1. Thermochemistry Problems
Dealing with Heat TCIS Style
A guide to solving our Thermochemistry Problems.

2. We want to solve several types of Heat Problems.
Introduction
We want to solve several types of Heat Problems.
Remember to use the 4 steps of problem solving that we have stressed all year.

3. Problem Solving Steps Reviewed
Step 1: Write down the given information
Step 2: Find the formula.
Step 3: Substitute the information into the formula.
Step 4: Solve

4. Q = heat in Joules or calories m = mass in grams
Problem Type 1:
Using Q = m ● ΔT ● Cp
Q = heat in Joules or calories
m = mass in grams
ΔT= change in Temp. (°C)
Cp= Specific Heat (J/g°C) or (cal/g°C)

5. Solving Problem Type 1:
How much heat is needed to raise the temperature of 50 grams of water from 18°C to 30°C ?
YOU NEED to know that the Specific heat of water is 4.18 J/g°C (or 1cal/g°C). You don’t need to memorize the specific heat values for every material – just water !

6. Solving Problem 1
Q = ? Mass = 50 grams
ΔT = 12°C Cp = 4.18 J/g°C
Q = m ● ΔT ● Cp
Q = 50 ● 12 ● 4.18
Q = 2508 J

7. Solving Problem 2
Similar Problem except that we are going to look for a different variable.
What is the Cp of a metal if 900 Joules are needed to raise 50 grams of it 48°C ?
Q = m ● ΔT ● Cp
Substitute and Solve
Cp = .375 J / g °C

8. Solving Problem Type 3
A more difficult type of this problem is to find the final temperature. Remember that ΔT is change in temperature and not the initial or final temperature. These type of problems are easier if you remember to solve for ΔT first and then work out the final temperature.

9. Solving Problem Type 3
What is the final temperature of a metal if 2750 Joules are added to 150 grams of it at an initial temperature of 20°C ? It has a Cp of .4 J / g°C.
Q = m ● ΔT ● Cp
Solve for ΔT ΔT = °C
Since it started at 20°C the final temp. is 20°C °C = °C

10. mLost ● ΔTLost ● Cp = mGain ● ΔTGain ● Cp
Solving Problem Type 4
A more difficult type of problem is one where one material is added to another.
Remember that HEAT LOST = HEAT GAINED (Law of Conservation of Energy) so we can adjust our formula as follows:
QLost = QGained and therefore
mLost ● ΔTLost ● Cp = mGain ● ΔTGain ● Cp

11. Problem Type 4 mL ● ΔTL ● Cp = mG ● ΔTG ● Cp
The Law of Conservation of Energy states that in any chemical or physical process, energy is neither created nor destroyed. Heat Energy can be transferred from one object to another.
We will be using the formula from the previous slide
mL ● ΔTL ● Cp = mG ● ΔTG ● Cp
If 400 grams of Aluminum (Cp = .9 J/g°C) is warmed to 95°C and then added to 1,000 grams of 20°C water, what is the final temperature ?

12. Solving Problem Type 4
Heat Lost = Heat Gained QLost = QGained
mL ● ΔTL ● Cp = mG ● ΔTG ● C
400 ● (95°C – Tf) ● .9 = ,000 ● (Tf – 20°C ) ● 4.18
360 (95°C – Tf) = 4,180 (Tf – 20°C )
34200 – 360Tf = 4,180Tf – 83,600
+ 83, Tf = + 360Tf + 83, ,800 = 4540 Tf
Tf = °C

13. Solving Problem Type 5 This type of problem is simple stoichiometry.
If you have a reaction and you know the Heat of Reaction (ΔHrxn) then you have a lot of information.
Example: In the equation below the ΔH = kJ
C + O2  CO2
So one mole of CO2 produced gives off kJ of heat.
However, as we have stated numerous times (remember our campfire example) that the more material reacted the more heat you have.
2 moles of CO2 produced will give off twice as much heat : 2 x kJ = -787 kJ of heat.

14. Solving Problem Type 5
How much heat is given off when 14 grams of CO2 is produced in the rxn. ?
C + O2  CO ΔH = kJ
It is best to simply set up a proportion: 1 mole of C gives off kJ of heat. Therefore:
1 n / kJ = x moles / y heat
WHAT VALUE DO WE HAVE ?

15. Solving Problem Type 5 y = - 125.13 kJ of heat
We have the number of grams of Carbon Dioxide (14 grams). Converting to moles gives us .318 moles.
Our proportion looks like this now:
1 n / kJ = .318 moles / y heat
y = kJ of heat

16. Hess’s Law (developed in 1840)
Germain Henri Hess ( )
Hess’s Law (developed in 1840)
If you add two or more
chemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction.
Called Hess’s Law of Heat Summation

17. How do we use Hess’s Law # 1
If you turn an equation around, you change the sign:
If H2(g) + 1/2 O2(g)  H2O(g) ΔH= kJ
then the reverse is:
H2O(g)  H2(g) + 1/2 O2(g) ΔH = kJ
Did you notice that the reactants are now the products and the products are now the reactants ?
Also notice that the heat has the same value but the sign is changed.

