverse:
- Uses work W to remove Qβ from cold reservoir
- Delivers Qβ = Qβ + W to hot reservoir
Coefficient of Performance: COP = Qβ/W
A good refrigerator has high COP (removes a lot of heat with little work). Unlike efficiency, COP can be greater than 1.
Entropy
Entropy (S) is a measure of disorder or randomness in a system.
- dS = dQ_reversible/T
- Entropy of the universe always increases in any irreversible process: ΞS_universe > 0
- In a reversible process: ΞS_universe = 0
- High entropy = high disorder (gases are higher entropy than liquids)
The second law can be restated: The total entropy of the universe never decreases.
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Frequently Asked Questions
What is the second law of thermodynamics Class 12?
Second law has two statements. Kelvin-Planck: no engine can convert all heat to work β some must always be rejected. Clausius: heat cannot spontaneously flow from cold to hot body. Both prove 100% efficiency is impossible for any real heat engine operating between two temperatures.
What is Carnot efficiency formula Class 12?
Carnot efficiency: Ξ· = 1 β Tβ/Tβ = (TββTβ)/Tβ, where Tβ is hot source temperature and Tβ is cold sink temperature, both in Kelvin. This is the maximum possible efficiency of any engine between two temperatures. Real engines are always less efficient than Carnot engine due to irreversibilities.
What is the difference between Otto cycle and Diesel cycle?
Otto cycle (petrol engine): spark ignition, compression ratio 8-10, efficiency Ξ· = 1β1/r^(Ξ³β1). Diesel cycle: compression ignition (no spark), higher compression ratio 14-22, more fuel efficient. Diesel is more efficient because of higher compression ratio. Both are modeled on Carnot concept but less efficient.
What is the difference between heat engine and refrigerator?
Heat engine absorbs Qβ from hot source, does work W, rejects Qβ to cold sink β efficiency = W/Qβ. Refrigerator runs in reverse β uses work W to move Qβ from cold to hot reservoir. Refrigerator performance is measured by COP = Qβ/W, which can be greater than 1 unlike efficiency.
What is entropy in Class 12 Physics simple explanation?
Entropy measures disorder or randomness in a system. In any irreversible process total entropy of the universe increases (ΞS > 0). In reversible process ΞS = 0. Entropy explains why heat flows from hot to cold naturally and why time moves forward β processes of increasing disorder are spontaneous.
What is COP of refrigerator formula Class 12?
COP of refrigerator = Qβ/W, where Qβ is heat removed from cold reservoir and W is work input. Since Qβ = Qβ + W, COP = Qβ/(QββQβ) = Tβ/(TββTβ) for ideal refrigerator. Unlike engine efficiency, COP can be greater than 1 and is typically 3-5 for household refrigerators.
What is Carnot cycle in Class 12 Physics?
Carnot cycle has four stages: isothermal expansion at Tβ (absorbs Qβ), adiabatic expansion (temperature drops to Tβ), isothermal compression at Tβ (rejects Qβ), adiabatic compression (temperature rises back to Tβ). The enclosed PV diagram area equals net work done per cycle. It is the most efficient possible cycle.
Which topics from second law are most important for exam?
Most important: Carnot efficiency derivation and formula (4 marks), statements of second law (2 marks), comparison of Otto and Diesel cycles (2 marks), refrigerator COP formula (2 marks), and entropy concept (2 marks). Carnot efficiency numerical problems appear in almost every NEB Class 12 Physics board paper.
