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Ch12.1 – Thermal Energy

Ch12.1 – Thermal Energy

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Ch12.1 – Thermal Energy

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  1. Ch12.1 – Thermal Energy Temperature – measure of the degree of hotness -explained by kinetic molecular theory: (PME) 1. Everything is made of tiny particles. 2. Particles are in constant motion. 3. All collisions are perfectly elastic (no energy lost.) Hot Cold Temperature really measures the amount of kinetic energy Temp = average KE How does a thermometer work? Hot water

  2. How cold is cold? Absolute zero – coldest temperature. No molecular motion 0oK = -273oC Exs) Body Temp = 37oC = ___ K 273oK = 0oC 400K = ___ oC 298oK = 25oC 373oK = 100oC Outer space 2-4 K Laboratory 0.00001 K .00000002K Since (-) #’s stink Kelvin designed a thermometer, modeled after Celsius, but put zero at coldest temp. Vol -300 0 100 Temp (°C) oC + 273 = K *Kelvin is always bigger

  3. How cold is cold? Absolute zero – coldest temperature. No molecular motion 0oK = -273oC Exs) Body Temp = 37oC = 310K 273oK = 0oC 400K = 127oC 298oK = 25oC 373oK = 100oC Outer space 2-4 K Laboratory 0.00001 K .00000002K Since (-) #’s stink Kelvin designed a thermometer, modeled after Celsius, but put zero at coldest temp. Vol -300 0 100 Temp (°C) oC + 273 = K *Kelvin is always bigger

  4. Heat (Q) – Energy that flows between 2 objects Q is (-) heat left object (feels hot) Q is (+) heat entered (feels cold) 3 ways for heat to transfer : Conduction – Objects in direct contact. Convection – involves the flow of fluids. Radiation – transfer of energy when no matter is present. (Sun Earth) Electromagnetic waves

  5. Specific Heat --different materials have different abilities to gain and lose energy. Heat Energy = mass ∙ specific heat ∙ change in temp Ex2) How much heat is required to raise the temp of 10.0kg of water 5˚C. Cp for water is 4.18 kJ/kg∙˚C Q = m∙Cp∙∆T

  6. Specific Heat --different materials have different abilities to gain and lose energy. Heat Energy = mass ∙ specific heat ∙ change in temp Ex2) How much heat is required to raise the temp of 10.0kg of water 5˚C. Cp for water is 4.18 kJ/kg∙˚C Q = m∙Cp∙∆T Q=? m = 10.0kg ∆T = 5˚ C Cp = 4.18 Q = (10.0 kg) (4.18kJ/kg∙˚C)(5˚C) =+209 kJ

  7. Ex2) A 0.400kg block of iron is heated from 295K to 325K. How much heat energy was transferred? Cp for Fe is .450kJ/kg∙˚C.

  8. Ex2) A 0.400kg block of iron is heated from 295K to 325K. How much heat energy was transferred? Cp for Fe is .450kJ/kg∙˚C. ∆T = 30 K 30˚C Q = m∙Cp∙∆T = (.4)(.45)(30) =5.4 kJ Ch12 HW#1

  9. Lab12.1 – Charles Law - due tomorrow - Ch12 HW#1 due at beginning of period

  10. Ch12 HW#1 1-6 (a,b) • 1) Make the following conversions. • 0C to Kelvin = • 0 K to Celsius = • 273C to Kelvin = • 273 K to Celsius = • 2) Convert the following Celsius temperatures to Kelvin temperatures. • a. 27C = • b. 150C = • c. 560C = • d. -50C = • e. -184C = • f. -300C = • 3) Convert the following Kelvin temperatures to Celsius temperatures. • a. 110 K = • b. 70 K = • c. 22K = • d. 402 K = • e. 323 K = • f. 212 K =

  11. Ch12 HW#1 1-6 (a,b) • 1) Make the following conversions. • 0C to Kelvin = 273K • 0 K to Celsius = -273°C • 273C to Kelvin= • 273 K to Celsius = • 2) Convert the following Celsius temperatures to Kelvin temperatures. • a. 27C = • b. 150C = • c. 560C = • d. -50C = • e. -184C = • f. -300C = • 3) Convert the following Kelvin temperatures to Celsius temperatures. • a. 110 K = • b. 70 K = • c. 22K = • d. 402 K = • e. 323 K = • f. 212 K =