18. How do we use Hess’s Law # 2
If you multiply the equation by a number, you multiply the heat by that number:
2 H2O(g) ® 2 H2(g) + O2(g) DH = kJ
The equation above is double the amount of both reactants and products so the heat is also doubled.

19. How do we use Hess’s Law # 3
Our third option is that the equation remains unchanged and you use the Heat of Reaction given.
Please note that you can use more than one ‘option’ at a time. For example you can reverse a reaction and double the reactants and products together.

20. Why is Hess’s Law important ?
The reason Hess’s Law is important is because there are millions of compounds in the world and naturally there are countless ways to combine them in chemical reactions. It would take way too much time to do every conceivable chemical reaction to see the heat change in them all.
Hess’s Law enables this to take already known heats of reactions and manipulate them to get the heats of reactions of previously unknown reactions without having to do the actual experiment.

21. Solved Hess’s Law Problem
Time to dive in and solve one a Hess’s Law problem.
We want to find the Heat of Reaction. Let’s look at the following reaction:
C + 2 H2  CH4
We want to find the Heat of Reaction and to find it we are going to use information from a few other reactions that already have the heat of reaction calculated.

22. Solved Hess’s Law Problem
Rxn. 1 H2(g) + 1/2 O2(g)  H2O(l)
ΔHrxn = kJ
Rxn C(graphite) + O2(g)  CO2(g)
ΔHrxn = kJ
Rxn CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
ΔHrxn = kJ

23. Solved Hess’s Law Problem
We want to look for ‘clues’ to decide which equation to manipulate and how to manipulate it.
So for our reaction: C + 2H2  CH4
I notice that CH4 is only found in equation 3 and it is on the reactant side so I have to reverse equation 3. Doing this is the only way to get CH4 on the product side which is where it is at in the problem. This gives us the following:
CO2(g) + 2H2O(l)  CH4(g) + 2O2(g) ΔH = kJ
Notice that I always include a sign with heat even if it is positive.

24. Solved Hess’s Law Problem
Next I notice that C is needed on the reactant side and equation 2 already has it on the reactant side so in this case I can just leave the equation as is.
C(graphite) + O2(g)  CO2(g) ΔHrxn = kJ

25. Solved Hess’s Law Problem
I have now included O2 into the equation which is not in the final desired equation and I also have to get H2 in the equation. One clue for these types of equations is that if you are given an equation it will be used somewhere. This leaves us equation 1.
H2(g) + 1/2 O2(g)  H2O(l) ΔHrxn = kJ
I notice that in our desired equation that I need 2 moles of H2 as a product. If I double the H2 in this equation, I have 2 moles and I also create 1 mole of O2 that will cancel out the mole of O2 as a product that I had to include when I used equation 3. This gives me.
2 H2(g) + O2(g)  2 H2O(l) ΔHrxn = kJ

26. Solved Hess’s Law Problem
Now I combine the three equations that I manipulated.
CO2(g) + 2H2O(l)  CH4(g) + 2O2(g) ΔH = kJ
C(graphite) + O2(g)  CO2(g) ΔH = kJ
2 H2(g) + O2(g)  2 H2O(l) ΔH = kJ
Canceling out (for instance the oxygen gas) and combining the three equations gives me the following:
C + 2H2  CH4
I also combine the heat term to give me kJ
So ΔHrxn = kJ

27. Specific Heat
For water, Cp = 4.18 J/g oC (Joules), and Cp = 1.00 cal/g oC (calories).
Water takes a long time to heat up AND
Water takes a long time to cool down
Water is used as a coolant in cars for this very reason. It is also why the climate in cities next to large bodies of water is so different than the climate in cities that are inland.

28. What is Calorimetry ?
Calorimetry - the measurement of the heat into or out of a system for chemical and physical processes.
Based on the fact that the heat released = the heat absorbed
The device used to measure the absorption or release of heat in chemical or physical processes is called a calorimeter

29. Calorimetry
Ordinary foam cups are excellent heat insulators, and are commonly used as simple calorimeters.
This is why they are used for hot drinks like coffee and hot chocolate.

30. A simple Coffee-cup calorimeter
Simple coffee – cup calorimeter: 2 cups are nested together for better insulation

31. Calorimetry
Calorimetry experiments can be performed at a constant volume using a device called a “bomb calorimeter” - a closed system
Used by nutritionists to measure energy content of food.

