OBJECTIVE: Introduction to Thermochemistry,
energy units + energy unit conversion math,
review of Table B with descriptions,
review of Table I and descriptions, and the
inclusion of heat emergy to the Stoichiometry
of balanced chemical equations.
Get out reference tables,
and a calculator
1. Thermo = heat, the energy changes that happen in a chemical reaction.
2. Sometimes heat is given off which is exothermic
3. Sometimes energy is absorbed to make the reaction happen, this is endothermic.
4. There are 4 Energy Units. Some you “know”, some will sound odd, that’s okay.
5A. Calories are “food” calories. That’s a loose word that really means
1000 calories (with a small “c”)
5B. 1 Calorie = 1000 calories
5C. A Calorie is a kilo-calorie or 1000 calories.
6. In our class we will use CAPITAL C for the food or kilo calories.
With a lower case “c”, calories are the “science” calories, a smaller amount of energy.
There are 1000 calories in 1 Calorie.
7.
A calorie is the amount of energy required
to raise the temperature of one gram of
pure water by exactly 1 Kelvin (or 1.0°C)
1000 cal = 1 C = 1 kcal
8. A Hershey bar has 230 Calories, how many calories is that?
This is a one step unit conversion problem (do it now…)
8. A Hershey bar has 230 Calories, how many calories is that?
This is a one step unit conversion problem (do it now…)
230 C x 1000 cal
= 230,000 cal
1
1C
9. Another energy unit is the JOULE,
It’s named after J.P. Joule, the English physicist who
studied energy & work. His ideas led to the development
of the theory (now a law) of conservation of energy in
a chemical reactions and physical changes.
Joules + calories are related
with this equality:
4.18 Joules = 1.0 calories
10.
The last unit is kilo-Joule or kJ
11.
1000 Joules = 1 kilo-Joule
12. 1.0 C = 1000 cal = 4180 j = 4.180 kJ
Converting back and forth between units will
take some practice, but not much.
Table B (let’s take a lookie-look at what gifts we have there!)
14. The title of table B is…
15. Three Physical Constants of Water are on table B.
These 3 constants are for water.
Every substance will have these 3 constants, although
their numbers will be different than water’s numbers.
Table B
16. The Heat of Fusion symbol is: HF
17. The heat of fusion for water means the amount of energy required to be
added to one gram of ice at the melting point, to melt it into one gram of
water, without changing the temperature.
18. This means, 273 K solid to 273 K liquid, NO TEMPERATURE CHANGE
This is only phase change
It’s the cold phase change energy for water.
19. It also means the reverse: how much energy needs to be removed
from one gram of liquid water at the freezing point to turn it into a
solid.
20. Heat of fusion for water is 334 Joules/gram
This is ~1 GRAM of ICE (CUBE)
= 1 gram of H2O(S)
To MELT this gram of ice it takes a
CERTAIN AMOUNT of energy added in,
called the HEAT OF FUSION for water.
21. To MELT 1 gram of
ICE into water it requires
334 Joules of energy
be added
= about 1 gram of H2O(L)
To FREEZE this gram of ice it takes a
CERTAIN AMOUNT of energy removed,
called the HEAT OF FUSION for water.
22. To turn 1 gram of water
into an ICE CUBE it requires
334 J of energy
be removed
23. The Heat of Vaporization symbol is HV
24. The heat of vaporization for water means the amount of energy
required to be added to one gram of water at the boiling point, to
vaporize it into one gram of steam gas without changing the
temperature. This means, 373 K liquid to 373 K gas.
It’s the hot phase change energy for water.
25. It also means the reverse: how much energy needs to be removed
from one gram of steam at the condensing point to turn it into a liquid.
26. The Heat of vaporization for water is 2260 J/g
-----------------------------------------------
For water…
HF = 334 J/g
HV = 2260 J/g
27. The last constant for water here is the specific heat capacity who’s symbol is C
28. The Specific Heat Capacity is the amount of energy required to change the
temperature of one gram of water by one Kelvin (or one degree centigrade).
This change can be heating, or cooling. Energy added will heat the water,
energy removed will cool the water.
29. This constant can only be used when H2O is liquid phase, and
there is a change in temperature.