  12. Ch12 HW#1 1-6 (a,b) • 1) Make the following conversions. • 0C to Kelvin = 273K • 0 K to Celsius = -273°C • 273C to Kelvin= • 273 K to Celsius = 0°C • 2) Convert the following Celsius temperatures to Kelvin temperatures. • a. 27C = 300 K • b. 150C = 423 K • c. 560C= • d. -50C = • e. -184C = • f. -300C = • 3) Convert the following Kelvin temperatures to Celsius temperatures. • a. 110 K = • b. 70 K = • c. 22K = • d. 402 K = • e. 323 K = • f. 212 K =

  13. Ch12 HW#1 1-6 (a,b) • 1) Make the following conversions. • 0C to Kelvin = 273K • 0 K to Celsius = -273°C • 273C to Kelvin= • 273 K to Celsius = • 2) Convert the following Celsius temperatures to Kelvin temperatures. • a. 27C = 300 K • b. 150C = 423 K • c. 560C= • d. -50C = • e. -184C = • f. -300C = • 3) Convert the following Kelvin temperatures to Celsius temperatures. • a. 110 K = -163°C • b. 70 K = -203°C • c. 22K = • d. 402 K = • e. 323 K = • f. 212 K =

  14. 4) Make a guess at the Celsius temps, then convert to Kelvin. a. Room temperature = 20°C = b. Refrigerator temperature = 10°C = c. Typical hot summer day = d. Typical winter night = 5) How much heat is absorbed by 0.060 g of carbon when its temperature is raised from 20.0C to 80.0C? Cp for carbon is 0.710 kJ/kg∙K. 6) The cooling system of a car engine contains 20.0 L of water (1 L of water has a mass of 1kg). What is the change in temp of the water if the engine operates until 836.0 kJ of heat are added? Cp for water is 4.18 kJ/kg∙K.

  15. 4) Make a guess at the Celsius temps, then convert to Kelvin. a. Room temperature = 20°C = 293 K b. Refrigerator temperature = 10°C = 283 K c. Typical hot summer day = d. Typical winter night = 5) How much heat is absorbed by 0.060 g of carbon when its temperature is raised from 20.0C to 80.0C? Cp for carbon is 0.710 kJ/kg∙K. 6) The cooling system of a car engine contains 20.0 L of water (1 L of water has a mass of 1kg). What is the change in temp of the water if the engine operates until 836.0 kJ of heat are added? Cp for water is 4.18 kJ/kg∙K.

  16. 4) Make a guess at the Celsius temps, then convert to Kelvin. a. Room temperature = 20°C = 293 K b. Refrigerator temperature = 10°C = 283 K c. Typical hot summer day = d. Typical winter night = 5) How much heat is absorbed by 0.060 g of carbon when its temperature is raised from 20.0C to 80.0C? Cp for carbon is 0.710 kJ/kg∙K. 6) The cooling system of a car engine contains 20.0 L of water (1 L of water has a mass of 1kg). What is the change in temp of the water if the engine operates until 836.0 kJ of heat are added? Cp for water is 4.18 kJ/kg∙K. Q = m ∙ Cp ∙ ∆T = (.00006 kg)(.710 )(60K) = kJ kg ∙ K

  17. 4) Make a guess at the Celsius temps, then convert to Kelvin. a. Room temperature = 20°C = 293 K b. Refrigerator temperature = 10°C = 283 K c. Typical hot summer day = d. Typical winter night = 5) How much heat is absorbed by 0.060 g of carbon when its temperature is raised from 20.0C to 80.0C? Cp for carbon is 0.710 kJ/kg∙K. 6) The cooling system of a car engine contains 20.0 L of water (1 L of water has a mass of 1kg). What is the change in temp of the water if the engine operates until 836.0 kJ of heat are added? Cp for water is 4.18 kJ/kg∙K. Q = m ∙ Cp ∙ ∆T = (.00006 kg)(.710 )(60K) = kJ kg ∙ K Q m ∙ Cp 836 kJ (20kg)(4.18 ) ∆T = = = kJ kg∙ K

