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
How cold is cold?
Vol
-300
oC
+ 273 = K
0 100
Temp (°C)
Since (-) #’s stink
Kelvin designed
a thermometer,
modeled after
Celsius, but put zero
at coldest temp.
*Kelvin is always bigger
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
How cold is cold?
Vol
-300
oC
+ 273 = K
0 100
Temp (°C)
Since (-) #’s stink
Kelvin designed
a thermometer,
modeled after
Celsius, but put zero
at coldest temp.
*Kelvin is always bigger
Absolute zero – coldest temperature. No molecular motion
0oK = -273oC
Exs) Body Temp = 37oC = 310K
273oK = 0oC
400K = 127 oC
298oK = 25oC
373oK = 100oC
Outer space
2-4 K
Laboratory
0.00001 K
.00000002K
Heat (Q) – Energy that flows between 2 objects
Q is (-)
Q is (+)
heat left object (feels hot)
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
Specific Heat
--different materials have different abilities to gain and lose energy.
Heat Energy = mass ∙ specific heat ∙ change in temp
Q = m∙Cp∙∆T
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
Specific Heat
--different materials have different abilities to gain and lose energy.
Heat Energy = mass ∙ specific heat ∙ change in temp
Q = m∙Cp∙∆T
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 = 10.0kg
Q = (10.0 kg) (4.18kJ/kg∙˚C)(5˚C)
∆T = 5˚ C
Cp = 4.18
=+209 kJ
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.
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
Lab12.1 – Charles Law
- due tomorrow
- Ch12 HW#1 due at beginning of period
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. -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 =
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. -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 =
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. -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 =
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. -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 =
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.
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.
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.
Q = m ∙ Cp ∙ ∆T
kJ
= (.00006 kg)(.710
)(60K)
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.
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.
Q = m ∙ Cp ∙ ∆T
kJ
= (.00006 kg)(.710
)(60K)
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.
∆T =
Q
m ∙ Cp
=
836 kJ
=
(20kg)(4.18 kJ )
kg∙ K
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?
kJ
Cp of water is 4.18
kg ∙ °C
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?
kJ
Cp of water is 4.18
kg ∙ °C
∆T = Tf – Ti
= 23 – 96.5
= -73.5
Metal
Water
-Qlost
Qgained
-[m ∙ Cp ∙ ∆T]
= [m ∙ Cp ∙ ∆T]
-[(.02)∙ Cp ∙(-73.5)] = [(.08)∙ (4.18) ∙(3)]
Cp = .682
∆T = Tf – Ti
= 23 – 20
=3
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?
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
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
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?
∆T = Tf – Ti
∆T = Tf – Ti
#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?
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?
Iron
Water
∆T = Tf – Ti
-Qlost
Qgained
∆T = Tf – Ti
-[m ∙ Cp ∙ ∆T]
=
[m ∙ Cp ∙ ∆T]
-[(2) ∙ Cp ∙ (-187)] = [(2) ∙ (4.18) ∙ (20)]
#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 = T – T
f
i
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?
Iron
Water
∆T = Tf – Ti
-Qlost
Qgained
∆T = Tf – Ti
-[m ∙ Cp ∙ ∆T]
=
[m ∙ Cp ∙ ∆T]
-[(2) ∙ Cp ∙ (-187)] = [(2) ∙ (4.18) ∙ (20)]
#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?
Lead
Water
∆T = Tf – Ti
∆T = Tf – Ti
-Qlost
Qgained
-[m ∙ Cp ∙ ∆T] =
[m ∙ Cp ∙ ∆T]
-[(.4)∙ Cp ∙(-69.9)] = [(.378)∙ (4.18) ∙(2.3)]
#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?
∆T = Tf – Ti
∆T = Tf – Ti
#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?
#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?
Brass
Water
-Qlost
Qgained
∆T = Tf – Ti
∆T = Tf – Ti
-[m ∙ Cp ∙ ∆T]
= [m ∙ Cp ∙ ∆T]
-[(.1)∙ Cp ∙(-64.4)] = [(.103)∙ (4.18) ∙(5.6)]
#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
#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?