32. A Bomb Calorimeter A bomb calorimeter

33. How to determine Exothermic v. Endothermic
FOR EXOTHERMIC REACTIONS:
H is negative: H = -100 kJ
H is written as a product such as:
A + B  C + heat
Look at the next slide to see how the energy diagram appears for an exothermic reaction. Notice that the energy of the products are lower than the energy of the reactants.

34. Energy Reactants Products ® C + O2 CO2 C + O2 → CO2 + 393.5 kJ
393.5kJ given off
CO2

35. How to determine Exothermic v. Endothermic
FOR ENDOTHERMIC REACTIONS:
H is positive: H = kJ
H is written as a product such as:
A + B + heat  C + D
Look at the next slide to see how the energy diagram appears for an exothermic reaction. Notice that the energy of the products are higher than the energy of the reactants.

36. CaCO3 + 176 kJ → CaO + CO2 Energy Reactants Products ® CaO + CO2
176 kJ absorbed
CaCO3

37. Heat in Changes of State: Boiling and Freezing
When heat is added to a sample one of two things can happen.
1. The temperature changes as we saw with the formula Q = m ● ΔT ● Cp OR
2. The phase of the material can change. The next slide summarizes these changes and gives the formula that we will use.

38. Heat in Changes of State: Freezing
Molar Heat of Fusion (Hfus.) = the heat absorbed by one mole of a substance in melting from a solid to a liquid q = mol x Hfus. (no temperature change) Molar Heat of Solidification (Hsolid.) = the heat lost when one mole of liquid solidifies (or freezes) to a solid q = mol x Hsolid. (no temperature change)

39. Heat in Changes of State: Freezing
Note: You may also have the value of these equations as: q = mass x H
This is because some textbooks give the value of H as kJ/gram, instead of kJ/mol

40. Heat in Changes of State: Freezing
Heat absorbed by a melting solid equals the heat lost when a liquid solidifies.
Hfus. = - Hsolid.

41. Heat in Changes of State: Boiling
When liquids absorb heat at their boiling points, they become a vapor.
Molar Heat of Vaporization (Hvap.) = the amount of heat necessary to vaporize one mole of a given liquid.
q = mol x Hvap. (no temperature change)

42. Heat in Changes of State: Boiling
Condensation is the opposite of vaporization.
Molar Heat of Condensation (Hcond.) = amount of heat released when one mole of vapor condenses to a liquid
q = mol x Hcond. (no temperature change)
Hvap. = - Hcond.

43. Heat in Changes of State: Boiling
The large values for water Hvap. and Hcond. is the reason hot vapors such as steam are very dangerous!
You can receive a scalding burn from steam when the heat of condensation is released!
H2O(g)  H2O(l) Hcond. = kJ/mol

44. Heating Curves The following slides shows a heating curve.
Notice that there are various parts of the curve. Some are showing a change in temperature and others are showing a change in phase.
The parts of the curve showing a change in phase are the parts that are flat lines. These parts use the formulas found in the previous few slides concerning Heats of vaporization and heats of fusion, etc.
The next several slides will break down a heating curve for you. This is a heating curve for water. We will assume a 50 gram sample of water.

45. Leg ‘A’ of a 50 gram sample of water
The solid temperature is rising from -20 to 0 oC (use q = mass x ΔT x C) and we get 2100 Joules of heat needed. (note that the Cp of Ice is 2.1 J/g C and would be a given) ←

46. Leg ‘B’ of a 50 gram sample of water
The solid is melting at 0o C; no temperature change (use q = mass x ΔHfus.) and we get 16,700 J (Note that the ΔHfus. for water is 334 J/g ) ←

47. Leg ‘C’ of a 50 gram sample of water
The liquid temperature is rising from 0 to 100 oC (use q = mass x ΔT x C) and we get 20,920 Joules. ←

48. Leg ‘D’ of a 50 gram sample of water↑
The liquid is boiling at 100o C; no temperature change (use q = mass x ΔHvap.) we get 113,000 Joules. The ΔHvap. of water is 2260 J/g

49. Leg ‘E’ of a 50 gram sample of water↑
The gas temperature is rising from 100 to 120 oC (use q = mol x ΔT x C) so we get 2,000 Joules. Note that the Cp of water vapor is 2.0 J/g °C

50. Let’s finish the heating curve problem.
We need to add up all of the heat terms from each part of the heating curve.
Heat change from -20 to J
Heat change during melting J
Heat change from 0 to J
Heat change during boiling J
Heat change from 100 to J
TOTAL HEAT NEEDED J ( kJ)

51. Special Note about Heating Curve Problems
The above example was a complete heating curve. A problem doesn’t have to have all 5 parts. For example, the water can start at 30 degrees in which you will only need the last 3 parts.
It all comes down to reading the problem carefully.