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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

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

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

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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

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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. C p for water is 4.18 kJ / kg∙˚C Q = m∙C p ∙∆T

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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. C p for water is 4.18 kJ / kg∙˚C Q = m∙C p ∙∆T Q=? m = 10.0kg ∆T = 5˚ C C p = 4.18 Q = (10.0 kg) (4.18 kJ / kg∙˚C )(5˚C) =+209 kJ

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

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

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

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Ch12 HW#1 1-6 (a,b) 1) Make the following conversions. a.0 C to Kelvin = b.0 K to Celsius = c. 273 C to Kelvin = d. 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 C = f 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 =

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Ch12 HW#1 1-6 (a,b) 1) Make the following conversions. a.0 C to Kelvin = 273K b.0 K to Celsius = -273°C c. 273 C to Kelvin = d. 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 C = f 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 =

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Ch12 HW#1 1-6 (a,b) 1) Make the following conversions. a.0 C to Kelvin = 273K b.0 K to Celsius = -273°C c. 273 C to Kelvin = d. 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 C = f 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 =

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Ch12 HW#1 1-6 (a,b) 1) Make the following conversions. a.0 C to Kelvin = 273K b.0 K to Celsius = -273°C c. 273 C to Kelvin = d. 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 C = f 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 =

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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 g of carbon when its temperature is raised from 20.0 C to 80.0 C? C p for carbon is 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 kJ of heat are added? C p for water is 4.18 kJ/kg ∙K.

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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 g of carbon when its temperature is raised from 20.0 C to 80.0 C? C p for carbon is 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 kJ of heat are added? C p for water is 4.18 kJ/kg ∙K.

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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 g of carbon when its temperature is raised from 20.0 C to 80.0 C? C p for carbon is 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 kJ of heat are added? C p for water is 4.18 kJ/kg ∙K. Q = m ∙ C p ∙ ∆T = ( kg)(.710 )(60K) = kJ kg ∙ K

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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 g of carbon when its temperature is raised from 20.0 C to 80.0 C? C p for carbon is 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 kJ of heat are added? C p for water is 4.18 kJ/kg ∙K. Q = m ∙ C p ∙ ∆T = ( kg)(.710 )(60K) = kJ kg ∙ K ∆T = = = Q m ∙ C p 836 kJ (20kg)(4.18 ) kJ kg∙ K

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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 C p of this unknown sample? C p of water is 4.18 kJ kg ∙ °C

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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 C p of this unknown sample? C p of water is 4.18 kJ kg ∙ °C Metal -Q lost -[m ∙ C p ∙ ∆T] = -[(.02)∙ C p ∙(-73.5)] = ∆T = T f – T i = 23 – 96.5 = Water Q gained [m ∙ C p ∙ ∆T] [(.08)∙ (4.18) ∙(3)] ∆T = T f – T i = 23 – 20 = 3 C p =.682

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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 C p of zinc?

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Metal -Q lost -[m ∙ C p ∙ ∆T] = -[(.04)∙ C p ∙(-99)] = Water -Q gained [m ∙ C p ∙ ∆T] [(.5) ∙ (4.18) ∙(1)] C p =.527 Ch12 HW#2 7 – 10

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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

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Ch12 HW# #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 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 = T f – T i

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Iron -Q lost -[m ∙ C p ∙ ∆T] = -[(2) ∙ C p ∙ (-187)] = Water Q gained [m ∙ C p ∙ ∆T] [(2) ∙ (4.18) ∙ (20)] Ch12 HW# #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 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 = T f – T i

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Iron -Q lost -[m ∙ C p ∙ ∆T] = -[(2) ∙ C p ∙ (-187)] = Water Q gained [m ∙ C p ∙ ∆T] [(2) ∙ (4.18) ∙ (20)] Ch12 HW# #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 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? Lead -Q lost -[m ∙ C p ∙ ∆T] = -[(.4)∙ C p ∙(-69.9)] = Water Q gained [m ∙ C p ∙ ∆T] [(.378)∙ (4.18) ∙(2.3)] ∆T = T f – T i