Brass
Water
-Qlost
Qgained
-[m ∙ Cp ∙ ∆T]
= [m ∙ Cp ∙ ∆T]
-[(.1)∙ Cp ∙(-64.4)] = [(.103)∙ (4.18) ∙(5.6)]
#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?
Aluminum
-Qlost
-[m ∙ Cp ∙ ∆T]
-[(.100)∙ Cp ∙(-75)]
Water
Qgained
= [m ∙ Cp ∙ ∆T]
= [(.100)∙ (4.18) ∙(15)]
Ch12.3 – Heat & Changes of State
120
For H2O:
100
Temp
(°C)
0
-10
Time
Ch12.3 – Heat & Changes of State
120
5) Raise temp of gas:
l
g
Q = m ∙Cp ∙ ∆T
4) Vaporize it: Q = m ∙ Hv
For H2O:
100
Temp
(°C)
0
-10
3) Raise temp of liquid:
Q = m ∙Cp ∙ ∆T
s
l
2) Melt it: Q = m ∙ Hf
1) Raise temp of solid: Q = m ∙Cp ∙ ∆T
Time
6) Add steps
Heat of Fusion – amount of heat needed to melt a substance.
Temp levels as all energy goes to break solid apart.
Q = m ∙ Hf
( for water , Hf = 334
kJ
)
kg
Heat of vaporization – amt of heat needed to vaporize.
Temp levels as all energy goes to break bonds.
Q = m ∙ Hv
( for water , Hv = 2260
Cp for ice = 2.1
kJ
kg ∙ K
Cp for steam = 2.02
kJ
kg ∙ K
kJ
)
kg
Ex) How much heat is required to turn 63kg of ice at -50.4°C to vapor at
124.3°C?
Temp
(°C)
Time
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
1) Heat ice Q = m ∙Cp ∙ ∆T = (63kg)(2.1
)(50.4°C) = 6.7 kJ
kg ∙ °C
kJ
2) Melt
Q = m ∙Hf
= (63kg)(334
) = 21.1 kJ
kg
3) Heat water Q = m ∙Cp ∙ ∆T = (63kg)(4.18 kJ )(100°C) = 26.5 kJ
kg ∙ °C
kJ
4) Boil
Q = m ∙ Hv
= (63kg)(2260
) = 142.9 kJ
kg
5) Heat gas Q = m ∙Cp ∙ ∆T = (63kg)(2.02 kJ )(24.3°C) = 3.1 kJ
kg ∙ °C
6) Add :
200.2 kJ
Ch12 HW#3
Temperature of sample (°C)
Temperature of water versus time as thermal energy is removed
100
50
Vapor Changing
into a liquid
Liquid
0
Vapor
Time elapsed
Liquid
changing into
a solid
solid
Lab 12.3 – Heat of Fusion
- due tomorrow
- Ch12 HW#3 due at beginning of period
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?
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?
1)Heat ice Q = m∙Cp∙∆T = (.10kg)(2.1)(20°C) =
2)Melt
Q = m ∙Hf = (.10)(334) =
3) Add:
12. A 0.20kg sample of water at 60°C is heated to steam at 140.0°C.
How much heat is absorbed?
1)Heat water Q = m∙Cp∙∆T = (.20kg)(4.18)(40°C) =
2)Vaporize Q = m ∙Hv = (.20)(2260) =
3)Heat gas Q = m∙Cp∙∆T = (.20kg)(2.02)(40°C) =
4) Add:
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 = (.30kg)(2.1)(30°C) =
2)Melt Q = m ∙Hf = (.30)(334) =
3)Heat water Q = m∙Cp∙∆T = (.30kg)(4.18)(100°C) =
4)Vaporize Q = m ∙Hv = (.30)(2260) =
5)Heat gas Q = m∙Cp∙∆T = (.30kg)(2.02)(30°C) =
6) Add:
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
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.”
