Thermodynamics Lecture 11 Refrigerator & Heat Pump COP, Carnot Refrigerator & Numericals CBSE 11 &
Автор: CBSE & JEE Physics | Dr Kedar Pathak
Загружено: 2026-01-01
Просмотров: 4
Описание:
In Thermodynamics Lecture 11, we apply the second law to refrigerators and heat pumps, define coefficient of performance (COP), derive the COP of a Carnot refrigerator, and solve key numericals. This lecture is targeted at CBSE Class 11 Physics and JEE Main aspirants.
Covered in this video:
• Refrigerator as reverse of a heat engine
• Reverse Carnot cycle: adiabatic + isothermal steps for an ideal refrigerator
• Source, sink and working substance (refrigerant)
• Energy flow: (Q_{2}) from cold body, work (W) in, (Q_{1}) to hot body
• Coefficient of performance (COP) of refrigerator: (\beta = Q_{2}/W)
• Carnot refrigerator COP: (\beta = T_{2}/(T_{1} - T_{2}))
• Relation between COP and engine efficiency: (\beta = (1 - \eta)/\eta)
• Heat pump vs refrigerator: which heat quantity is “useful”
• Q1: COP of a Carnot refrigerator between (-10^{\circ}C) and (30^{\circ}C)
• Q2: Carnot refrigerator used to freeze water – mass of ice formed per minute
Use this lecture to master COP formulas and typical refrigerator/heat pump questions asked in CBSE Boards and JEE Main.
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