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#9) A 0.10 kg brass block at 90.0 C is placed in a plastic foam cup containing 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 C p of Al? ∆T = T f – T i

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Brass -Q lost -[m ∙ C p ∙ ∆T] = -[(.1)∙ C p ∙(-64.4)] = Water Q gained [m ∙ C p ∙ ∆T] [(.103)∙ (4.18) ∙(5.6)] #9) A 0.10 kg brass block at 90.0 C is placed in a plastic foam cup containing 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 C p of Al? ∆T = T f – T i

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Brass -Q lost -[m ∙ C p ∙ ∆T] = -[(.1)∙ C p ∙(-64.4)] = Water Q gained [m ∙ C p ∙ ∆T] [(.103)∙ (4.18) ∙(5.6)] Aluminum -Q lost -[m ∙ C p ∙ ∆T] = -[(.100)∙ C p ∙(-75)] = Water Q gained [m ∙ C p ∙ ∆T] [(.100)∙ (4.18) ∙(15)] #9) A 0.10 kg brass block at 90.0 C is placed in a plastic foam cup containing 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 C p of Al?

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Ch12.3 – Heat & Changes of State Time Temp (°C) For H 2 O:

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Ch12.3 – Heat & Changes of State Time Temp (°C) s l 1) Raise temp of solid: Q = m ∙C p ∙ ∆T l g 2) Melt it: Q = m ∙ H f 3) Raise temp of liquid: Q = m ∙C p ∙ ∆T 4) Vaporize it: Q = m ∙ H v 5) Raise temp of gas: Q = m ∙C p ∙ ∆T For H 2 O: 6) Add steps

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

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

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Ex) How much heat is required to turn 63kg of ice at -50.4°C to vapor at 124.3°C? Temp (°C) Time add2 1) Heat ice Q = m ∙Cp ∙ ∆T = (63kg)(2.1 )(50.4°C) = 6.7 kJ kJ kg ∙ °C kJ kg kJ kg ∙ °C kJ kg ∙ °C kJ kg 4) Boil Q = m ∙ H v = (63kg)(2260 ) = kJ 5) Heat gas Q = m ∙Cp ∙ ∆T = (63kg)(2.02 )(24.3°C) = 3.1 kJ 3) Heat water Q = m ∙Cp ∙ ∆T = (63kg)(4.18 )(100°C) = 26.5 kJ 2) Melt Q = m ∙H f = (63kg)(334 ) = 21.1 kJ 6) Add : kJ Ch12 HW#3

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

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

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Ch12 HW# 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?

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Ch12 HW# 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 ice Q = m∙Cp∙∆T = (.10kg)(2.1)(20°C) = 2)Melt Q = m ∙H f = (.10)(334) = 3) Add: 1)Heat water Q = m∙Cp∙∆T = (.20kg)(4.18)(40°C) = 2)Vaporize Q = m ∙H v = (.20)(2260) = 3)Heat gas Q = m∙Cp∙∆T = (.20kg)(2.02)(40°C) = 4) Add: 1)Heat ice Q = m∙Cp∙∆T = (.30kg)(2.1)(30°C) = 2)Melt Q = m ∙H f = (.30)(334) = 3)Heat water Q = m∙Cp∙∆T = (.30kg)(4.18)(100°C) = 4)Vaporize Q = m ∙H v = (.30)(2260) = 5)Heat gas Q = m∙Cp∙∆T = (.30kg)(2.02)(30°C) = 6) Add:

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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 : Q high 2. Gases expand pushing piston down : Work 3. Gases cool as expand : Q low

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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.”

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

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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?

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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 = -1100J W = +850J ∆U = ? Q = W + ∆U -1100J = +850J + ∆U ∆U = -1950J (decrease) Q = ? W = +350J ∆U = +50J Q = W + ∆U = +350J + (+50J) = +400J (heat added) C 12 HW #

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Ch12 HW# J 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? J 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? J 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 = J ∆U = ? Q = J W = -225 J ∆U = +100 J Q = ?