30. The specific heat capacity constant for water is
C = 4.18 Joules/gram x Kelvin (which we write as…)
31. C = 4.18 J/g·K
The “dot” means multiply
-----------------------------------------------
DO this NOW: Go to table B. Under it, in small letters, add:
1 C = 1000 cal
4.18 J = 1 cal
1000 J = 1 kJ
Objective Thermo Class #2: Heats of Reaction, ΔH
32. Table I shows us the HEATS OF REACTION for 25 different
balanced chemical reactions.
33. This means how much energy is released (exo) or absorbed (endo)
by these reactions when they occur at room temperature and
normal pressure. (this is not STP)
Let’s look over this first reaction on Table I Heats of Reaction
the combustion of methane gas.
CH4 + 2O2 ---> CO2 + 2H2O
we can read this as:
1 mole CH4 combusts with 2 moles O2
forming into 1 mole CO2 + 2 moles H2O
34. The ΔH for this reaction is -890.4 kJ/mole
35. This means: that when one mole of methane combusts, it
releases 890.4 kJ of energy.
36. (the - sign only means exothermic!)
37. There’s NO SUCH THING as negative energy.
All energy is energy.
38. So, this becomes...
CH4 + 2O2 ---> CO2 + 2H2O + 890.4 kJ
we can understand this as:
1 mole CH4 combusts with 2 moles oxygen,
into 1 mole CO2 and 2 moles water + 890.4 kJ energy
CH4 + 2O2 ---> CO2 + 2H2O + 890.4 kJ
You should be able to do this stoichiometry problem:
If 23.4 moles of methane combust, how many molecules of water form?
39. Well, now I’m telling you
that ENERGY is also in the mole ratio as well.
40. The mole ratio for the combustion of methane is 1:2:1:2:890.4 kJ
41. If 23.4 moles of methane combust, how much energy is released?
MR
CH4
energy
1
890.4 kJ
23.4
X kJ
X = (890.4)(23.4) = 20,835.36 kJ
X = 20,800 kJ with 3 SF
42. You eat about 1450 Calories per day.
Convert that amount of energy into
calories, joules, and kilojoules.
42. You eat about 1450 Calories per day.
Convert that amount of energy into
calories, joules, and kilojoules.
1450 C
1
X
1000 cal
1C
= 1,450,000 calories
Keep the 3 SF
1,450,000 calories x 4.18 Joules = 6,061,000 J
1 calorie
1
= 6,060,000 J
with 3 SF
1 kJ
6,060,000 J X
= 6,060 kJ
1000 J
1
keeping 3 SF
43. A teaspoon of sugar (sucrose, C12H22O11)
has 41 Calories. How many calories, joules
and kilojoules is in 2 teaspoons of sugar?
A normal amount in a cup of tea or coffee?
Show work for all three problems.
43. A teaspoon of sugar (sucrose, C12H22O11)
has 41 Calories. How many calories, joules
and kilojoules is in 2 teaspoons of sugar?
82 C X 1000 cal = 82,000 cal with 2 SF
1
1C
82,000 cal X 4.18 J = 342,760 J = 340,000 Joules with 2 SF
1
1 cal
340,000 J X 1 kJ = 340 kJ with 2 SF
1
1000 J
44.
Propane gas, C3H8 combusts according to Table I. How
much energy (in kilo joules) is released when
5.75 moles C3H8 combusts?
Hint: this is a simple mole to mole type stoich problem
44. Propane gas, C3H8 combusts according to Table I
How much energy (in kilo joules) is released when 5.75 moles C3H8 combusts?
C3H8 + 5 O2 ---> 3CO2 + 4 H2O
ΔH = -2219.2 kJ
Which means: C3H8 + 5 O2 ---> 3CO2 + 4 H2O + 2219.2 kJ
You need to get right now, that when 1 mole C3H8 combusts
that 2219.2 kJ of energy is released –ΔH indicates
exothermic. You also need to grasp that energy is part of the
actual mole ratio. So,
MR
C3H8
energy
1
2219.2 kJ
5.75
X kJ
X = 12760.4 kJ
X = 12,800 kJ with 3 SF
45. How much energy is absorbed by the reaction of
99.0 moles of HI gas forming? (hint, look on table I)
45.
