1. Entry Task: Block 2 Sept 27 Question:
The specific heat of copper metal is J/g-K. How many joules of heat energy are necessary to raise the temperature of a 5.0 g block of copper from 25.0oCto 88.5oC?
You have 5 minutes!!

2. Agenda:
Sign off on Ch. 5 sec 1-5 notes
HW: Calorimetric problems

3. I can…
Describe the relationship between energy (E), work (w) and heat (q).
Define the state of function and how it relates to enthalpy
Calculate the enthalpy change in a reaction
Use the calorimetry equation to solve for q.
Solve for q for solutions reactions.

5. Energy The ability to do work or transfer heat.
Work: Energy used to cause an object that has mass to move.
Heat: Energy used to cause the temperature of an object to rise.

6. Potential Energy
Energy an object possesses by virtue of its position or chemical composition.

7. Kinetic Energy
Energy an object possesses by virtue of its motion. 1 2
KE =  mv2

8. Units of Energy The SI unit of energy is the joule (J). kg m2
An older, non-SI unit is still in widespread use: The calorie (cal).
1 cal = J
1 J = 1 
kg m2 s2

9. System and Surroundings
The system includes the molecules we want to study (here, the hydrogen and oxygen molecules).
The surroundings are everything else (here, the cylinder and piston).

10. Work Energy used to move an object over some distance. w = F  d,
where w is work, F is the force, and d is the distance over which the force is exerted.

11. Heat Energy can also be transferred as heat.
Heat flows from warmer objects to cooler objects.

12. Transferal of Energy
The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall.

13. Transferal of Energy
The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall.
As the ball falls, its potential energy is converted to kinetic energy.

14. Transferal of Energy
The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall.
As the ball falls, its potential energy is converted to kinetic energy.
When it hits the ground, its kinetic energy falls to zero (since it is no longer moving); some of the energy does work on the ball, the rest is dissipated as heat.

15. First Law of Thermodynamics
Energy is neither created nor destroyed.
In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa.

16. Internal Energy
The internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E.

17. Internal Energy
By definition, the change in internal energy, E, is the final energy of the system minus the initial energy of the system:
E = Efinal − Einitial

18. Changes in Internal Energy
If E > 0, Efinal > Einitial
Therefore, the system absorbed energy from the surroundings.
This energy change is called endothermic

19. Changes in Internal Energy
If E < 0, Efinal < Einitial
Therefore, the system released energy to the surroundings.
This energy change is called exothermic.

20. Changes in Internal Energy
When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w).
That is, E = q + w.

21. E, q, w, and Their Signs

22. Exchange of Heat between System and Surroundings
When heat is absorbed by the system from the surroundings, the process is endothermic.

23. Exchange of Heat between System and Surroundings
When heat is absorbed by the system from the surroundings, the process is endothermic.
When heat is released by the system to the surroundings, the process is exothermic.

24. State Functions
Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem.

25. State Functions
However, we do know that the internal energy of a system is independent of the path by which the system achieved that state.
In the system below, the water could have reached room temperature from either direction.

26. State Functions Therefore, internal energy is a state function.
It depends only on the present state of the system, not on the path by which the system arrived at that state.
And so, E depends only on Einitial and Efinal.

27. State Functions However, q and w are not state functions.
Whether the battery is shorted out or is discharged by running the fan, its E is the same.
But q and w are different in the two cases.

28. Heat (Enthalpy) change DH
Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.
DH = H (products) – H (reactants)
DH = heat given off or absorbed during a reaction at constant pressure

29. Enthalpy Hproducts < Hreactants Hproducts > Hreactants DH < 0

30. Endothermic and Exothermic
A process is endothermic when H is positive.
6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.
H2O (s) H2O (l)
DH = 6.01 kJ

31. Endothermic and Exothermic
A process is exothermic when H is negative.
890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)
DH = kJ

32. Enthalpy of Reaction H = Hproducts − Hreactants
The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants:
H = Hproducts − Hreactants

33. Enthalpy of Reaction
This quantity, H, is called the enthalpy of reaction, or the heat of reaction.

34. The Truth about Enthalpy
Enthalpy is an extensive property.
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) ∆H of -890kJ
2CH4(g) + 4O2(g)  2CO2(g) + 4H2O(l) ∆H of kJ
H for a reaction in the forward direction is equal in size, but opposite in sign, to H for the reverse reaction.
CO2(g) + 2H2O (l)  2O2(g) + CH4(g) ∆H of +890kJ
H for a reaction depends on the state of the products and the state of the reactants.
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) ∆H of -802kJ