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Ch12 HW# J 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? J 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? J 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 = J ∆U = ? Q = J Q = W + ∆U = ∆U ∆U = W = -225 J ∆U = +100 J Q = ? Q = W + ∆U = -225J + (+100J) Q =

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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 = J W = J ∆U = ? W = ? Q = J ∆U = +300 J Ch12 HW#

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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 = J W = J ∆U = ? Q = W + ∆U -850J = -250J + ∆U ∆U = W = ? Q = J ∆U = +300 J Q = W + ∆U +525J = W + (+300J) W = Ch12 HW#

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Ch12.5 – Efficiency Efficiency of a heat engine:eff = x 100% T H – T L T H Temps must be in Kelvins! (C = K) Q W + ∆U goodbad 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?

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Ch12.5 – Efficiency Efficiency of a heat engine:eff = x 100% T H – T L T H Temps must be in Kelvins! (C = K) Q W + ∆U goodbad 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 eff = x 100% = 55% 823K – 373K 823K

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Ch12 HW# Calculate the efficiency of a heat engine that operates between 350 o C and 50 O C. 20. A heat engine has an input temp of 3250 o C and an exhaust temp of 1125 o C. What is the its ideal efficiency? T H = 350°C = 623K T L = 50°C = 323K T H = 3250°C = 3523K T L = 1125°C = 1398K

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Ch12 HW# Calculate the efficiency of a heat engine that operates between 350 o C and 50 O C. 20. A heat engine has an input temp of 3250 o C and an exhaust temp of 1125 o C. What is the its ideal efficiency? T H = 350°C = 623K T L = 50°C = 323K T H = 3250°C = 3523K T L = 1125°C = 1398K 623K – 323K 623K eff = x 100% = eff = x 100% T H – T L T H 3523K – 1398K 3523K eff = x 100% = eff = x 100% T H – T L T H

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21. How much work is done on/by a system that has 1000J of heat added to it while its internal energy increases by 800J? 22. If the system from #21 has an input temp of 500K and an output temp of 400K, what is its efficiency? W = ? Q = +1000J ∆U = +800J

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21. How much work is done on/by a system that has 1000J of heat added to it while its internal energy increases by 800J? 22. If the system from #21 has an input temp of 500K and an output temp of 400K, what is its efficiency? W = ? Q = +1000J ∆U = +800J Q = W + ∆U +1000J = W + (+800J) W = 1000K – 800K 1000K eff = x 100% = eff = x 100% T H – T L T H

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23. A system does 500J of work on its surroundings while its internal energy increases by 1500J. How much heat energy was added to/taken away? W = +500J ∆U = +1500J Q = ?

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23. A system does 500J of work on its surroundings while its internal energy increases by 1500J. How much heat energy was added to/taken away? W = +500J ∆U = +1500J Q = ? Q = W + ∆U Q = (+500J) + (+1500J) Q =

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Ch13.1 – Properties of Matter Pressure = Force Area P = FAFA (units: ) Standard Atmospheric Pressure:

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Ch13.1 – Properties of Matter Pascal’s Principle – if a pressure is exerted on a fluid, that same pressure gets exerted throughout the fluid. Fluids – substances that flow. ( Liquids & Gases) Input piston Output piston Pressure = Force Area P = FAFA (units: )(pascals) Standard Atmospheric Pressure: 101.3kPa / 14.7psi / 1 atm / 760 mm Hg N m 2

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Ch13.1 – Pressure Pascal’s Principle – if a pressure is exerted on a fluid, that same pressure gets exerted throughout the fluid. Fluids – substances that flow. ( Liquids & Gases) F in F out Input piston Output piston Pressure = Force Area P = FAFA (units: )(pascals) Standard Atmospheric Pressure: 101.3kPa / 14.7psi / 1 atm / 760 mm Hg N m 2 F in A in F out A out = same pressure =

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P = FAFA Ex1) Your head and shoulders have an area of about 0.08 m 2 as seen from above. If air pressure is kPa, What is the force of the air on your body?