H2(G) + I2(G) ---> 2HI (G)
ΔH = +53.0 kJ
Which means, when 2 moles HI gas forms, then 53.0 kJ of energy is absorbed.
Use the mole ratio to do this problem for 99.0 moles HI gas.
MR
HI
energy
2
53.0 kJ
99.0
X kJ
2X = 5247 kJ
X = 2623.5 kJ
X = 2620 kJ
with 3 SF
OBJECTIVE: Using table B constants
with energy changes in
real life problems
(get a calculator)
You have a 35.0 g ice cube at 273 K. If you put it into your
mouth to melt it, how much energy will it take to convert it
into liquid water without temperature change?
To do this we need a formula from table T, the heat of fusion formula. In this case
we need to “un-fuse” ice into liquid water.
46. Write the formula, then under it, fill in the blanks, with units!
q = mHF
47. q is the amount of heat in joules
m is the mass in grams of H2O
HF is the specific heat capacity constant = 334 J/g
which is how much energy will it take to change
1 g of ice → 1 g water when ΔT is zero
48. You have 35.0 g of ice at 273 K.
How much energy to melt it without
raising the temperature?
q = mHF
48. You have 35.0 g of ice at 273 K.
How much energy to melt it without
raising the temp?
q = mHF
q = (35.0 g) x (334 J/g)
q = 11,690 J = 11,700 J
with 3 SF
49.
Calculate how much energy
must be removed to freeze
1550 mL of water (in Joules)
q = mHF
49. Calculate how much energy must be removed to
freeze 1550 mL of water (in kJ)
q = mHF
q = (1550 g) x (334 J/g)
q = 517,700 Joules
q = 518,000 J
with 3 SF
50. How many joules of energy will it take to
vaporize 550. grams of water at the boiling
point into gas at the same temperature?
Hint: this time we’ll use the heat of vaporization formula
with the HV constant.
This is not the fusing or unfusing of water at 273 K,
This is the HOT phase change at 373 Kelvin.
50. First , write the formula
q = mHV
50. How many joules of energy will it take to
vaporize 550. grams of water at the
boiling point into gas at the same temp?
q = mHV
q = (550. g) x (2260 J/g)
q = 1,243,000 Joules = 1,240,000 J
with 3 SF
51.
When you fill a large tub with hot
water it will cool down as energy is lost
to the air. If your tub holds 62.5 L of
water, and it is at 318 Kelvin, and after
15 minutes the water is cooled down to
301 K, how much energy has this water
lost? Calculate J and kJ
Hint: this is a change in temperature formula
problem, so you cannot use the
HF or HV formula. Use
q = mCΔT
51. When you fill a large tub with hot water it will cool down
as energy is lost to the air. If your tub holds 62.5 L of water,
and it is at 318 Kelvin, and after 15 minutes the water is
cooled down to 301 K, how much energy has this water lost?
Calculate J and kJ
q = mCΔT
51. When you fill a large tub with hot water it will cool down
as energy is lost to the air. If your tub holds 62.5 L of water,
and it is at 318 Kelvin, and after 15 minutes the water is
cooled down to 301 K, how much energy has this water lost?
Calculate J and kJ
q = mCΔT
q = (62,500 g)(4.18 J/g.K)(17.0 K)
q = 4,441,250 J = 4,440,000 Joules
4,440,000 J
1
X
1 kJ
1000J
= 4,440 kJ
Note: grams and
Kelvin both cancel
out, leaving just
joules as a unit.
with 3 SF
with 3 SF
You know how when you go out to eat
and order an appetizer and a meal and a
dessert, and each part has it’s own price
on the check?
And then you order a soda too.
You also have to pay tax on top
of all this. You don’t get a little bill
for each of these parts, the server adds
it all up for you and you pay one price.
You’re smart enough
(and suave enough)
to realize how simple that makes things.
The same thing happens in
thermochem (really).
52. What if you put 155 g of ice at 273 K
in a room for 4 hours. The ice melts, then
the water warms up to 294 K.
How many joules did the H2O absorb?
This problem is like eating out, there are 2 smaller
energy events that must be added into one whole energy
total (no tax though).
155 g of ice at 273; the water reaches 294 K.
52. First, MELT the ice…
q = mHF
Then, warm up the water
q = mCΔT
SUM:
52.
