Thermochemistry Answers - Worksheet Number One
We will ignore any heats losses to the walls of the container and losses to the air. These is a typical position to take since, in a real experiment, both would have to be accounted for, making for much more complexity.
1. q = (20.0 g) (20.0 °C) (2.02 J/g °C). (Note Cp of gas is used.)
2. q = (15.0 g) (25.0 °C) (2.02 J/g °C)
3. q = (120.0 g) (22.0 °C) (4.184 J/g °C)
4. 800,000 J = (720.0 g) (x) (2.02 J/g °C). (Notice the conversion to J.)
5. q = (85.0 g) (190.0 °C) (0.129 J/g °C). Divide answer by 1000 to obtain kJ.
6. 41.72 J = (18.69 g) (17.0 °C) (x)
7. (333.51 J/g) (18.015 g/mol)
8. 41,840 = (x) (6.5 °C) (4.184 J/g °C)
9. Here's the graph:
The five number sections correspond to the following:
1) Δt = 15 °C as a solid
2) melting (no temperature change)
3) Δt = 100 °C as a liquid
4) boiling (no temperature change)
5) Δt = 20 °C as a gas
The five solutions are:
1) (50.0 g)(15 °C) (2.06 J/g °C)
2) (50.0 g / 18.0 g/mol) (6.02 kJ/mol)
3) (50.0 g) (100 °C) (4.184 J/g °C)
4) (50.0 g / 18.0 g/mol) (40.7 kJ/mol)
5) (50.0 g) (20 °C) (2.02 J/g °C)
10. Here's the graph:
The five number sections correspond to the following:
1) Δt = 10 °C as a gas
2) condensing (no temperature change)
3) Δt = 100 °C as a liquid
4) freezing (no temperature change)
5) Δt = 40 °C as a solid
The five solutions are:
1) (32.0 g) (10 °C) (2.02 J/g °C)
2) (32.0 g / 18.0 g/mol) (40.7 kJ/mol)
3) (32.0 g) (100 °C) (4.184 J/g °C)
4) (32.0 g / 18.0 g/mol) (6.02 kJ/mol)
5) (32.0 g) (40 °C) (2.06 J/g °C)
11. The five solutions are:
a) (23.0 g) (46 °C) (2.06 J/g °C)
b) (6.02 kJ/mol)) (23.0 g / 18.0 g/mol)
c) (23.0 g) (100 °C) (4.184 J/g °C)
d) (40.7 kJ/mol)) (23.0 g / 18.0 g/mol)
e) (23.0 g) (9 °C) (2.02 J/g °C)
12. The five solutions are:
a) (10.0 g) (20 °C) (2.02 J/g °C)
b) (40.7 kJ/mol)) (10.0 g / 18.0 g/mol)
c) (10.0 g) (100 °C) (4.184 J/g °C)
d) (6.02 kJ/mol)) (10.0 g / 18.0 g/mol)
e) (10.0 g) (20 °C) (2.06 J/g °C)
When I set up the answer to #12, I just copied the format from #11 and changed the numbers. Can you follow the changes? If you can, you are well on your way to seeing the pattern underlying this type of problem.
13. The two solutions (Δt = 80 °C and boil at 100 °C) are:
a) (100.0 g) (80 °C) (4.184 J/g °C)
b) (40.7 kJ/mol) (100.0 g / 18.0 g/mol)
14. Before doing the numbers, try to visualize what is happening:
a) 540.0 g of water at 25.0 °C is going to zero. It must lose energy to do so.
b) The lost heat energy goes and melts ice (assume no heat loss to container walls).
The question then becomes "how much ice got melted by the heat lost?" First calculate the heat lost by the 540.0 g of water:
q = (540.0 g) (25.0 °C) (4.184 J/g °C) = 56,484 J = 56.484 kJ
Now, determine how much ice can be melted by that amount of energy:
56.484 kJ = (6.02 kJ/mol) (x / 18.0 g/mol)
15. Before doing the numbers, try to visualize what is happening:
a) 18.0 g of steam at 100.0 °C goes through a phase change and then, as a liquid, goes through a Δt of 100 °C. Two calculations are required to determine the loss of energy.
b) The lost heat energy goes and melts ice (assume no heat loss to container walls).