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P = FAFA Ex1) Your head and shoulders have an area of about 0.08 m 2 as seen from above. If air pressure is kPa, What is the force of the air on your body? P = kPa = 101,300 Pa F = P ∙ A = (101, )(.08m 2 ) = 8104 N Nm2Nm2 Ex2) A women wears high heels. If she stands on the heels that have a total area of m 2, and she weighs 495N, what pressure is exerted on the floor?

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P = FAFA Ex1) Your head and shoulders have an area of about 0.08 m 2 as seen from above. If air pressure is kPa, What is the force of the air on your body? P = kPa = 101,300 Pa F = P ∙ A = (101, )(.08m 2 ) = 8104 N Nm2Nm2 Ex2) A women wears high heels. If she stands on the heels that have a total area of m 2, and she weighs 495N, what pressure is exerted on the floor? FAFA P = = = 2,475,000 Pa 495N m 2

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Ex3) A dentist’s chair uses a hydraulic-lift system. The input piston has a cross-sectional area of 72cm 2, and the output piston has an area of 1440cm 2. If the dentist wants to lift an 800N patient, what force must be applied?

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Ex4) A hydraulic floor jack consists of a handle attached to an input piston with an area of 3cm 2. A person applies an input force of 200N to lift a car that weighs 20,000N. What is the area of the output piston? Ch13 HW#1 1 – 6

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1. The atmospheric pressure at sea level is about 1.0x10 5 Pa. What is the force at sea level that air exerts on the top of a typical office desk, 1.52m long and 0.76 m wide? P = 1.0x10 5 Pa A = 1.52m x.76m = 1.16m 2 F = ? FAFA P = -----

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Ch13 HW#1 1 – 6 1. The atmospheric pressure at sea level is about 1.0x10 5 Pa. What is the force at sea level that air exerts on the top of a typical office desk, 1.52m long and 0.76 m wide? P = 1.0x10 5 Pa A = 1.52m x.76m = 1.16m 2 F = ? FAFA P = -----F = P ∙ A = (1.0 x )(1.16m 2 ) =115,520 N Nm2Nm2

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2. What is the force the air exerts on your right big toe nail (area of m 2 ) when air pressure is 1.0x10 5 Pa. 3. A car tire makes contact with the ground on a rectangular area of 0.12 m by 0.18 m. The car’s mass is 925 kg. What pressure does the car exert on the ground?

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2. What is the force the air exerts on your right big toe nail (area of m 2 ) when air pressure is 1.0x10 5 Pa. 3. A car tire makes contact with the ground on a rectangular area of 0.12 m by 0.18 m. The car’s mass is 925 kg. What pressure does the car exert on the ground? F = P ∙ A = (1.0 x )(0.0002m 2 ) = Nm2Nm2 F = m ∙ g = (925 kg)(9.8 ) = 9065 N A = 4 x (.12m x.18m) =.0864 m 2 m/s 2 P = = = 104,919 Pa FAFA 9065N.0864m 2

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4. The teacher’s chair uses a hydraulic-lift system. The input piston has an area of 2cm 2. The output piston has an area of 10cm 2. If the teacher wants to lift his 700N self, what force must he apply to the handle? Pascal’s Principle: Pressure is the same throughout the system. P in = P out F in F out A in A out 5. A ‘cherry-picker’ is used to pull engines out of cars. It consists of a handle attached to an input piston with an area of 5cm 2. A person applies an input force of 100N to lift an engine that weighs 1000N. What is the area of the output piston?