1st law formulas:
Qlow
(Entropy)
Q = W + ∆U
Heat
Work
transferred
done
to/from
by/on
system
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
Internal
Energy
lost/gained
by system
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?
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?
Q=?
Q = W + ∆U
W = -200J
= -200J + (+150J)
∆U = +150 J
= - 50J (taken away)
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?
Q = W + ∆U
Q = -1100J
-1100J = +850J + ∆U
W = +850J
∆U = -1950J (decrease)
∆U = ?
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 + ∆U
Q=?
= +350J + (+50J)
W = +350J
= +400J (heat added)
∆U = +50J
C 12 HW #4
14-18
Ch12 HW#4 14-18
14.134J 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?
W = -134 J
∆U = +93 J
Q=?
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?
W = +1230 J
∆U = ?
Q = +2050 J
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 = -225 J
∆U = +100 J
Q=?
Ch12 HW#4 14-18
14.134J 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?
W = -134 J
Q = -134 J + (+93J)
∆U = +93 J
Q=
Q=?
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?
W = +1230 J
Q = W + ∆U
∆U = ?
+2050 = +1230 + ∆U
Q = +2050 J
∆U =
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?
Q = W + ∆U
W = -225 J
= -225J + (+100J)
∆U = +100 J
Q=
Q=?
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?
Q = - 850 J
W = - 250 J
∆U = ?
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?
W=?
Q = + 525 J
∆U = +300 J
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?
Q = - 850 J
Q = W + ∆U
W = - 250 J
-850J = -250J + ∆U
∆U = ?
∆U =
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?
W=?
Q = + 525 J
∆U = +300 J
Q = W + ∆U
+525J = W + (+300J)
W=
Ch12.5 – Efficiency
Efficiency of a heat engine:
TH – TL
eff = ---------x 100%
TH
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?
Ch12.5 – Efficiency
Efficiency of a heat engine:
TH – TL
eff = ---------x 100%
TH
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
eff = -------------- x 100% = 21%
473K
Ex 2) A heat engine has an input temp of 550°C and an
exhaust temp of 100°C. What is its ideal efficiency?
eff =
Ch12 HW#5 19 – 23
823K – 373K
-------------- x 100% = 55%
823K
Ch12 HW#5 19-23
19. Calculate the efficiency of a heat engine that operates between 350oC
and 50OC.
TH = 350°C + 273 = 623K
TL = 50°C + 273 = 323K
20. A heat engine has an input temp of 3250oC and an exhaust temp
of 1125oC. What is the its ideal efficiency?
TH = 3250°C + 273 = 3523K
TL = 1125°C + 273 = 1398K
Ch12 HW#5 19-23
19. Calculate the efficiency of a heat engine that operates between 350oC
and 50OC.
H – TL
TH = 350°C + 273 = 623K
eff =T---------x 100%
TH
TL = 50°C + 273 = 323K
– 323K x 100% =
eff =623K
-------------623K
20. A heat engine has an input temp of 3250oC and an exhaust temp
of 1125oC. What is the its ideal efficiency?
TH = 3250°C + 273 = 3523K
TL = 1125°C + 273 = 1398K
H – TL
eff =T---------x 100%
TH
3523K – 1398K
eff = -------------- x 100% =
3523K
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?
W=?
Q = +1000J
∆U = +800J
22. If the system from #21 has an input temp of 500K and an output temp
of 400K, what is its efficiency?
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?
W=?
Q = +1000J
∆U = +800J
Q = W + ∆U
+1000J = W + (+800J)
W=
22. If the system from #21 has an input temp of 500K and an output temp
of 400K, what is its efficiency?
H – TL
eff =T---------x 100%
TH
– 800K x 100% =
eff = 1000K
-------------1000K
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=?