35. 5.3 problem
Hydrogen peroxide can decompose to water and oxygen by the reaction 2 H2O2(l)  2 H2O(l) + O2(g) H = –196 kJ Calculate the quantity of heat released when 5.00 g of H2O2(l) decomposes at constant pressure.
We have to find out how many moles of H2O2 are in 5.00g.
5.00 g
1 mole H2O2
= mol H2O2
34 g H2O2
And there are 2 moles of H2O2 to release -196 kJ of energy.
mol H2O2
-196 kJ
= kJ
2 mol H2O2

36. 5.28 Consider the reaction: CH3OH(g)  CO(g) + 2H2(g) ∆H = +90.7
a) Exothermic or endothermic? Its endothermic (pos ∆H)
b) Calculate amount of energy when 45.0 g is transferred.
45.0 g
1 mole CH3OH
= 1.40 mol CH3OH
32.04 g CH3OH
1.40 mol CH3OH
+90.7 kJ
= 127 kJ
1 mol CH3OH

37. 5.28 Consider the reaction: CH3OH(g)  CO(g) + 2H2(g) ∆H = +90.7
c) If the enthalpy is changed to 16.5 kJ, how many grams of hydrogen gas are produced?
16.5 kJ
2 mole H2
=0.364mol H2
+90.7 kJ
0.364 mol H2
2 g H2
= 0.727g
1 mol H2

38. 5.28 Consider the reaction: CH3OH(g)  CO(g) + 2H2(g) ∆H = +90.7
d)How many joules of heat are released when 10.0g of CO reacts completely with hydrogen to form CH3OH at constant pressure
10.0 g CO
1 mole Co
=0.357mol CO
28 g CO
0.357 mol CO
+90.7 kJ
= kJ
1 mol CO

39. Calorimetry
Since we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow.

40. Heat Capacity and Specific Heat
The amount of energy required to raise the temperature of a substance by 1 K (1C) is its heat capacity.

41. Heat Capacity and Specific Heat
We define specific heat capacity (or simply specific heat) as the amount of energy required to raise the temperature of 1 g of a substance by 1 K (or 1 C).

42. Heat Capacity and Specific Heat
Specific heat, then, is
Specific heat =
heat transferred
mass  temperature change
Specific Heat (c) = q
m  T

43. Since we know the specific heat of most substances, we will rearrange the equation to solve for the amount of heat (q) released or absorbed which is what we don’t know.

44. q = c x m x ΔT q = the heat absorbed or released
c= the specific heat of water = Joules
m= mass of the sample in grams
ΔT= the temperature difference (Tf - Ti)

45. q=cmT 4515 J or 4.515 x 103 J q = the heat absorbed or released
If the temperature of 34.4g of ethanol increases from 25.0˚C to 78.8˚C, how much heat has been absorbed by ethanol? Specific heat of ethanol is 2.44 J/g C
q=cmT
q = the heat absorbed or released
c= the specific heat of ethanol= 2.44 J/g C
m= 34.4 g
ΔT= 53.8
4515 J or x 103 J

46. A 4. 50 g nugget of pure gold absorbs 276 J of heat
A 4.50 g nugget of pure gold absorbs 276 J of heat. What was the final temperature of gold if the initial temperature was 25.0˚C?
q=cmT
q = 276 J
c= the specific heat of gold= Joules
m= 4.5 g
ΔT= 25-X
T = q/cm =
276J
= 475°C
0.129 J/(gC)
x 4.5 g
Since the temperature started at 25.0°C, the final temperature is 500°C.

47. q=cmT c= q/Tm = q = 5696J c= the specific heat = X Joules m= 155 g
A 155 g sample of an unknown substance was heated from 25.0˚C to 40.0˚C. In the process, the substance absorbs 5696 J of energy. What is the specific heat of the substance? What is this substance?
q=cmT
q = 5696J
c= the specific heat = X Joules
m= 155 g
ΔT= 25-40=15
c= q/Tm =
5696J
= 2.44 J/(gC)
15 C
x 155g

48. 5.4 problem
q=cmT
(a) Large beds of rocks are used in some solar-heated homes to store heat. Assume that the specific heat of the rocks is 0.82 J/g–K. Calculate the quantity of heat absorbed by 50.0 kg of rocks if their temperature increases by 12.0 C. (b) What temperature change would these rocks undergo if they emitted 450 kJ of heat?
a. (0.82J/g-K)(50000g)(12K) = x 105 J
b. ( kJ/g-K)(50000g)(X) = 450 kJ
(X) = kJ
(41)
= or 11.0 temp increase

49. WHAT ABOUT REACTIONS?
By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, one can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter.

