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IB Physics Topic 3 & 10 Mr. Jean May 7 th, 2014. The plan: Video clip of the day Thermodynamics Carnot Cycle Second Law of Thermodynamics Refrigeration.

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Presentation on theme: "IB Physics Topic 3 & 10 Mr. Jean May 7 th, 2014. The plan: Video clip of the day Thermodynamics Carnot Cycle Second Law of Thermodynamics Refrigeration."— Presentation transcript:

1 IB Physics Topic 3 & 10 Mr. Jean May 7 th, 2014

2 The plan: Video clip of the day Thermodynamics Carnot Cycle Second Law of Thermodynamics Refrigeration

3 Recap 1 st Law of Thermodynamics energy conservation Q = U + W Heat flow into system Increase in internal energy of system Work done by system V P l U depends only on T (U = nRT = PV) l point on PV plot completely specifies state of system (PV = nRT) l work done is area under curve l for complete cycle U = 0 Q = W

4 What do the cycles apply to? THTH TCTC QHQH QCQC W HEAT ENGINE THTH TCTC QHQH QCQC W REFRIGERATOR system l system taken in closed cycle U system = 0 l therefore, net heat absorbed = work done Q H - Q C = W (engine) Q C - Q H = -W (refrigerator) energy into blue blob = energy leaving bluegreen blob

5 Heat Engine: Efficiency THTH TCTC QHQH QCQC W HEAT ENGINE Goal: Get work from thermal energy in the hot reservoir 1 st Law: Q H - Q C = W, ( U = 0 for cycle) Define efficiency as work done per thermal energy used e What is the best we can do? Solved by Sadi Carnot in 1824 with the Carnot Cycle W QHQH

6 Carnot Cycle Adiabat Q = 0 P V Isotherm Q H = W H Isotherm Q C = W C Designed by Sadi Carnot in 1824, maximally efficient Q H enters from 1-2 at constant T H and Q C leaves from 3-4 at constant T L Work done W net = W H – W C = Q H – Q C = W Efficiency is W / Q H = ( Q H – Q C ) / Q H Since U T then Q – W is also proportional to T but from (1-2) and (3-4) Q = W so Q T Efficiency is W / T H = ( T H – T C ) / T H e max = 1 – QHQH QCQC TCTC THTH /

7 Heat Engine: Entropy We can define a useful new quantity Entropy, S Entropy measures the disorder of a system Only changes in S matter to us S = T Q Change in entropy depends on thermal energy flow (heat) at temperature T THTH TCTC QHQH QCQC W HEAT ENGINE

8 Heat Engine: Entropy Entropy, Smeasures the disorder of a system changes in S matter S = If = as in the Carnot Cycle T Q THTH QHQH TCTC QCQC … then there is no net change in entropy for the cycle and efficiency is a maximum, … because we do as much work as is possible THTH TCTC QHQH QCQC W HEAT ENGINE

9 2 nd Law of Thermodynamics Heat flows from hot to cold naturally One cannot convert a quantity of thermal energy entirely to useful work (Kelvin) The entropy, disorder, always increases in closed systems In closed systems, S > 0 for all real processes One cannot transfer thermal energy from a cold reservoir to hot reservoir without doing work (Clausius) Only in the ideal case of maximum efficiency would S = 0

10 Does the apparent order of life on Earth imply the 2 nd law is wrong or that some supernatural being is directing things? EXAMPLE No. The second law applies to closed systems, those with no energy coming in or going out. As long as the Sun shines more energy falls on the Earth, and more work can be done by the plants to build new mass, release oxygen, grow, metabolize.

11 What is happening to the Universe? EXAMPLE The universe is slowly coming to an end. When the entire universe is at the same temperature, then no work will be possible, and no life and no change … billions and billions and billions of years from now … Heat Death

12 Consider a hypothetical device that takes 1000 J of heat from a hot reservoir at 300K, ejects 200 J of heat to a cold reservoir at 100K, and produces 800 J of work. Is this possible? EXAMPLE The maximum efficiency is e max = 1 – T L /T H = 67%, but the proposed efficiency is e prop = W/Q H = 80%. This violates the 2 nd law – do not buy shares in the company designing this engine!

13 Consider a hypothetical refrigerator that takes 1000 J of heat from a cold reservoir at 100K and ejects 1200 J of heat to a hot reservoir at 300K. Is this possible? EXAMPLE The entropy of the cold reservoir decreases by S C = 1000 J / 100 K = 10 J/K The entropy of the heat reservoir increases by S H = 1200 J / 300 K = 4 J/K There would be a net decrease in entropy which would violate the 2 nd Law, so this refrigerator is not possible What is the minimum work needed? 2000 J, so that S H becomes at least 10 J/K

14 Air Conditioners Uses a working fluid (freon or other nicer gas) to carry heat from cool room to hot surroundings – same as a refrigerator, moving Q from inside fridge to your kitchen, which you must then air condition!

15 Air Conditioners

16 Evaporator located in room air transfers heat from room air to fluid Compressor located in outside air does work on fluid and heats it further Condenser located in outside air transfers heat from fluid to outside air Then the fluid reenters room for next cycle

17 Evaporator Fluid nears evaporator as a high pressure liquid near room temperature A constriction reduces the fluid pressure Fluid enters evaporator as a low pressure liquid near room temperature Heat exchanger made from a long metal pipe Working fluid evaporates in the evaporator – requires energy L V to separate molecules, so fluid cools & Q flows from room to fluid Fluid leaves evaporator as a low pressure gas near room temperature, taking thermal energy with it, leaving the room cooler!

18 Compressor Working fluid enters compressor as a low pressure gas near room temperature Gas is compressed (PV work) so gas T rises (1 st Law, T U & U when PV work is done) Compressing gas forces Q out of it into surroundings (open air) Fluid leaves compressor as hot, high pressure gas

19 Condenser Fluid enters condenser (heat exchanger made from long metal pipe) as a hot, high pressure gas Q flows from fluid to outside air Gas releases energy across heat exchanger to air and condenses forming bonds releases energy L V – thermal energy & fluid becomes hotter liquid so even more heat flows from fluid into outside air Fluid leaves condenser as high pressure liquid near room temperature to repeat the cycle

20 Summary Condenser – in outside air transfers heat from fluid to outside air, including thermal energy extracted from inside air and thermal energy added by compressor Evaporator – in room transfers heat from room air to working fluid Compressor – outside does work on fluid, so fluid gets hotter Entropy of room has decreased but entropy of outside has increased by more than enough to compensate – order to disorder


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