1. Thermodynamics

2. Energy is... The ability to do work. Conserved. Made of heat and work.
A state function: values that depend on the state of the substance, and not on how that state was reached. 
Independent of the path, or how you get from point A to B.
Work is a force acting over a distance.
Heat is energy transferred between objects because of temperature difference.

3. The Universe Is divided into two halves.
The system and the surroundings.
The system is the part you are concerned with.
The surroundings are the rest.
Exothermic processes release energy to the surroundings.
Endothermic processes absorb energy from the surroundings.

4. Units of Heat Calorie (cal) Joule (J) 1 cal = 4.184 J
The quantity of heat required to change the temperature of one gram of water by one degree Celsius.
Joule (J)
SI unit for heat
1 cal = J

5. Direction Every energy measurement has three parts.
A unit ( Joules or calories).
A number how many (magnitude).
A sign to tell direction.
negative - exothermic
positive- endothermic

6. Surroundings
System
Energy
DE <0

7. Surroundings
System
Energy
DE >0

8. First Law of Thermodynamics
The energy of the universe is constant.
Law of conservation of energy.
q = heat
Take the system’s point of view to decide signs.
q is negative when the system loses heat energy
q is positive when the system absorbs heat energy

9. Heat Capacity
The quantity of heat required to change the temperature of a system by one degree.
Three “types” with different systems:
Specific Heat Capacity
Molar Heat Capacity (we won’t cover this one in Physics)
Heat Capacity

10. Heat Capacity Types q = (m)(C)(T) Specific heat capacity
System is one gram of substance
Units are J/goC
q = heat change
m = mass
C = Specific Heat Capacity
DT = final temperature – initial temperature
q = (m)(C)(T)

11. Heat Capacity Types q = CT Heat capacity of an object
Specific for a given amount or unit
so it is not mass dependent
Units are J/oC
q = heat
C = heat capacity
DT = final temperature – initial temperature
q = CT

12. Heat Capacity Types So how do you know which one to use when?
Look at the known information
Select the value with appropriate units to cancel

13. Example
The specific heat of graphite is 0.71 J/gºC. Calculate the energy needed to raise the temperature of 75 kg of graphite from 294 K to 348 K.

14. Example
The specific heat of graphite is 0.71 J/goC. Calculate the energy needed to raise the temperature of 75 kg of graphite from 294 K to 348 K.
q = mCΔT = (75,000 g)(0.71 J/goC)( )K
= 2.88 x 106 J
Note that ΔT oC and ΔT K are interchangeable, since they are the same “size”.

15. Example
When a piece of copper (5.0 g) is heated for 2.0 seconds, and 100 J of heat energy is transferred to the copper, the temperature increases from 20.0 ˚C to 71.9 ˚C. What is the specific heat of the copper? m = 5.0 g; ΔT = (71.9 – 20.0) = 51.9˚C; q = 100 J; t = 2.0 s q = mCΔT; q = C = 100 = J/goC mΔT (5.0)(51.9) time is irrelevant.

16. Example
If 10.0 g of Cu is heated for 2.0 seconds from 20.0 ˚C and 200 J of heat are absorbed, what is the final temperature of the block? (Let’s assume time is again irrelevant.) m = 10.0 g; Ti = 20.0 oC; Tf = ?; CCu = J/goC; q = 200 J q = mCΔT; ΔT = q/mC = (Tf – Ti); Tf = q/mC + Ti Tf = [200/(10*0.385)] + 20 = 71.9oC

17. Calorimetry Measuring temperature changes to calculate heat changes.
Use a calorimeter.
Two kinds
Constant pressure calorimeter (called a coffee cup calorimeter)
Constant volume calorimeter (called a bomb calorimeter)

18. Calorimetry
A coffee cup calorimeter measures temperature and calculates q.
An insulated cup, full of water (constant pressure, variable volume)
Water is the surroundings in which the system changes
Calculate the heat change of water.
The specific heat of water is J /goC
Heat of water qH2O = CH2O mH2O DT

19. qsystem + qsurroundings = 0
qH2O + qrxn = 0
qH2O = ‒ qrxn
In interactions between a system and its surroundings the total energy remains constant— energy is neither created nor destroyed.
Law of Conservation of Energy

20. Coffee Cup Calorimeter
A simple calorimeter.
Well insulated and therefore isolated.
Measure temperature change.
qrxn = ‒ qcal

21. Determination of Specific Heat

22. Example
Determining Specific Heat from Experimental Data.
Use the data presented on the last slide to calculate the specific heat of lead.
q Pb = ‒ q H2O
q H2O = mcT = (50.0 g)(4.184 J/g °C)(28.8 ‒ 22.0)°C
q H2O = 1.4x103 J
q Pb = ‒1.4x103 J = mcT = (150.0 g)(c)(28.8 ‒ 100.0)°C
c Pb = 0.13 Jg‒1°C‒1

23. Example
Equal masses of liquid A, initially at 100 ˚C, and liquid B, initially at 50 ˚C, are combined in an insulated container. The final temperature of the mixture is 80 ˚C. Which has the larger specific heat capacity, A or B? mA = mB = m; TiA = 100˚C; TiB = 50˚C; Tf = 80˚C qA = qB; mCAΔTA = mCBΔTB; CAΔTA = CBΔTB; CA/CB = ΔTB/ΔTA CA/CB = (80–50)/(100–80) = 1.5. Since CA/CB > 1, CA > CB

24. You do one…
When 86.7 grams of water at a temperature of 73.0 ˚C is mixed with an unknown mass of water at a temperature of 22.3 ˚C the final temperature of the resulting mixture is 61.7 ˚C. What was the mass of the second sample of water?
a g
c g
b. 302 g
d. 419 g

25. States of Matter

26. General Heating Curve

27. Changes of State for Water
Heat of vaporization (liquid <----> gas)
H2O(l) ↔ H2O(g)
Hv = 2,258 J/g
Heat of fusion (solid <----> liquid )
H2O(s) ↔ H2O(l)
Hf = 333 J/g

28. Heating Curve for Water
Total heat absorbed by the water as it is warmed can be determined by finding the heat involved in each “step” of the process.
q1+ q2 + q3 …. = qtotal
Note:
Cice= Csteam = 2.09 J/goC
q = mCT
qphase change = Hphase change

29. Example
Calculate q for the process in which 50.0 g of water is converted from liquid at 10.0°C to vapor at 125.0°C.

30. Example
What is the heat of fusion of lead in J/g if 6.30 kilojoules of heat are required to convert 255 grams of solid lead at its melting point into a liquid?
a J/g
c J/g
b J/g
d J/g

31. Example
What quantity of heat is required to heat 1.00 g of lead from 25 ˚C to the melting point (327 ˚C) and melt all of it? (The specific heat capacity of lead is J/g • K and it requires 24.7 J/g to convert lead from the solid to the liquid state.)
a J
c J
b J
d J