PHYSICS 231 Mini Review Engines and fridges PHY 231 1 For a general engine turns water to steam W Qh Qc efficiency W Qh e 1 Q 1 Qc Qh heat reservoir Th Qh the steam moves the piston engine Qc work W the steam is condensed cold reservoir Tc work is done by steam For a reversible engine Qc Qh Tc Th efficiency 1 Tcold l Thot PHY 231 2 Reverse direction the fridge refrigerator air conditioner heat is expelled to outside Coefficient of performance COP Qc W h t reservoir heat s i Th Qh a piston compresses the coolant engine Qc the fridge is cooled work is done on coolant W work k cold reservoir Tc PHY 231 3 The laws of thermodynamics 0th law If objects A and B are both in thermal equilibrium with an object C than A and B are also in thermal equilibrium 1st law U Q W In a cyclic process U 0 U 0 then W Q it cannot do more work than the amount of energy heat put inside 2nd law It is impossible to construct an engine that can get 100 efficiency producing no other effect than the absorption of energy from a reservoir and the performance of an equal amount of work W cannott gett 100 efficiency We ffi i PHY 231 4 Zeroth law of thermodynamics If objects A and B are both in thermal equilibrium with an object j C than A and B are also in thermal equilibrium There is no transfer of energy between A B and C PHY 231 5 First Law of thermodynamics U Uf Ui Q W U change in internal energy Q energy Q gy transfer through g heat if heat is transferred to the system W energy transfer through work if work is done on the system This law is a general rule for conservation of energy that includes heat PHY 231 6 2nd law of thermodynamics rephrased 2nd law It is impossible to construct an engine that operating ti iin a cycle l produces d no other th effect ff t th than th the absorption of energy from a reservoir and the performance of an equal q amount of work we cannot g get 100 efficiency y 2nd law rephrased p The total entropy py of the universe increases when an irreversible process occurs 2nd law rephrased It is not possible to convert heat to work at a constant temperature PHY 231 7 Internal Energy In a monatomic gas the translational K E is the only type of energy the molecules can have 3 U nRT 2 U is the internal energy of the gas In a polyatomic gas additional possibilities for contributions to the internal energy are rotational and vibrational energy in the molecules PHY 231 8 Non isobaric compression In general the pressure can change when lowering the piston The work done on the gas when going from an initial state i to a final state t t f iis th the area under d th the P P V V diagram The work done by the gas is the opposite of the work done on the gas PV nRT PHY 231 9 Work done on gas signs Work Area under the curve on P V diagram V Vf Vi If the arrow goes from right to left V 0 positive work is done on the gas If the arrow goes from left to right V 0 negative work is done on the gas the gas has done positive work on the piston PHY 231 10 P A P Quiz B v P C v v I which In hi h case iis th the work kd done by b the th gas largest l t Case C area under curve is larger g and arrow goes from left to right positive work is done by the gas PHY 231 11 engines and fridges engine work was done by the gas while heat was added to the gas Wby gas enclosed area Convert heat to mechanical work fridge Net work was done on the h gas to extract out heat h Won gas enclosed area Use work to extract out heat PHY 231 12 Cyclic processes An Engine The system returns to its original state state Therefore Therefore the internal energy must be the same after completion of f the th cycle l U 0 U 0 U Uf Ui Q W U change in internal energy Q heat positive if heat is transferred to the system W work positive if work is done on the system PHY 231 13 Cyclic Process step by step 1 Process A B P A B Negative work is done on the gas the g gas is doing g positive p work W Area under P V diagram 50 10 10 3 1 0 0 0 105 50 10 10 3 5 0 1 0 105 W 12000 J work done on gas U 3 2nR T 3 2 PBVB P PAVA 1 5 1E 5 50E 03 5E 5 10E 03 0 The internal energy has not changed U Q W so Q U W 12000 J Heat that was added to the 14 system was used to do thePHY work 231 Cyclic process step by step 2 Process B C W Area under P V diagram 3 1 0 0 0 10 5 10 50 10 W 4000 J Work was done on the gas U 3 2nR T 3 2 PCVC PBVB 1 5 1E 5 10E 3 1E 5 50E 3 6000 J The internal energy has decreased by 6000 J U Q W so Q U W 6000 4000 J 10000 J 10000 J of energy has been transferred out of the system PHY 231 15 Cyclic process step by step 3 Process C A W Area under P V diagram W 0J W 0 No work was done on the gas U 3 2nR T 3 2 PAVA PCVC 1 5 5E 5 10E 3 1E 5 10E 3 6000 J The internal energy has increased by 6000 J U Q W so Q U W 6000 0 J 6000 J 6000 J of energy has been transferred into the system PHY 231 16 Summary of the process AREA A B Quantity Work W Heat Q Process U A B A B 12000 12000 J 12000 J 0 B C 4000 J 10000 J 6000 C A 0J 6000 J 6000 SUM 8000 J 8000 J 0 B C C A PHY 231 17 What did we do The gas performed net work 8000J Area while heat was supplied 8000J We have built an engine Convert heat to mechanical work What if the p process was done in the reverse way Net work was performed on the gas and heat extracted from the gas gas We have built a heat pump A fridge Use work to extract out heat PHY 231 18 example The efficiency of a Carnot engine is 30 The engine absorbs 800 J of energy per cycle by heat from a hot reservoir at 500 K Determine 5 D …
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