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4. The teacher’s chair uses a hydraulic-lift system. The input piston has an area of 2cm 2. The output piston has an area of 10cm 2. If the teacher wants to lift his 700N self, what force must he apply to the handle? Pascal’s Principle: Pressure is the same throughout the system. P in = P out F in F out F in 700N A in A out 2cm 2 10cm 2 5. A ‘cherry-picker’ is used to pull engines out of cars. It consists of a handle attached to an input piston with an area of 5cm 2. A person applies an input force of 100N to lift an engine that weighs 1000N. What is the area of the output piston? P in = P out F in F out A in A out = = =

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5. A ‘cherry-picker’ is used to pull engines out of cars. It consists of a handle attached to an input piston with an area of 5cm 2. A person applies an input force of 100N to lift an engine that weighs 1000N. What is the area of the output piston? P in = P out F in F out 100N 1000N A in A out 5cm 2 A out 6. A hydraulic crane consists of an arm attached to a motor that applies an input force of 1000N to lift a beam that weighs 50,000N. If the area of the output piston is 2000cm 2, what is the area of the input piston? P in = P out F in F out A in A out == =

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6. A hydraulic crane consists of an arm attached to a motor that applies an input force of 1000N to lift a beam that weighs 50,000N. If the area of the output piston is 2000cm 2, what is the area of the input piston? P in = P out F in F out 1000N 50,000N A in A out A in 2000cm 2 = =

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Ch13.2 – Fluid Pressure Continuity Equation – when a fluid passes thru a smaller area, its speed increases. A1A1 A2A2 A 1. vel 1 = A 2. vel 2

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Continuity Equation – when a fluid passes thru a smaller area, its speed increases. Bernoulli’s Principle – as the velocity of a fluid increases, the pressure the fluid exerts sideways, decreases. A1A1 A2A2 A 1. vel 1 = A 2. vel 2 fluid Higher velocity Lower velocity

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Bernoulli’s Principle – as the velocity of a fluid increases, the pressure the fluid exerts sideways, decreases. Density mass m Volume V Density = D = or: m = D. V

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Static Fluid Pressure – the weight of a fluid creates pressure on objects. FAFA P = ----

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Static Fluid Pressure – the weight of a fluid creates pressure on objects. Ex1) What is the pressure at the bottom of my 80cm tube? F m. g (D. V). g D. (πr 2 h)g A A πr 2 πr 2 P = ---- = = = (Density of water P = D. g. h = 1000kg/m 3 ) Pressure = (Density)x(gravity)x(height of fluid)

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Bernoulli Equation – A pressure on a fluid can cause a velocity where the fluid escapes. Pressure due to height = Pressure at the outlet P height = P hole D. g. h = ½D. v 2 (This isnt completely true.) Ex2) A tube 50cm tall is filled with water. It has a small hole drilled into it,10 cm from its base. What is the velocity of the fluid as it approaches the hole? Ch13 HW#2 7 – 12

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7. What is the pressure at the bottom of a swimming pool 3m deep? 8. Behind my house is a water tank on the hill. Using the surveying equipment from the beginning of the year, I found the top of the tank to be 50m above my sink. What water pressure could my sink have? 9. Based on #8, what speed does the water travel through the pipes?

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Ch13 HW#2 7 – What is the pressure at the bottom of a swimming pool 3m deep? P = Dgh = (1000kg/m 3 )(9.8m/s 2 )(3m) = 8. Behind my house is a water tank on the hill. Using the surveying equipment from the beginning of the year, I found the top of the tank to be 50m above my sink. What water pressure could my sink have? P height = P hole D. g. h = ½D. v 2 (1000)(9.8)(50) = 490,000 Pa = ½(1000)v 2 v = 9. Based on #8, what speed does the water travel through the pipes?

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10. How tall would a column of water have to be to provide 1 atmosphere of pressure? (1atm = 101,300Pa) P = D. g. h 11. In our next lab, it will be determined that water is flowing from a hole in a pipe at a speed of about 2.5m/s. What height of water is needed to achieve this? 12. If a tank full of water is 10m tall, how fast will water flow towards a hole at the bottom?