What if you had say 155 g of ice at 273 K and let it sit in a room
for four hours. Then the water has reached 294 K.
How many joules of energy did this H2O absorb to do this?
q = mHF
q = (155 g)(334 J/g) = 51770 J = 51,800 J
q = mCΔT
q = (155 g)(4.18 J/g.K)(21.0 K) = 13605.9 J = 13,600 J
ANSWER = 51,800 J + 13,600 J = 65,400 J
with 3 SF
53. Convert that energy into food
Calories?
ANSWER = 51,800 J + 13,600 J = 65,400 J
with 3 SF
53. Convert that energy into Calories?
65,400 J
1 cal
x
1
4.18 J
= 15,645.933 cal
= 15,600 cal
15,600 cal x
1
1 Calorie
1000 cal
= 15.6 C
54.
If 16.3 grams of steam at exactly 373 K
condenses onto a kitchen window, then cools quickly to
305 Kelvin. How much energy was released?
(condensing is exothermic since energy is released)
Hint: this first must condense through the hot phase change
(heat of vaporization) then cool with a change in temperature
formula as well.
q = mHV
q = mCΔT
54. If 16.3 grams of steam at exactly 373 K condenses
onto a kitchen window, then cools quickly to 305
Kelvin. How much energy was released?
(condensing is exothermic since energy is released)
q = mHV
q = (16.3 g)(2260 J/g) = 36,838 J = 36,800 J
q = mCΔT
q = (16.3 g)(4.18 J/g.K)(68.0 K) = 4633.112 J = 4630 J
with 3 SF
ANSWER: 36,800 J + 4630 J = 41430 J =
41,400 J
with 3 SF
ThermoChem Class #4 – Calorimeters
OB: how is energy in food measured?
(It’s the “bomb”!)
You will need a calculator and
need to put your thinking hat now.
By Law, all food sold in the
USA has to have a food
Nutrition Label, with one of
the “facts” being the Caloric
Content, how much energy
are you eating, put onto the
package.
It takes a special machine,
and careful measurements,
and fancy mathematics and
conversions, to calculate this
data.
In our Dorito’s Lab, we will
attempt to measure the
energy content in the chips,
and compare it to the
nutritional label on the
package.
55. If we find that the water absorbs 8.85 C,
how many Joules of energy came out of the chip?
(assume no energy loss).
55. If we find that the water absorbs 8.85 C,
how many Joules of energy came out of the chip?
(assume no energy loss).
8.85C
1
885 cal
1
X
1000 cal = 885 cal
1C
X
4.18 J
1 cal
= 4248 Joules
= 4250 J
(3SF)
56. There is no easy way to directly measure the energy that is in food. An
indirect way has been well figured out, using a machine called a calorimeter.
The food is burned up in a sealed box containing oxygen, and the heat
that it gives off, the energy that is inside the food, is used to heat
pure water. This water is carefully measured, and the energy gained
by the water changes its temperature. This temperature change is
pushed through our q = mC∆T formula, and you can calculate the
number of joules it took to heat up the water, and then convert the
joules into cal, then into calories.
It seems harder than it is, but it is easy enough to do tomorrow.
In our Dorito’s Lab we will use a five cent calorimter machine. Real
calorimeters are much more complex, and insulated. They measure all of
the energy gain, we will “do” the same thing, but MOST of our energy
will get lost to the air.
We’ll do the same process, just get poor results. Terrible measuring,
great learning!
57.
Let’s draw and label a calorimeter now. Picture on next slide.
Food goes into the cup. It’s
sealed in the “bomb” with
oxygen. The electricity leads
are attached. This is placed
into a certain mass of pure
water, at measured
temperature. The stirrer
moves the water around and
makes sure it’s all at the same
temperature. At some point
you spark and burn the food.
Heat is released into the
water and the temperature
rises.
Using q = mC∆T, calculate the
number joules required to
make that temperature
change, convert to cal, then
Calories.
58. Let’s assume that there is exactly 2120. mL of water in our
bomb calorimeter and it’s at exactly 295.0 Kelvin. After burning up
our food sample, the temperature of the water rises to 354.5 Kelvin.
How many Calories of energy are in this food sample?
