Temperature
•the average kinetic energy of an object
•measured in degrees Celsius (oC)
• as molecules move or vibratecollide
• more collisionshigher temperature.
Heat•The total kinetic energy in an object
• measured in Joules (J)
• transferred between two objects
because of a temperature difference.
Temperature vs Heat
Thermal Equilibrium
•occurs when 2 objects have the same temperature, thus no heat (energy) flows between them.
•heat only flows from a hot object to a cold object.
Temperature vs Heat
Example: Compare a glass of water at 20oC to a bathtub of water at 20oC.
Both have the same temperature, but the bathtub has more heat because it has more molecules of water at that temperature.
Conduction: Heat is transferred through direct contact.
Example: holding ice in your hand
Convection: Heat
transferred due to density differences.
Example: cooking a hot dog above a flame
Radiation: Heat transferred by waves electromagnetic
radiation (Infrared)
Example: Warming your hands next to a flame
Absolute zero: Coldest possible temperature.
The Kinetic energy of the object would equal 0.
0 Kelvin = -273.15oC
TC = TK – 273.15 TK = Kelvin Temperature (K) TC = Celsius Temperature (oC)
TF = (9/5)TC + 32
TF = Fahrenheit Temperature (oF) TC = Celsius Temperature (oC)
1. Convert the following temperatures:
a) 50 oC into Fahrenheit b) 300 oF into Celsius
c) 350 oC into Kelvin d) 5 K into Celsius
e) 40 oF into Kelvin
Practice Problems
122 oF 149 oC 623.15 K -268.15 oC 277.6 K
A method of transferring energy to a system due to a temperature difference.
Calorie: 1 calorie is the amount of energy required to raise 1 g of water by 1 oC.
1 calorie = 4.186 J Joule: SI unit of energy
Units of Energy:
The amount of heat required to raise 1 g of a substance by 1oC.
Every substance has a unique specific heat.
(you can find them on p. 372 in your text) Specific Heat of Water
4.186 J/g•oC 4186 J/kg•oC
1 cal/g•oC Therefore, 1 cal = 4.186 J
Q = mcΔT
Q = Heat gain or loss (J or cal) m = mass (kg or g)
c = specific heat (units vary)
ΔT = change in temperature (oC) Note: When temperature decreases, then ΔT is negative and thus Q is negative, and heat is flowing out of the object heat is lost.
When temperature increases, then Q is
positive and heat is flowing into the object heat is gained.
Heat Gain or Loss
The heat (energy) lost by one object in a
system must be absorbed by another object in the system. Energy (heat) cannot be
created or destroyed.
•Energy is transferred from one object to another by heating.
(Conservation of Energy) 1st Law of Thermodynamics
A 0.05 kg piece of metal is heated to 200oC and then dropped into a beaker containing 0.4 kg of water that is initially at 20oC. If the final
equilibrium temperature of the mixed system is 22.4oC, find the specific heat of the metal.
Heat lost by metal + Heat gained by water = 0 Qlost by metal + Qgained by water = 0
This term will
be negative This term will be positive
mmetalcmetalΔTmetal + mwatercwaterΔTwater = 0
(0.05 kg)cm(22.4oC – 200oC) +
(0.4 kg)(4186 J/kg•oC)(22.4oC – 20oC) = 0 cm = 453 J/kg•oC
Compare to known values (p.372)
most likely Iron (Fe)
Qlost by metal + Qgained by water = 0
mmcm(Tf – Tm) + mwcw(Tf – Tw) = 0
1) How much heat is gained or lost when:
a) 30 g of water goes from 30 oC to 50 oC
b) 30 g of water goes from 25 oC to 5 oC
Practice Questions
2512 J
-2512 J
2a) 120 ml of water gains 1000 J of energy, what was its change in temperature?
**Note: the density of water is 1 g/mL, so:
1 g of water = 1 mL of water
b) 120 g of water gains 500 calories of
energy, what was its change in temperature?
1.99 oC
4.17 oC
Practice Questions
3) A 0.2 kg piece of 150 oC metal is placed in
500 ml of water initially at 15 oC. If the final equilibrium temperature of the mixture is 25o C what is the specific heat of the metal?
4) A 2 kg piece of 140 oC copper is placed in
600 mL of water initially at 20 oC. What is the final equilibrium temperature of the mixture?
(ccopper = 387 J/kg•oC)
0.84 J/g•oC
48.3 oC
Practice Questions
Objects A and B are brought into close
thermal contact with each other, but they are well isolated from their surroundings. Initially TA = 0°C and TB = 100°C. The specific heat of A is more than the specific heat of B. The two objects will soon reach a common final
temperature Tf. The final temperature is 1. Tf > 50°C.
2. Tf = 50°C.
3. Tf < 50°C.
The fact that it takes a lot of energy to change the temperature of water is due to water’s
1. Low density.
2. High density.
3. Low specific heat.
4. High specific heat.
The temperature of a glass of water increases
from 20°C to 30°C. What is ∆T in units of Kelvin?
1. 303 K 2. 293 K 3. 283 K 4. 10 K
Phase Change: When a substance undergoes a physical alteration from one form to another (solid to liquid, liquid to gas, etc…)
Heat required to change the phase of a substance (not the temperature):
Q = Heat (J or cal) m = mass (kg)
L = Latent Heat (J/kg)
Q = mL
The Latent heat is a property of a
substance. (values found on p. 379 in text)
Latent Heat of Fusion (Lf): term used when changing from solid to liquid. (or liquid to solid)
Latent Heat of Vaporization (Lv): term used when changing from liquid to gas. (or gas to
liquid)
When a solid melts into a liquid…its temperature DOES NOT CHANGE!!!