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10. How tall would a column of water have to be to provide 1 atmosphere of pressure? (1atm = 101,300Pa) P = D. g. h 101,300 = (1000)(9.8)h h = 11. In our next lab, it will be determined that water is flowing from a hole in a pipe at a speed of about 2.5m/s. What height of water is needed to achieve this? D. g. h = ½D. v 2 (9.8)h = ½(2.5) 2 h = 12. If a tank full of water is 20m tall, how fast will water flow towards a hole at the bottom? D. g. h = ½D. v 2 (9.8)(20) = ½(v) 2 v =

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Archimede’s Principle – Place an object in the fluid, there is a buoyant force that pushes up on an object. (makes it feel lighter.) Ch 13.3 Buoyant Force Ex1) A rock weighs 4.0N. When lowered into water, it weighs 2.5N. What is the buoyant force acting on it?

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Archimede’s Principle – Place an object in the fluid, there is a buoyant force that pushes up on an object. (makes it feel lighter.) Ch 13.2 Buoyant Force Ex1) A rock weighs 4.0N. When lowered into water, it weighs 2.5N. What is the buoyant force acting on it? FBFB FgFg F NET = F g – F B 2.5N = 4.0N - F B F B = 1.5N The buoyant force is equal to the weight of the fluid displaced.

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- find the volume (V) of fluid displaced, then use its density to find its mass D = mVmV m = D ∙ V The buoyant force is equal to the weight of the fluid displaced.

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- find the volume (V) of fluid displaced, then use its density to find its mass D = mVmV m = D ∙ V F g = m ∙ g F B = D ∙ V ∙ g Weight of fluid displaced is the buoyant force ∆V -density of water : 1 g/mL or 1000kg/m 3 The buoyant force is equal to the weight of the fluid displaced.

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Ex2) A mass is dropped into water causing the water level to rise from 500mL to 625mL. What is the buoyant force acting on the mass? Ex3) A piece of Aluminum has a volume of 0.002m 3. It is submerged in water. a) What is the magnitude of the buoyant force? b) If the density of aluminum is 2700 kg/m 3, what is its apparent weight? ∆V=125mL

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Ex2) A mass is dropped into water causing the water level to rise from 500mL to 625mL. What is the buoyant force acting on the mass? Ex3) A piece of Aluminum has a volume of 0.002m 3. It is submerged in water. a) What is the magnitude of the buoyant force? b) If the density of aluminum is 2700 kg/m 3, what is its apparent weight? F net = F g – F B = m. g – 20N = 53N – 20N = 33N Ch13 HW#3 13 – 17 F B = D ∙ V ∙ g = [(1 g/mL)(125mL)](9.8m/s 2 ) = [125g](9.8m/s 2 ) [.125kg](9.8m/s 2 ) = 1.25N F B = D ∙ V ∙ g = [(1000 kg/m 3 )(0.002m 3 )](9.8m/s 2 ) = 20N

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Ch13 HW#3 13 – A rock, weighed with a spring scale, weighs 6.0 N in air. It is lowered into an aquarium, and the spring scale reads 4.0 N. What is the buoyant force of the water acting on the rock? 14. A rock is lowered into a beaker of water and the water level rises by 250 mL. If the density of water is 1.0 g/mL, what is the buoyant force acting on the rock?

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13. A rock, weighed with a spring scale, weighs 6.0 N in air. It is lowered into an aquarium, and the spring scale reads 4.0 N. What is the buoyant force of the water acting on the rock? 14. A rock is lowered into a beaker of water and the water level rises by 250 mL. If the density of water is 1.0 g/mL, what is the buoyant force acting on the rock? F NET = F g – F B 4.0N = 6.0N - F B F B = D ∙ V ∙ g = [(1 g/mL)(250mL)](9.8m/s 2 ) = [250g](9.8m/s 2 ) [.250kg](9.8m/s 2 ) =