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=
Ch13.1 – Properties of Matter
Force
Pressure = ---------Area
Standard Atmospheric Pressure:
F
P = ----A
(units:
)
Ch13.1 – Properties of Matter
Force
Pressure = ---------Area
F
P = ----A
N
(units: -----(pascals) )
m2
Standard Atmospheric Pressure: 101.3kPa / 14.7psi / 1 atm / 760 mm Hg
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
Ch13.1 – Pressure
Force
Pressure = ---------Area
F
P = ----A
N
(units: -----(pascals) )
m2
Standard Atmospheric Pressure: 101.3kPa / 14.7psi / 1 atm / 760 mm Hg
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)
Fin
Fout
Input piston
Output piston
Fout
Fin
Ain = same pressure = Aout
F
P = ----A
Ex1) Your head and shoulders have an area of about 0.08 m2 as
seen from above. If air pressure is 101.3 kPa,
What is the force of the air on your body?
F
P = ----A
Ex1) Your head and shoulders have an area of about 0.08 m2 as
seen from above. If air pressure is 101.3 kPa,
What is the force of the air on your body?
P = 101.3 kPa
F=P∙A
N
= 101,300 Pa
2)
= (101,300 -----)(.08m
= 8104 N
m2
Ex2) A women wears high heels. If she stands on the heels that have a
total area of 0.0002m2, and she weighs 495N, what pressure is exerted
on the floor?
F
P = ----A
Ex1) Your head and shoulders have an area of about 0.08 m2 as
seen from above. If air pressure is 101.3 kPa,
What is the force of the air on your body?
P = 101.3 kPa
F=P∙A
N
= 101,300 Pa
2)
= (101,300 -----)(.08m
= 8104 N
m2
Ex2) A women wears high heels. If she stands on the heels that have a
total area of 0.0002m2, and she weighs 495N, what pressure is exerted
on the floor?
F = ---------495N
P = ----= 2,475,000 Pa
2
A
0.0002m
Ex3) A dentist’s chair uses a hydraulic-lift system. The input piston
has a cross-sectional area of 72cm2, and the output piston has an area
of 1440cm2. If the dentist wants to lift an 800N patient, what force must
be applied?
Ex4) A hydraulic floor jack consists of a handle attached to an input
piston with an area of 3cm2. 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
Ch13 HW#1 1 – 6
1. The atmospheric pressure at sea level is about 1.0x105 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 = 1.0x105 Pa
A = 1.52m x .76m = 1.16m2
F=?
F
P = ----A
Ch13 HW#1 1 – 6
1. The atmospheric pressure at sea level is about 1.0x105 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 = 1.0x105 Pa
A = 1.52m x .76m = 1.16m2
F=?
F
P = ----A
N )(1.16m2)
F = P ∙ A = (1.0 x 105---m2
=115,520 N
2. What is the force the air exerts on your right big toe nail
(area of 0.0002 m2) when air pressure is 1.0x105 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?
2. What is the force the air exerts on your right big toe nail
(area of 0.0002 m2) when air pressure is 1.0x105 Pa.
N )(0.0002m2) =
F = P ∙ A = (1.0 x 103 -----m2
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 = m ∙ g = (925 kg)(9.8 m/s2 ) = 9065 N
A = 4 x (.12m x .18m) = .0864 m2
F = --------9065N = 104,919 Pa
P = ----A
.0864m2
4. The teacher’s chair uses a hydraulic-lift system. The input piston has
an area of 2cm2. The output piston has an area of 10cm2. 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.
Pin = Pout
Fin
Ain
Fout
Aout
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 5cm2. A person
applies an input force of 100N to lift an engine that weighs 1000N. What
is the area of the output piston?
4. The teacher’s chair uses a hydraulic-lift system. The input piston has
an area of 2cm2. The output piston has an area of 10cm2. 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.
Pin = Pout
Fin Fout
Ain = Aout
Fin
700N
2cm2 =10cm2
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 5cm2. A person
applies an input force of 100N to lift an engine that weighs 1000N. What
is the area of the output piston?
Pin = Pout
Fin Fout
=
Ain Aout
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 5cm2. A person
applies an input force of 100N to lift an engine that weighs 1000N. What
is the area of the output piston?
Pin = Pout
Fin Fout
Ain = Aout
100N 1000N
5cm2 = Aout
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 2000cm2 , what is the area of the input
piston?