  18. Ch12.2 – Calorimetry Calorimeter – a device used to measure changes in thermal energy. Ex1) An unknown metal sample weighing .020kg was placed in boiling water, at 96.5°C. It was then removed and placed in a calorimeter with .080kg of distilled water, at 20°C. The temp of the calorimeter leveled at: 23°C. What is the Cp of this unknown sample? Cp of water is 4.18 kJ kg ∙ °C

  19. Ch12.2 – Calorimetry Calorimeter – a device used to measure changes in thermal energy. Ex1) An unknown metal sample weighing .020kg was placed in boiling water, at 96.5°C. It was then removed and placed in a calorimeter with .080kg of distilled water, at 20°C. The temp of the calorimeter leveled at: 23°C. What is the Cp of this unknown sample? Cp of water is 4.18 kJ kg ∙ °C • Metal • -Qlost • -[m ∙ Cp ∙ ∆T] = • [(.02)∙ Cp ∙(-73.5)] = Water Qgained [m ∙ Cp ∙ ∆T] [(.08)∙ (4.18) ∙(3)] ∆T = Tf – Ti = 23 – 96.5 = -73.5 ∆T = Tf – Ti = 23 – 20 = 3 Cp = .682

  20. Ex2) A calorimeter contains 0.50 kg of water at 15°C. A 40 g piece of zinc at 115°C is placed in water. The final temperature leveled at 16°C. What is the Cp of zinc?

  21. Ex2) A calorimeter contains 0.50 kg of water at 15°C. A 40 g piece of zinc at 115°C is placed in water. The final temperature leveled at 16°C. What is the Cp of zinc? Metal -Qlost -[m ∙ Cp ∙ ∆T] = -[(.04)∙ Cp ∙(-99)] = Water -Qgained [m ∙ Cp ∙ ∆T] [(.5) ∙ (4.18) ∙(1)] Cp = .527 Ch12 HW#2 7 – 10

  22. Lab12.2 – Calorimetry - Lab write up due in 2 days - Unknown due at the end of the period - Ch12 HW#2 due at beginning of period

  23. Ch12 HW#2 7-10 #7) (Mod) A 2.0 kg sample of iron at an initial temperature of 500K is dropped into a bucket containing 2.0 kg of water at an initial temperature of 293K. The two reach thermal equilibrium at 313K. What is the specific heat capacity of the Iron? #8) (Mod) A 0.40 kg sample of lead at 92.3 C is dropped into a 0.378 kg beaker of water at 20.1C. The temp of the mixture levels at 22.4C. What is the specific heat capacity of the lead? ∆T = Tf – Ti ∆T = Tf – Ti

  24. Ch12 HW#2 7-10 #7) (Mod) A 2.0 kg sample of iron at an initial temperature of 500K is dropped into a bucket containing 2.0 kg of water at an initial temperature of 293K. The two reach thermal equilibrium at 313K. What is the specific heat capacity of the Iron? #8) (Mod) A 0.40 kg sample of lead at 92.3 C is dropped into a 0.378 kg beaker of water at 20.1C. The temp of the mixture levels at 22.4C. What is the specific heat capacity of the lead? • Iron • -Qlost • -[m ∙ Cp ∙ ∆T] = • [(2) ∙ Cp ∙ (-187)] = Water Qgained [m ∙ Cp ∙ ∆T] [(2) ∙ (4.18) ∙ (20)] ∆T = Tf – Ti ∆T = Tf – Ti ∆T = Tf – Ti ∆T = Tf – Ti

  25. Ch12 HW#2 7-10 #7) (Mod) A 2.0 kg sample of iron at an initial temperature of 500K is dropped into a bucket containing 2.0 kg of water at an initial temperature of 293K. The two reach thermal equilibrium at 313K. What is the specific heat capacity of the Iron? #8) (Mod) A 0.40 kg sample of lead at 92.3 C is dropped into a 0.378 kg beaker of water at 20.1C. The temp of the mixture levels at 22.4C. What is the specific heat capacity of the lead? • Iron • -Qlost • -[m ∙ Cp ∙ ∆T] = • [(2) ∙ Cp ∙ (-187)] = Water Qgained [m ∙ Cp ∙ ∆T] [(2) ∙ (4.18) ∙ (20)] ∆T = Tf – Ti ∆T = Tf – Ti • Lead • -Qlost • -[m ∙ Cp ∙ ∆T] = • [(.4)∙ Cp ∙(-69.9)] = Water Qgained [m ∙ Cp ∙ ∆T] [(.378)∙ (4.18) ∙(2.3)] ∆T = Tf – Ti ∆T = Tf – Ti

