Solve 19 1 cycle The heat input is calculated as follows a The engine has a thermal efficiency of 40 0 40 and a work output of 100 J per b Because W Q Q C H out the heat exhausted is 0 40 Q H 250 J out W Q H 100 J Q H Q Q W H C out 250 J 100 J 150 J Solve During each cycle the work done by the engine is W out 20 J and the engine exhausts 30 J of heat energy Because W Q Q C H out 19 2 Q C Thus the efficiency of the engine is Q W H out Q C 20 J 30 J 50 J 1 1 0 40 Q C Q H 30 J 50 J 19 3 Solve a During each cycle the heat transferred into the engine is Q H 55 kJ and the heat exhausted is Q C 40 kJ The thermal efficiency of the heat engine is 1 1 0 27 27 Q C Q H 40 kJ 55 kJ W out Q Q C H 55 kJ 40 kJ 15 kJ b The work done by the engine per cycle is 19 4 Solve The coefficient of performance of the refrigerator is K Q C W in Q W H in W in 50 J 20 J 20 J 1 5 19 5 Solve a The heat extracted from the cold reservoir is calculated as follows b The heat exhausted to the hot reservoir is K 4 0 Q C 200 J Q C W in Q C 50 J Q H Q W in C 200 J 50 J 250 J 19 6 Solve Model Assume that the car engine follows a closed cycle a Since 2400 rpm is 40 cycles per second the work output of the car engine per cycle is W out 500 kJ s 1 s 40 cycles 12 5 kJ cycle Q H 62 5 kJ out W Q H 12 5 kJ 0 20 Q Q W in H C 62 5 kJ 12 5 kJ 50 kJ b The heat input per cycle is calculated as follows The heat exhausted per cycle is 19 7 Solve The amount of heat discharged per second is calculated as follows out W Q H W out Q W C out C Q W out 1 900 MW 1 913 10 W 9 1 1 0 32 1 That is each second the electric power plant discharges American house needs 2 0 10 J with the waste heat is 9 1 913 10 J 2 0 10 J 1 913 10 J of energy into the ocean Since a typical of energy per second for heating the number of houses that could be heated 96 000 4 4 9 19 8 Solve The amount of heat removed from the water in cooling it down in 1 hour is The mass of the water is Q m c water water C T m water V water water 1000 kg m 1 L 3 3 100 kg m 10 m 3 3 1 0 kg Q C 1 0 kg 4190 J kg K 20 C 5 C 6 285 10 J 4 The rate of heat removal from the refrigerator is The refrigerator does work refrigerator is W 8 0 W 8 0 J s to remove this heat Thus the performance coefficient of the Q C 4 6 285 10 J 3600 s 17 46 J s K 17 46 J s 8 0 J s 2 2 Model Process A is isochoric process B is isothermal process C is adiabatic and process D is 19 9 isobaric Solve Process A is isochoric so the increase in pressure increases the temperature and hence the thermal Q increases for process A Process B is isothermal so T is constant energy Because and 0 J E Q W s th W s th 0 E J The work done Ws is positive because the gas expands Because Q is and hence positive for process B Process C is adiabatic so Q 0 J Ws is positive because of the increase in volume Since is negative for process C Process D is isobaric so the decrease in volume leads to a Q decrease in temperature and hence a decrease in the thermal energy Due to the decrease in volume Ws is negative Because Q also decreases for process D Q W s thE 0 J W s E E E th th Q W s th A B C D Eth 0 Ws 0 Q 0 19 10 Model Process A is adiabatic process B is isothermal and process C is isochoric Solve Process A is adiabatic so Q 0 J Work Ws is positive as the gas expands Since Q Ws Eth 0 J Eth must be negative The temperature falls during an adiabatic expansion Process B is isothermal so T 0 and Eth 0 J The gas is compressed so Ws is negative Q Ws for an isothermal process so Q is negative Heat energy is withdrawn during the compression to keep the temperature constant Process C is isochoric No work is done Ws 0 J and Q is positive as heat energy is added to raise the temperature Eth positive Eth 0 Ws 0 Q 0 A B C 19 11 inside the triangle is Solve The work done by the gas per cycle is the area inside the closed p versus V curve The area W out 1 2 3 atm 1 atm 600 10 m 200 10 m 6 3 6 1 013 10 Pa 1 atm 5 1 2 2 atm 400 10 m 6 3 40 5 J 3 19 12 Solve The work done by the gas per cycle is the area enclosed within the pV curve We have 60 J 1 2 p max 100 kPa 800 cm 200 cm 3 3 p max 1 0 10 Pa 5 2 60 J 6 600 10 m 3 p max 3 0 10 Pa 300 kPa 5 19 13 consists of three individual processes Solve Model The heat engine follows a closed cycle starting and ending in the original state The cycle a The work done by the heat engine per cycle is the area enclosed by the p versus V graph We get The heat energy transferred into the engine is QH 120 J Because W Q Q C H out the heat energy exhausted is b The thermal efficiency of the engine is W out 1 2 200 kPa 100 10 m 6 3 10 J Q Q W H C out 120 J 10 J 110 J out W Q H 10 J 120 J 0 0833 Assess Practical engines have thermal efficiencies in the range 0 1 0 4 19 14 Solve Model The heat engine follows a closed cycle which consists of four individual processes a The work done by the heat engine per cycle is the area enclosed by the p versus V graph We get 400 kPa 100 kPa 100 10 …
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