The question then becomes "how much ice got melted by the heat lost?" First calculate the total heat lost by the 18.0 g of water:
a) (40.7 kJ/mol) (18.0 g / 18.0 g/mol)
b) (18.0 g) (100 °C) (4.184 J/g °C)
Second, calculate the grams of ice melted using the same technique as in problem #14. Good luck!!
16. In this problem no phase changes are involved, so only the specific heat equation will be involved. Also, it should be apparent that the lead will go down in temperature as it loses energy to the colder water, which will go up in temperature.
The first key to the solution lies in thinking about the relationship of the starting temperatures to the final temperature. So let's start by calling the final, ending temperature 'x.' Keep in mind that BOTH the water and the lead will wind up at the temperature we are calling 'x.' Also, make sure you understand that the 'x' we are using IS NOT the Δt, but the FINAL temperature.
The lead goes down from to 95 to x, so this means its Δt equals 95 minus x. The water goes up in temperature, so its Δt equals x minus 25.
The second key is to see that the energy amount going out of the lead is equal to the energy amount going into the water. This means qlost = qgain. Remember, q stands for the amount of heat involved.
So, by substitution, we have:
(150) (95 - x) (0.444) = (500) (x - 25) (4.18)
Solve for x
Another way to explain this uses simultaneous equations:
Set Δtlead = x and Δtwater = y. The first equation then is:
(150) (x) (0.444) = (500) (y) (4.18)
The second equation is x + y = 70.
The solution from there is left to the reader.
17. Note that in this question the 'x' will be the specific heat of the metal.
The heat lost by the metal is equal to the heat gained by the water. When put as an equation, it is:
qlost = qgain
By substitution, we then have:
(mass) (Δt) (Cp) = (mass) (Δt) (Cp)
with the metal values on the left and the water values on the right.
Plugging in the numbers gives:
(80.0 g) (54.0 °C) (x) = (100.0 g) (6.0 °C) (4.184 J / g °C)
You may finish the solution.
18. Heat energy is transferred from the metal to the water. The amount lost is equal to the amount gained, so we can set the equations equal to each other
(55.0 g) (75.3 °C) (x) = (225 g) (1.7 °C) (4.184 J / g °C)
The 75.3°C Δt results from it going from 99 to 23.7 (NOT 22).
19. I think I'll be mean and nasty and not provide a solution!!
20. Note that even though the water is at zero degrees, we will assume it to be liquid. In a problem like this, we would have to have an explicit statement that it was a solid at zero.
The cold water (at zero) will go up in temperature while the warm water (at 40) will go down. Once again, the following can be determined:
a. let x = the final temperature
b. therefore Δt0 = x - 0 and Δt40 = 40 - x
c. qlost = qgain
From this we get:
(40.0) (40 - x) (4.18) = (15.0) (x - 0) (4.18)
I put the side losing energy on the left side, but it really doesn't matter.
Solving for x gives a final temperature of 29.1 °C
The simultaneous equation approach is left entirely to the reader.
21. The graph below shows a pure substance which is heated by a constant source of heat supplying 2000.0 joules per minute. Identify the area described in the questions below and complete the necessary calculations.
UV = 0.36 min, VW = 3.6 min, WX = 3.6 min, XY = 19.4 min, YZ = 0.6 min
a. being warmed as a solid ___________
b. being warmed as a liquid __________
c. being warmed as a gas ____________
d. changing from a solid to a liquid _____
e. changing from a liquid to a gas ______
f. What is its boiling temperature? _________________
g. What is its melting temperature? _________________
h. How many joules were needed to change the liquid to a gas? ____________
i. Where on the curve do the molecules have the highest kinetic energy? ______
j. If the sample weighs 10.0 g, what is its heat of vaporization in J/g? ______
22. The number of Joules needed to raise the temperature of 100 grams of water 10 °C. is the same as the number of Joules needed to raise the temperature of 1000 grams of water
a. 1 °C b. 0.1 °C c. 10 °C d. 100 °C
23. 10.0 g of a fuel are burned under a calorimeter containing 200.0 g of H2O. The temperature of the water increases from 15.0 °C to 55.0 °C. Calculate the total heat produced (in joules) and the heat of combustion per gram of fuel.
24. Why does moisture condense on the outside of a glass of cold water?
25. Is it possible for a cup of water to completely evaporate in a room with a constant temperature?
26. Why does alcohol at room temperature feel cooler to the touch than does water at the same temperature?
27. If you put a very shallow dish of water in a pan of alcohol and blow air over it by means of a electric fan, the water will freeze...why?