To do this, start with the q = mC∆T formula. Solve for q.
58. Let’s assume that there is exactly 2120. mL of water in our bomb calorimeter and it’s
at exactly 295.0 Kelvin. After burning up our food sample, the temperature of the water
rises to 354.5 Kelvin. How many Calories of energy are in this food sample?
To do this, start with the q = mC∆T formula. Solve for q.
q = mC∆T
q = (2120. g)(4.18 J/g·K)(59.50 K)
q = 527,265.2 J
527,265.2 J
1
X
1 cal
= 126,140 cal
4.18 J
126,140 cal = 126.1 Calories
(4 SF)
59.
In a different calorimeter you measure some
food energy content. You have exactly 4005 mL
of pure water at 274.2 Kelvin and it warms up to
365.4 Kelvin when you burn up 50.00 grams of
chocolate delight dessert. What is the Calorie
content of the dessert? Round to the nearest 10th C
59. In a different calorimeter you measure some food energy content. You
have exactly 4005 mL of pure water at 274.2 Kelvin and it warms up to 365.4
Kelvin when you burn up 50.00 grams of chocolate delight dessert. What is
the Calorie content of the dessert?
q = mC∆T
q = (4005. g)(4.18 J/g·K)(91.20 K)
q = 1,526,770 Joules
1,526,770 J
1
X
1 cal
= 365,255 cal
4.18 J
365,255 cal = 365.3 Calories
(4 SF)
60. Calculate the specific heat capacity constant for
aluminum if your hunk of pure aluminum has mass of 147.2 g
and the temperature changes from 279 K to 365 K when
you impart 11,355 Joules of energy onto it.
WRITE THE FORMULA, what are you solving for?
60. Calculate the specific heat capacity constant for
aluminum if your hunk of pure aluminum has mass of 147.2 g
and the temperature changes from 279 K to 365 K when you
impart 11,393 Joules of energy onto it.
q = mCΔT
11,355 J = (147.2g)(C)(86.0K)
11,355 J
(147.2g)(86.0K)
=C
0.897 J/g·K = C
NOTE: metals have a fairly low specific heat capacity, water’s is 4.18 J/g·K
Water is a common enough substance in your life that the regents will not remind
you what phase changes it might go through in any problem. You will have to
determine that from the temperatures in the problem.
Ice can be colder than 273 K (0°C) and can be made colder or warmer while
staying a solid. Water phase changes at 273 K.
When ever there is a temperature change (ice can change temp) we use
the basic heat formula: q = mCΔT
If water is liquid, we use the 4.18 J/g·K constant.
When the H2O is solid ice, it has a different constant,
61.
CICE = 2.10 J/g·K
When you are given information like this, you must use it in your
problems, like the next slide will do. Write the ice constant near table B
in your reference table now.
62. For fun, you obtain a snowball (255 g) at -5.00°C
and put it in the back of your friend’s shirt. It melts
then warms up to
body temperature of 36.0°C.
How much energy does that take? (do math on next slide)
q = mHF
268 K
q=mCΔT
-5.00°C
q=mCΔT
309 K
0°C
36.0°C
C = 2.10 J/g·K
C = 4.18 J/g·K
62. For fun, you obtain a snowball (255 g) at -5.00°C
and put it in the back of your friend’s shirt. It melts
then warms up to
body temperature of 36.0°C.
How much energy does that take?
This is a 3 step thermochem problem, ice warms, ice melts, water warms.
Make note to use the proper “C” value for your H2O in the proper place.
Add up the three sets of Joules for one total answer.
62. For fun, you obtain a snowball (255 g) at -5.00°C and put it in the back
of your friend’s shirt. It melts then warms up to
36.0°C. How much energy does that take?
body temperature of
q=mCΔT = (255g)(2.10J/g·K)(5.00 K) = 2677.5 J
q = mHF =
(255g)(334J/g) = 85,170 J
q=mCΔT = (255g)(4.18J/g·K)(36.0K) = 38,372.4 J
Thermochem is like eating out in
a fancy restaurant, multi courses
with ONE total bill.
Add up the parts - sum.for total
126,219.9 J
1,260,000 J
with 3 SF
63. Compare how much energy it takes to condense
15.0 g steam into water - to the amount of energy it
takes to vaporize 15.0 g water into steam.