  26. #9) A 0.10 kg brass block at 90.0C is placed in a plastic foam cup containing 0.103 kg of water at 20.0C. No heat is lost to the surroundings. The final temp of the mix is 25.6C. What is the specific heat capacity of brass? #10) (Mod) A 100g Aluminum slug at 100C is placed in 100 g of water at 10C. The final temp is 25C. What is the Cp of Al? ∆T = Tf – Ti ∆T = Tf – Ti

  27. #9) A 0.10 kg brass block at 90.0C is placed in a plastic foam cup containing 0.103 kg of water at 20.0C. No heat is lost to the surroundings. The final temp of the mix is 25.6C. What is the specific heat capacity of brass? #10) (Mod) A 100g Aluminum slug at 100C is placed in 100 g of water at 10C. The final temp is 25C. What is the Cp of Al? • Brass • -Qlost • -[m ∙ Cp ∙ ∆T] = • [(.1)∙ Cp ∙(-64.4)] = Water Qgained [m ∙ Cp ∙ ∆T] [(.103)∙ (4.18) ∙(5.6)] ∆T = Tf – Ti ∆T = Tf – Ti ∆T = Tf – Ti ∆T = Tf – Ti

  28. #9) A 0.10 kg brass block at 90.0C is placed in a plastic foam cup containing 0.103 kg of water at 20.0C. No heat is lost to the surroundings. The final temp of the mix is 25.6C. What is the specific heat capacity of brass? #10) (Mod) A 100g Aluminum slug at 100C is placed in 100 g of water at 10C. The final temp is 25C. What is the Cp of Al? • Brass • -Qlost • -[m ∙ Cp ∙ ∆T] = • [(.1)∙ Cp ∙(-64.4)] = Water Qgained [m ∙ Cp ∙ ∆T] [(.103)∙ (4.18) ∙(5.6)] • Aluminum • -Qlost • -[m ∙ Cp ∙ ∆T] = • [(.100)∙ Cp ∙(-75)] = Water Qgained [m ∙ Cp ∙ ∆T] [(.100)∙ (4.18) ∙(15)]

  29. Ch12.3 – Heat & Changes of State 120 For H2O: 100 Temp (°C) 0 -10 Time

  30. Ch12.3 – Heat & Changes of State 5) Raise temp of gas: Q = m ∙Cp ∙ ∆T 120 For H2O: l g 100 4) Vaporize it: Q = m ∙ Hv Temp (°C) 3) Raise temp of liquid: Q = m ∙Cp ∙ ∆T s l 2) Melt it: Q = m ∙ Hf 0 1) Raise temp of solid: Q = m ∙Cp ∙ ∆T -10 Time 6) Add steps

  31. Heat of Fusion – amount of heat needed to melt a substance. Temp levels as all energy goes to break solid apart. Q = m ∙ Hf kJ kg ( for water , Hf = 334 ) Heat of vaporization – amt of heat needed to vaporize. Temp levels as all energy goes to break bonds. Q = m ∙ Hv kJ kg ( for water , Hv = 2260 ) kJ kg ∙ K Cp for ice = 2.1 Cp for steam = 2.02 kJ kg ∙ K

  32. Ex) How much heat is required to turn 63kg of ice at -50.4°C to vapor at 124.3°C? Temp (°C) Time

  33. Ex) How much heat is required to turn 63kg of ice at -50.4°C to vapor at 124.3°C? 5 4 Temp (°C) 3 2 6 add 1 Time kJ kg ∙ °C 1) Heat iceQ = m ∙Cp ∙ ∆T = (63kg)(2.1 )(50.4°C) = 6.7 kJ kJ kg 2) MeltQ = m ∙Hf= (63kg)(334 ) = 21.1 kJ kJ kg ∙ °C 3) Heat waterQ = m ∙Cp ∙ ∆T = (63kg)(4.18 )(100°C) = 26.5 kJ kJ kg 4) BoilQ = m ∙ Hv= (63kg)(2260 ) = 142.9 kJ kJ kg ∙ °C 5) Heat gasQ = m ∙Cp ∙ ∆T = (63kg)(2.02 )(24.3°C) = 3.1 kJ 6) Add : 200.2 kJ Ch12 HW#3