Pin = Pout
Fin Fout
=
Ain Aout
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 2000cm2, what is the area of the input
piston?
Pin = Pout
Fin Fout
Ain = Aout
1000N 50,000N
Ain = 2000cm2
Ch13.2 – Fluid Pressure
Continuity Equation – when a fluid passes thru a smaller area,
its speed increases.
A1.vel1 = A2.vel2
A1
A2
Continuity Equation – when a fluid passes thru a smaller area,
its speed increases.
A1.vel1 = A2.vel2
Bernoulli’s Principle – as the velocity of a fluid increases, the pressure
the fluid exerts sideways, decreases.
A1
Lower
velocity
Higher velocity
fluid
A2
Bernoulli’s Principle – as the velocity of a fluid increases, the pressure
the fluid exerts sideways, decreases.
Density
mass
Density = ----------Volume
m
D = ----V
or: m = D.V
Static Fluid Pressure – the weight of a fluid creates pressure on objects.
F
P = ---A
Static Fluid Pressure – the weight of a fluid creates pressure on objects.
.g
.V).g
.(πr2h)g
F
m
(D
D
P = ---- = -------- = --------= ------------2
A
A
πr
πr2
P = D.g.h
(Density of water
= 1000kg/m3)
Pressure = (Density)x(gravity)x(height of fluid)
Ex1) What is the pressure at the bottom of my 80cm tube?
Bernoulli Equation – A pressure on a fluid can cause a velocity
where the fluid escapes.
Pressure due to height = Pressure at the outlet
Pheight = Phole
D.g.h = ½D.v2
(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
Ch13 HW#2 7 – 12
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?
Ch13 HW#2 7 – 12
7. What is the pressure at the bottom of a swimming pool 3m deep?
P = Dgh = (1000kg/m3)(9.8m/s2)(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?
Pheight = Phole
D.g.h = ½D.v2
(1000)(9.8)(50) =
490,000 Pa = ½(1000)v2
v=
9. Based on #8, what speed does the water travel through the pipes?
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?
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.v2
(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.v2
(9.8)(20) = ½(v)2
v=
Ch 13.3 Buoyant Force
Archimede’s Principle – Place an object in the fluid, there is a
buoyant force that pushes up on an object. (makes it feel lighter.)
Ex1) A rock weighs 4.0N. When lowered into water, it weighs 2.5N.
What is the buoyant force acting on it?
Ch 13.2 Buoyant Force
Archimede’s Principle – Place an object in the fluid, there is a
buoyant force that pushes up on an object. (makes it feel lighter.)
Ex1) A rock weighs 4.0N. When lowered into water, it weighs 2.5N.
What is the buoyant force acting on it?
FNET = Fg – FB
2.5N = 4.0N - FB
FB = 1.5N
FB
Fg
The buoyant force is equal to the weight of the fluid displaced.
The buoyant force is equal to the weight of the fluid displaced.
- find the volume (V) of fluid displaced,
then use its density to find its mass
m
D = ----m=D∙V
V
The buoyant force is equal to the weight of the fluid displaced.
- find the volume (V) of fluid displaced,
then use its density to find its mass
m
D = ----m=D∙V
V
Weight of fluid
displaced
is the buoyant force
Fg = m ∙ g
FB = D ∙ V ∙ g
∆V
-density of water : 1 g/mL or 1000kg/m3
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?
∆V=125mL
Ex3) A piece of Aluminum has a volume of 0.002m3. It is submerged in water.
a) What is the magnitude of the buoyant force?
b) If the density of aluminum is 2700 kg/m3, what is its apparent weight?
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?
FB = D ∙ V ∙ g
= [(1 g/mL)(125mL)](9.8m/s2)
= [125g](9.8m/s2) [.125kg](9.8m/s2)
= 1.25N
Ex3) A piece of Aluminum has a volume of 0.002m3. It is submerged in water.
a) What is the magnitude of the buoyant force?