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15. An iron cylinder has a volume of m 3. It is submerged in water. a. What buoyant force acts on it? b. If iron has a density of 7900 kg/m 3, what is its apparent weight in water? F net = F g – F B = 16. A titanium cylinder has a volume of m 3. It is submerged in water. a. What buoyant force acts on it? b. If titaniium has a density of 4500 kg/m3, what is its apparent weight in water? F B = D ∙ V ∙ g =

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15. An iron cylinder has a volume of m 3. It is submerged in water. a. What buoyant force acts on it? b. If iron has a density of 7900 kg/m 3, what is its apparent weight in water? F net = F g – F B = m. g – 20N = 387N – 49N = 16. A titanium cylinder has a volume of m 3. It is submerged in water. a. What buoyant force acts on it? b. If titaniium has a density of 4500 kg/m3, what is its apparent weight in water? F B = D ∙ V ∙ g = [(1000 kg/m 3 )(0.005m 3 )](9.8m/s 2 ) = 49N

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16. A titanium cylinder has a volume of m 3. It is submerged in water. a. What buoyant force acts on it? b. If titaniium has a density of 4500 kg/m3, what is its apparent weight in water? F net = F g – F B = m. g – 20N = 387N – 49N = 17. A girl weighs 600N. If she is floating in a freshwater lake. What is the buoyant force acting on her? How much water does she displace? F B = D ∙ V ∙ g = [(1000 kg/m 3 )(0.010m 3 )](9.8m/s 2 ) = 98N

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16. A titanium cylinder has a volume of m 3. It is submerged in water. a. What buoyant force acts on it? b. If titaniium has a density of 4500 kg/m3, what is its apparent weight in water? F net = F g – F B = m. g – 20N = 387N – 49N = 17. A girl weighs 600N. If she is floating in a freshwater lake. What is the buoyant force acting on her? How much water does she displace? F B = D ∙ V ∙ g = [(1000 kg/m 3 )(0.010m 3 )](9.8m/s 2 ) = 98N F net = F g – F B 0 = 600N – F B F B = 600N F B = D ∙ V ∙ g 600N = (1000 kg/m 3 )(V)(9.8m/s 2 ) V =

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Ch12,13 Rev 1. Make the following conversions o C to Kelvins _______ 2. 0 K to o C ______ o C to Kelvins _______ o C to K______ 5. 70K to o C _______ K to o C ______ 2. How much heat is absorbed by kg of copper when its temp is raised from 20.0 o C to 80.0 o C ? (C p for copper when is.385 kJ/kg K) 3. A 0.100kg aluminum block at C is placed in 0.100kg of water at 10.0 o C. The final temp of the mixture is 25.0 o C. What is the specific heat of the aluminum? ( C pwater = kJ/kg K)

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4. A kg sample of water at 50.0 C is heated to steam at C. How much heat is absorbed? (C pwater = 4.18 kJ/kg. C, C psteam = 2.01 kJ/kg. C, H vap for H 2 O is 2260 kJ/kg) 5. How much pressure does a car exert on frozen ice if it has a weight of 10,000 N, exerted over the four tires that have a combined area of 0.5 m 2 ?

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kJ of heat are put into a system that in turn has its internal energy raised by 15 kJ. How much work is done on or by the system? 7. The Karman Line is considered to be the boundary line for the atmosphere by some. It is 100,000km above the earth’s surface. If the average density of air is 1.225kg/m 3, what pressure does it exert at the earth’s surface?

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8. A hydraulic floor jack consists of an arm attached to an input piston. If the area of the input piston is 4cm 2 and you apply an input force of 50N to lift a quad that weighs 5000N, what is the area of the output piston? 9. Water emerges from the bottom of a water tower at a speed of 25m/s. How tall is the column of water in the tower?

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10. A 45 N iron block measures 10 cm by 5 cm by 5 cm. What is its apparent weight in water? 11. A kg piece of zinc at C is placed in a calorimeter full of kg of water at 20.0 C. The temp levels at 21.1 C. Find the C p of zinc.

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