  34. Temperature of water versus time as thermal energy is removed 100 Vapor Changing into a liquid Temperature of sample (°C) 50 Liquid changing into a solid 0 Liquid Time elapsed Vapor solid

  35. Lab 12.3 – Heat of Fusion - due tomorrow - Ch12 HW#3 due at beginning of period

  36. Ch12 HW#3 11 -13 11. How much heat is absorbed by 0.10kg of ice at -20°C to become water at 0°C? 12. A 0.20kg sample of water at 60°C is heated to steam at 140.0°C. How much heat is absorbed? 13. How much heat is needed to change 0.30kg of ice at -30°C to steam at 130°C?

  37. Ch12 HW#3 11 -13 11. How much heat is absorbed by 0.10kg of ice at -20°C to become water at 0°C? 12. A 0.20kg sample of water at 60°C is heated to steam at 140.0°C. How much heat is absorbed? 13. How much heat is needed to change 0.30kg of ice at -30°C to steam at 130°C? 1)Heat iceQ = m∙Cp∙∆T = (.10kg)(2.1)(20°C) = 2)MeltQ = m ∙Hf= (.10)(334) = 3) Add: 1)Heat waterQ = m∙Cp∙∆T = (.20kg)(4.18)(40°C) = 2)VaporizeQ = m ∙Hv= (.20)(2260) = 3)Heat gasQ = m∙Cp∙∆T = (.20kg)(2.02)(40°C) = 4) Add: 1)Heat iceQ = m∙Cp∙∆T = (.30kg)(2.1)(30°C) = 2)MeltQ = m ∙Hf= (.30)(334) = 3)Heat waterQ = m∙Cp∙∆T = (.30kg)(4.18)(100°C) = 4)VaporizeQ = m ∙Hv= (.30)(2260) = 5)Heat gasQ = m∙Cp∙∆T = (.30kg)(2.02)(30°C) = 6) Add:

  38. Ch12.4 – Thermodynamics First Law of Thermodynamics - (Conservation of Energy) Energy can’t be created or destroyed, it only changes form. - in the end, energy turns into thermal energy - it’s hard to turn thermal energy back into more useful forms of energy (like work), but heat engines attempt to do this Heat Engine - takes a high temp heat source, converts some of the thermal energy to work, then exhausts the lower temp heat that remains. - combustion engines (cars), steam engines, heat pumps, refrigerators, etc. Internal Combustion Engine 1. Chemical reaction creates high temps : Qhigh 2. Gases expand pushing piston down : Work 3. Gases cool as expand : Qlow

  39. Second Law of Thermodynamics - Natural Processes tend to go in a direction that increases the total amount of entropy of the universe. Entropy – a measure of the amount of disorder “Entropy is a game you can’t win, you can’t break even, and you can’t even get out of the game.”

  40. 1st law formulas: Qlow (Entropy) Q = W + ∆U Heat transferred to/from system Internal Energy lost/gained by system Work done by/on system Heat lost by system = - Q Heat gained by system = +Q Work done by the system = +W Work done on the system = - W Internal Energy lost = - ∆U Internal Energy gained = + ∆U

  41. Ex1) 200 J of work are done on a system while its internal energy increases by 150 J. How much heat was added to or taken from the system? Ex2) 1100 J of heat are transferred from a system when the system does 850 J of work on its surroundings. What is the change in internal energy on the system? Ex3) 350 J of work is done by a system while its internal energy is made to increase by 50 J. How much heat was transferred to/from the system?