When a liquid evaporates into a gas…its temperature DOES NOT CHANGE!!!
Lv = 2.26 x 106 J/kg For Water: Lf = 3.33 x 105 J/kg
For Water:
Phase changes can be viewed as a rearrangement of molecules when heat is added or subtracted from a substance.
Example: Phase change from solid to liquid.
Solid: molecules are close together with
strong internal forces between them.
Liquid: Molecules are farther apart and have weaker forces in between molecules.
Energy was needed to separate molecules
Temperature vs. Heat added (for ice)
Q = mcΔT Q = mLf
Q = mLv
How much energy (in Joules) must be
transferred to convert 10 g of ice at –5oC into water at 15oC?
(the specific heat of ice is 2.090 J/g•oC) 1.Water must warm up from –5o C ice
to 0o C ice (add energy) Q = mciceΔT
Q = (10 g)(2.090 J/g•oC)(0 oC – (-5 oC)) Q = 104.5 J
2.Must melt all of the ice from 0 oC ice to 0 oC water. (add energy)
(NO TEMPERATURE CHANGE!!!)
Q = mLf
Q = (0.010 kg)(3.33 x 105 J/kg) Q = 3330 J
Note: this would be negative if we were
going from water to ice, since we would be losing energy in that case.
3.Must convert 0 oC water to 15 oC water.
(add energy)
Q = mcwaterΔT
Q = (10 g)(4.186 J/g•oC)(15 oC – 0 oC) Q = 627.9 J
Qtotal = mciceΔT + mLf + mcwaterΔT 4. Add up all the energies.
Qtotal = 104.5 J + 3330 J + 627.9 J Qtotal = 4062.4 J
5. How much energy is “lost” when 2 kg of 175 oC steam is condensed into 50oC water?
-5.24 x 106 J cwater = 4186 J/kg•oC
csteam = 2010 J/kg•oC Lv = 2.26 x 106 J/kg For Water:
Practice Questions
6. How much energy must be transferred to 500 g of –30 oC ice to change it to 125 oC steam?
1.56 x 106 J cice = 2090 J/kg•oC
cwater = 4186 J/kg•oC csteam = 2010 J/kg•oC For Water:
Lv = 2.26 x 106 J/kg Lf = 3.33 x 105 J/kg
Practice Questions
7. A 15 g piece of 0 oC ice is placed in 50 g of water initially at 65 oC. What is the final temperature of the mixture?
Tf = 31.6 oC
Practice Questions
Lf = 3.33 x 105 J/kg = 333 J/g For Water:
(cwater = 4186 J/kg•oC = 4.186 J/g•oC)
Heat gain by ice to change to
water
+
Heat gain by water toraise its
temperature
+
Heat loss by water tolower its
temperature
=
0(15 g)(333 J/g) + (15 g)(4.186 J/g•oC)(Tf - 0oC)
(50 g)(4.186 J/g•oC)(Tf - 65oC)
+
= 0
4995 J + 62.79(Tf) J + 209.3(Tf) J – 13604.5 J = 0
272.09(Tf) J = 8609.5 J
1. Heat flows from areas of higher
temperature to areas of lower temperature.
Heat never flows from a cooler to a hotter body on its own.
2. Heat can never be taken from a reservoir without something else happening in the system. (conservation of energy)
3. The entropy (randomness or disorder) of a system can increase or decrease, but the total entropy of the universe is always increasing.
Entropy
Nature (or the universe) is always trying to become more disorganized. Work must be done (energy added) to make a system
more organized.
ex. Deck of cards Ice melting
Random arrangement of trees
Is entropy of the system increasing or decreasing in the following examples?
• shuffle a deck of cards
• fold laundry
increasing decreasing
• open a soda bottle and
hear a fizz sound increasing
• you eat a piece of pizza decreasing
Heat Engine:
A device that
converts thermal energy to other
useful forms, such as electrical and mechanical energy.
The net work, W, done by a heat engine equals the net heat flowing through it.
W = Qh - Qc
h h c
h Q
Q Q
Q
e W
Efficiency of a Heat Engine Heat Engine
A device that converts thermal energy to other
useful forms, such as electrical & mechanical energy.
Refrigerators
Opposite of a heat engine Uses electricity to power a motor which causes a temperature difference between two reservoirs
A heat engine takes in 500 J of energy and expels 300 J of energy.
W = Qh - Qc
W = 500 J – 300 J W = 200 J
a) How much work was done by the heat engine?
b) What is the efficiency of the heat engine?
Qh
e W
J 500
J
200 0.4 40%
Rank in order, from largest to smallest, the work Wout performed by these 4 heat engines.
1. Wb > Wa > Wc > Wd 2. Wb > Wa > Wb > Wc 3. Wb > Wa > Wb = Wc 4. Wd > Wa = Wb > Wc 5. Wd > Wa > Wb > Wc
What, if
anything, is wrong with this
refrigerator?
1. It violates the first law of thermodynamics.
2. It violates the second law of thermodynamics.
3. It violates the third law of thermodynamics.
4. Nothing is wrong.
What is the generic name for a cyclical device that transforms heat energy into work.
1. Refrigerators
2. Thermal Motors 3. Heat Engines
4. Carnot Cycles
5. Otto processors
It’s a really hot day and your air conditioner is broken. Your roommate says, “Let’s open the refrigerator door and cool this place off.” Will this work?
1. Yes.
2. It might, but it will depend on how hot the room is.
3. No.