50. Reactions in Solutions
Because the specific heat for water is well known (4.184 J/g-K), we can measure H for the reaction with this equation:
q = c  m  T

51. Reactions in Solutions
Heat produced from a reaction is entirely absorbed by the solution and does not escape.
Exothermic rxn- heat is lost by the reaction and absorbed by the solution- temperature rises

52. Reactions in Solutions
Heat gained by solution, q(soln) is therefore equal in magnitude and opposite in sign from reaction
q(soln)=
(specific heat of solution) (grams of solution)(∆T) = - q (rxn)
q(soln) = c (soln) (m soln)(∆ T) = -q(rxn)

53. Enthalpy of solution
Enthalpy of solution is the ∆H = qp (constant pressure)
Problem steps
What does solution temperature do after reaction
Gets warmer (solution is absorbing energy) the reaction is opposite is releasing energy ∆H is negative
Gets colder (solution is releasing heat) the reaction is opposite and is absorbing energy ∆H is positive.
Plug in your numbers qrxn=(c)(m)(∆T)
Change the sign to reflect concept

54. Enthalpy of solution
Enthalpy of solution is the ∆H = qp (constant pressure)
Problem steps
If it asks to calculate the enthalpy change for the reaction in kJ/mol of a substance
Your given grams, convert to moles then divide the ∆H by the moles.
Your given molarity (moles/1L = M) rearrange to get moles then divide the ∆H by the moles.

55. 5.4 problem
When 50.0 mL of M AgNO3 and 50.0 mL of M HCl are mixed in a constant-pressure calorimeter, the temperature of the mixture increases from C to C. The temperature increase is caused by the following reaction:
AgNO3(aq) + HCl(aq)  AgCl(s) + HNO3(aq)
Calculate H for this reaction in AgNO3, assuming that the combined solution has a mass of g and a specific heat of 4.18 J/g C.
First find the amount of energy THEN we can calculate the # of moles of a specific molarity!
q = (100g)(4.18J/g-C)(-0.81) = J
(0.050L)(0.100M) = mol of AgNO3 J
= J or kJ
0.005 mol of AgNO3

56. 5.42 problem
When a 4.25 g sample of solid ammonium nitrate dissolves in 60.0g of water in a coffee-cup calorimeter, the temperature drops from 22.0 C to 16.9 C. Calculate the ∆H (in kJ/mol of NH4NO3) for the solution process.
NH4NO3(aq)  NH4 + (aq) NO3- (aq)
Assume that the specific heat of solution is 4.18 J/g C.
First find the amount of energy of SOLUTION THEN we can calculate the # of moles of a specific molarity!
q = (64.25g)(4.18J/g-C)(5.1) = J OR 1.37 kJ
How many moles of NH4NO3 are in 4.25 g
4.25 g
1 mol NH4NO3
= mol NH4NO3
80.0 g NH4NO3

57. 5.42 problem
When a 4.25 g sample of solid ammonium nitrate dissolves in 60.0g of water in a coffee-cup calorimeter, the temperature drops from 22.0 C to 16.9 C. Calculate the ∆H (in kJ/mol of NH4NO3) for the solution process.
NH4NO3(aq)  NH4 + (aq) NO3- (aq)
Assume that the specific heat of solution is 4.18 J/g C.
So mol NH4NO3 are in 4.25 g and 1.37 kJ of energy is released. Divide mol into 1.37 kJ to find the number of kJ per mol.
1.37 kJ
= 25.8 kJ or kJ
0.053 mol of NH4NO3

58. 5.44 problem
When a g sample of octane (C8H18) was burned in a bomb calorimeter whose total heat capacity is kJ/C. The temperature of the calorimeter plus contents increased from 21.36C to 28.78C What is the heat of combustion per gram of octane? Per mole of octane? 2C8H18 (g) + 17O2 (g)  16CO2 (g) H2O (g)
First find the amount of energy of SOLUTION THEN we can calculate the # of moles of a specific molarity!
q = (64.25g)(4.18J/g-C)(5.1) = J OR 1.37 kJ
How many moles of NH4NO3 are in 4.25 g
4.25 g
1 mol NH4NO3
= mol NH4NO3
80.0 g NH4NO3

59. 5.42 problem
When a 4.25 g sample of solid ammonium nitrate dissolves in 60.0g of water in a coffee-cup calorimeter, the temperature drops from 22.0 C to 16.9 C. Calculate the ∆H (in kJ/mol of NH4NO3) for the solution process.
NH4NO3(aq)  NH4 + (aq) NO3- (aq)
Assume that the specific heat of solution is 4.18 J/g C.
So mol NH4NO3 are in 4.25 g and 1.37 kJ of energy is released. Divide mol into 1.37 kJ to find the number of kJ per mol.
1.37 kJ
= 25.8 kJ or kJ
0.053 mol of NH4NO3