Vaporize 15.0 g water
q = mHV
Condense 15.0 g water
q = mHV
q = (15.0g)(2260J/g)
q = (15.0g)(2260J/g)
q = 33,900 J
q = 33,900 J
These are obviously the same. What’s
important to understand is which is
EXOTHERMIC & which is ENDOTHERMIC?
63. Compare how much energy it takes to condense
15.0 g steam into water - to the amount of energy it
takes to vaporize 15.0 g water into steam.
Vaporize 15.0 g water
q = mHV
Condense 15.0 g water
q = mHV
q = (15.0g)(2260J/g)
q = (15.0g)(2260J/g)
q = 33,900 J
q = 33,900 J
This is the endothermic
one, add heat energy
to vaporize water.
This one is exothermic,
remove heat energy
to condense steam.
64.
When 56.0 grams of carbon and sufficient hydrogen
synthesize completely into ethane (C2H6) gas, how
much energy is released (or absorbed) in JOULES?
64.
When 56.0 grams of carbon and sufficient hydrogen
synthesize completely into ethane (C2H6) gas, how
much energy is released (or absorbed) in JOULES?
NOTE: ΔH is a negative number, that means energy is a product,
which we can write this way…
2C + 3H2
56.0 g C
1
MR
X
carbon
energy
C2H6 + 84.0 kJ
1 mole C = 4.67 moles C
12 g C
1
84.0
4.67
x
2X = 392.28 kJ
X = 196 kJ
66.
A
Cooling curve for Chromium metal
C
1
B
K
E
L
V
I
N
A. What temps are 1 + 2?
B. What’s PE doing BC and CD?
C. What’s KE doing AB and DE
D. Why is BC longer than DE?
E. Which thermochem formula do you use for BC?
2
F. How about for EF?
E
D
F
Heat removed at a constant rate (time)
66.
Cooling curve for Chromium metal
A
KE
PE
2944
B
K
E
L
V
I
N
C
q = mHV
PE
Why is BC longer than DE?
BC represents the condensation phase
change for chromium gas to liquid,
which is a “bigger” energy event that the
freezing of chromium.
KE
2180
E
D
q = mCΔT
Heat removed at a constant rate (time)
F
Converting balanced chemical equations into balanced thermochemical equations.
67. Look at table I, choose the second equation: propane combusts.
Write the balanced chemical equation with the ΔH
68. Write a balanced thermochemical equation
69. Find the most endothermic equation on Table I, write the balanced
chemical equation with the ΔH
70. Write the balanced thermochemical equation for this reaction as well.
Converting balanced chemical equations into balanced thermochemical equations.
Look at table I, choose the second equation, propane combusts. Write the
balanced chemical equation with the ΔH, then, write a balanced
thermochemical equation below it (properly).
67.
C3H8 + 5O2
68.
C3H8 + 5O2
3CO2 + 4H20
ΔH= -2219.2 kJ
3CO2 + 4H20 + 2219.2 kJ
Energy is a PRODUCT in an exothermic reaction
Then, write the most endothermic equation of all, with the ΔH, then
below that one, the balanced thermochemical equation properly.
69.
2C + H2
C2H 2
70.
2C + H2 + 227.4 kJ
ΔH = +227.4 kJ
C2 H 2
Energy is a REACTANT in an endothermic reaction
OBJECTIVE: A review of all types
of thermochemistry problems
Celebration of Knowledge
in
Thermochemistry
Is on
____________________________
71. You decide to warm up some water for oatmeal. How
much energy does it take to warm up 354 mL of water
from 24.5°C to the boiling point and to let 7.50 grams
vaporize making the teapot whistle.
Be careful and do good calculations.
71. You decide to warm up some water for oatmeal. How
much energy does it take to warm up 354 mL of water
from 24.5°C to the boiling point and to let 7.50 grams
vaporize making the teapot whistle.
Be careful and do good calculations.
q = mCΔT = (354 g)(4.18 J/g·K)(75.5 K) = 111,718.86 J
q = mHV
= (7.50 g)(2260 J/g) = 16950 J
111,718.86 J + 16960J = 128,668.86 J
129,000 Joules with 3SF
72. Write the balanced thermochemical equation for the
synthesis of aluminum oxide, then calculate how much
energy is released (or absorbed) when you form 21.0 moles
of Al2O3.