FB = D ∙ V ∙ g
= [(1000 kg/m3)(0.002m3)](9.8m/s2)
= 20N
b) If the density of aluminum is 2700 kg/m3, what is its apparent weight?
Fnet = Fg – FB
= m.g – 20N
= 53N – 20N = 33N
Ch13 HW#3 13 – 17
m
 m  D V
V
 (2700 kg 3 )( 0.002m 3 )
m
 5.3kg
DAl 
Ch13 HW#3 13 – 17
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?
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?
FNET = Fg – FB
4.0N = 6.0N - FB
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?
FB = D ∙ V ∙ g
= [(1 g/mL)(250mL)](9.8m/s2)
= [250g](9.8m/s2) [.250kg](9.8m/s2)
=
15. An iron cylinder has a volume of 0.005 m3. It is submerged in water.
a. What buoyant force acts on it?
b. If iron has a density of 7900 kg/m3, what is its apparent
weight in water?
m
D

 m  D V
Fe
FB = D ∙ V ∙ g
V
=
 (7900 kg 3 )(0.005m 3 )
m
Fnet = Fg – FB
 39kg
=
16. A titanium cylinder has a volume of 0.010 m3. 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?
15. An iron cylinder has a volume of 0.005 m3. It is submerged in water.
a. What buoyant force acts on it?
b. If iron has a density of 7900 kg/m3, what is its apparent
weight in water?
m
D

 m  D V
Fe
FB = D ∙ V ∙ g
V
3
3
2
= [(1000 kg/m )(0.005m )](9.8m/s )
 (7900 kg 3 )(0.005m 3 )
= 49N
m
Fnet = Fg – FB
 39kg
= m.g – 20N
= 387N – 49N =
16. A titanium cylinder has a volume of 0.010 m3. 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?
16. A titanium cylinder has a volume of 0.010 m3. 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?
m
FB = D ∙ V ∙ g
DFe   m  D V
V
= [(1000 kg/m3)(0.010m3)](9.8m/s2)
kg )(0.010m3 )

(
4500
= 98N
m3
Fnet = Fg – FB
 45kg
= 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?
16. A titanium cylinder has a volume of 0.010 m3. 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?
m
FB = D ∙ V ∙ g
DFe   m  D V
V
= [(1000 kg/m3)(0.010m3)](9.8m/s2)
kg )(0.010m3 )

(
4500
3
= 98N
m
Fnet = Fg – FB
 45kg
.
= 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?
Fnet = Fg – FB
0 = 600N – FB
FB = 600N
FB = D ∙ V ∙ g
600N = (1000 kg/m3)(V)(9.8m/s2)
V=
Ch12,13 Rev
1. Make the following conversions.
1. 0oC to Kelvins
_______
2. 0 K to oC
______
3. 27oC to Kelvins
_______
4. -50oC to K ______
5. 70K to oC
_______
6. 323K to oC ______
2. How much heat is absorbed by 0.060 kg of copper when its temp is
raised from 20.0 oC to 80.0oC ? (Cp for copper when is .385 kJ/kg K)
3. A 0.100kg aluminum block at 100.00C is placed in 0.100kg of water
at 10.0oC. The final temp of the mixture is 25.0oC. What is
the specific heat of the aluminum? ( Cpwater = 4.180 kJ/kg K)
4. A 0.200 kg sample of water at 50.0C is heated to steam at 140.0C.
How much heat is absorbed?
(Cpwater = 4.18 kJ/kg.C, Cpsteam = 2.01 kJ/kg.C ,
Hvap for H2O 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 m2?
6. 20.5 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/m3,
what pressure does it exert at the earth’s surface?
8. A hydraulic floor jack consists of an arm attached to an input piston.
If the area of the input piston is 4cm2 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?
10. A 45 N iron block measures 10 cm by 5 cm by 5 cm.
What is its apparent weight in water?
11. A 0.020 kg piece of zinc at 100.0C is placed in a calorimeter full
of 0.150 kg of water at 20.0C. The temp levels at 21.1C.
Find the Cp of zinc.