  42. Ex1) 200 J of work are done on a system while its internal energy increases by 150 J. How much heat was added to or taken from the system? Ex2) 1100 J of heat are transferred from a system when the system does 850 J of work on its surroundings. What is the change in internal energy on the system? Ex3) 350 J of work is done by a system while its internal energy is made to increase by 50 J. How much heat was transferred to/from the system? Q = ? W = -200J ∆U = +150 J Q = W + ∆U = -200J + (+150J) = - 50J (taken away) Q = W + ∆U -1100J = +850J + ∆U ∆U = -1950J (decrease) Q = -1100J W = +850J ∆U = ? Q = W + ∆U = +350J + (+50J) = +400J (heat added) Q = ? W = +350J ∆U = +50J C 12 HW #4 14-18

  43. Ch12 HW#4 14-18 • 134J of work are done on a system while its internal energy increases by 93J. How much heat was added to or taken away from the system? • 15. 2050J of heat are transferred to a system when the system does 1230J of work on its surroundings. What is the change in the internal energy of the system? • 16. 225J of work is done on a system while its internal energy is made to increase by 100J. How much heat was transferred to/from the system? W = -134 J ∆U = +93 J Q = ? W = +1230 J ∆U = ? Q = +2050 J W = -225 J ∆U = +100 J Q = ?

  44. Ch12 HW#4 14-18 • 134J of work are done on a system while its internal energy increases by 93J. How much heat was added to or taken away from the system? • 15. 2050J of heat are transferred to a system when the system does 1230J of work on its surroundings. What is the change in the internal energy of the system? • 16. 225J of work is done on a system while its internal energy is made to increase by 100J. How much heat was transferred to/from the system? W = -134 J ∆U = +93 J Q = ? Q = -134 J + (+93J) Q = W = +1230 J ∆U = ? Q = +2050 J Q = W + ∆U +2050 = +1230 + ∆U ∆U = Q = W + ∆U = -225J + (+100J) Q = W = -225 J ∆U = +100 J Q = ?

  45. Ch12 HW#4 14-18 • 17. 850 J of heat is lost by a system while 250 J of work is done on it. What is its change in internal energy? • 18. How much work is done on/by a system that has 525 J of heat added to it while its internal energy increases by 300 J? Q = - 850 J W = - 250 J ∆U = ? W = ? Q = + 525 J ∆U = +300 J

  46. Ch12 HW#4 14-18 • 17. 850 J of heat is lost by a system while 250 J of work is done on it. What is its change in internal energy? • 18. How much work is done on/by a system that has 525 J of heat added to it while its internal energy increases by 300 J? Q = - 850 J W = - 250 J ∆U = ? Q = W + ∆U -850J = -250J + ∆U ∆U = W = ? Q = + 525 J ∆U = +300 J Q = W + ∆U +525J = W + (+300J) W =

  47. Ch12.5 – Efficiency TH – TL TH Efficiency of a heat engine: eff = ---------- x 100% Temps must be in Kelvins! (C + 273 = K) Q W + ∆U good bad Ex 1) Calculate the efficiency of a heat engine that operates between 200°C and 100°C. Ex 2) A heat engine has an input temp of 550°C and an exhaust temp of 100°C. What is its ideal efficiency?

  48. Ch12.5 – Efficiency TH – TL TH Efficiency of a heat engine: eff = ---------- x 100% Temps must be in Kelvins! (C + 273 = K) Q W + ∆U good bad Ex 1) Calculate the efficiency of a heat engine that operates between 200°C and 100°C. 473K – 373K 473K eff = -------------- x 100% = 21% Ex 2) A heat engine has an input temp of 550°C and an exhaust temp of 100°C. What is its ideal efficiency? Ch12 HW#5 19 – 23 823K – 373K 823K eff = -------------- x 100% = 55%

  49. Ch12 HW#5 19-23 • 19. Calculate the efficiency of a heat engine that operates between 350oC • and 50OC. • 20. A heat engine has an input temp of 3250oC and an exhaust temp • of 1125oC. What is the its ideal efficiency? TH = 350°C + 273 = 623K TL = 50°C + 273 = 323K TH = 3250°C + 273 = 3523K TL = 1125°C + 273 = 1398K

  50. Ch12 HW#5 19-23 • 19. Calculate the efficiency of a heat engine that operates between 350oC • and 50OC. • 20. A heat engine has an input temp of 3250oC and an exhaust temp • of 1125oC. What is the its ideal efficiency? TH – TL TH TH = 350°C + 273 = 623K TL = 50°C + 273 = 323K eff = ---------- x 100% 623K – 323K 623K eff = -------------- x 100% = TH = 3250°C + 273 = 3523K TL = 1125°C + 273 = 1398K TH – TL TH eff = ---------- x 100% 3523K – 1398K 3523K eff = -------------- x 100% =