72. Write the balanced thermochemical equation for the
synthesis of aluminum oxide, then calculate how much
energy is released (or absorbed) when you form 21.0 moles
of Al2O3.
4Al + 3O2 → 2Al2O3 + 3351 kJ
MR
Energy
Al2O3
2x = 70371
3351 kJ
2
x = 35,185.5 kJ
X kJ
21.0
35,200 kJ with 3 SF
73. Draw the heating curve for phosphorous.
Indicate the actual MP and BP. Label graph
points ABCDEF left to right.
What formula would you use
For BC
For CD
For DE
What happens to PE + KE at BC and CD?
73. Heating Curve for Phosphorous
F
D
554 K
E
BC use: q = mHF
CD use: q = mCΔT
B
317 K
C
A
DE use: q = mHV
Heat added at a constant rate →
BC: KE steady, PE increases
CD: KE increases, PE steady
74. 45.0 g of an unknown metal absorbs 1.51 kiloJoules of
heat. The temperature changes from 268 K to 345 K. What
is the specific heat capacity constant for this metal?
74. 45.0 g of an unknown metal absorbs 1.50 kiloJoules of
heat. The temperature changes from 268 K to 345 K. What
is the specific heat capacity constant for this metal?
q = mCΔT
1510 J = (45.0g)(C)(77.0K)
1510J
(45.0g)(77.0K)
3465g·K
=C
= 0.436 J/g·K
75. Which takes more energy?
Melting 50.0 g of ice or
vaporizing 50.0 g of water into steam?
75. Which takes more energy?
Melting 50.0 g of ice or vaporizing 50.0 g of water into steam?
No math needed, just think…
HF constant for water is 334 J/g
HV constant for water is 2260 J/g
It will take more than 7X the energy
to vaporize than melt
76. Which takes more energy?
Heating 23.0 mL of water from 274 K to 299 K
or
Heating 23.0 mL of water from 299 K to 323 K?
76. Which takes more energy?
Heating 23.0 mL of water from 274 K to 299 K, or,
Heating 23.0 mL of water from 299 K to 323 K?
Think. The ΔT for the first one is 25.0 K
The ΔT for the second one is only 24.0 K
NO PHASE CHANGES to think about, so the first one
requires slightly more energy (4.18 J/g because it’s
one more temperature unit of ΔT)
77. Which takes more energy?
Vaporizing 21.0 g water from 373 K liquid to gas
or
Changing the temp of 100.0 g H2O by 97.0 K?
77. Which takes more energy? [DO THE MATH THIS TIME]
Vaporizing 21.0 g water from 373 K liquid to gas, or,
100 g water from 274 K to 371 K?
q = mHV = (21.0g)(2260J/g) = 47,460 J = 47,500 J 3SF
q = mCΔT = (100.0g)(4.18J/g·K)(97.0K) = 40,500 J 3SF
The first one takes more energy!
78. Which has the LOWEST AVERAGE KINETIC ENERGY?
100 mL water at 51.0°C
100 mL water at 50.0°C
175 mL water at 49.0°C
175 mL water at 23.0°C
78. Which has the LOWEST AVERAGE KINETIC ENERGY?
100 mL water at 50.0°C
100 mL water at 50.0°C
175 mL water at 49.0°C
1750 mL water at 23.0°C
Lowest temp = lowest average KE always.
Same in reverse:
Highest temp = Highest Average KE
This horse
has a boo boo.
That’s an ICE PACK
on his leg.
79. Describe the
thermochem:
A. Heat flows from ice pack → leg
B. Cold flows from ice pack → leg
C. Heat flows from leg → ice pack
D. Cold flows from leg → ice pack
Energy can be transferred.
Heat is energy,
cold is THE LACK OF ENERGY.
Cold cannot flow.
What happens to you when
you jump into a cold pool?
Heat leaves your body and
you get chilly.
The COLD
DOES NOT FLOW
INTO YOU.
79. C. Heat flows from leg → ice pack
2nd Q ends ______________________________________
ALL BACK WORK DUE by _____________
Midterm is NEXT _______________________.
It counts 2x celebrations.