4-1 Fundamentals of Calorimetry Lab Report Dr Z Edits
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The Fundamentals of Calorimetry
Kristen Brown
11/17/2023
Data Table 1A
Time
(min)
Trial 1 Temp. °C
Trial 2 Temp. °C
1
45
45
2
44
45
3
44
44
4
43
43
5
43
43
6
42
42
7
41
42
8
41
41
9
40
40
10
40
40
Question 1: Create a spreadsheet and graph of the data from Data Table 1A, plotting Temperature vs. Time. Insert a trendline. Use the y intercept to find the temperature at time 0, (T
0
), when the two volumes of water are mixed. Insert a copy of your graph below. Be sure to display the equation of the line on the graph. Your name and date must be
included in the title on your graph
. Be sure the equations of the lines are displayed clearly and are legible. 0
2
4
6
8
10
12
37
38
39
40
41
42
43
44
45
46
f(x) = − 0.6 x + 45.8
f(x) = − 0.58 x + 45.47
Calorimeter Temperature
Kristen Brown 11/17/2023
Trial 1 Temp
Linear (Trial 1 Temp)
Trial 2 Temp
Linear (Trial 2 Temp)
Time (min)
Temperature (Celcius)
©2016 2 Carolina Biological Supply Company
Data Table 1B: You must include your work, with units, in the Data Table. You cannot complete this data table without the graph/equation of the line from question 1. Calculation
#
Trial 1
Trial 2
Mass of cold water
50g
50g
Initial temperature of cold water = T
cold (°C)
20(°C)
20(°C)
Initial temperature of warm water = T
hot (°C)
85(°C)
85(°C)
Temperature at time 0 from graph, T
0
, (°C)
80(°C)
80(°C)
2
Change in Temp Cold Water (
ΔT
Cold = T
0
-
T
cold
)
(
ΔT
Cold = T
0
-
T
cold
)
80(°C) – 20(°C) = 60(°C)
(
ΔT
Cold = T
0
-
T
cold
)
80(°C) – 20(°C) = 60(°C)
3
Change in Temp Hot Water (
ΔT
Hot
= T
0 -
T
Hot
)
(
ΔT
Hot
= T
0 -
T
Hot
)
80(°C) – 85(°C) = -5(°C)
(
ΔT
Hot
= T
0 -
T
Hot
)
80(°C) – 85(°C) = -5(°C)
4
Change in q
Cold Water (final unit should be J)
∆q
(cold water)
= C
water
x ΔT
Cold
x mass of water
∆q
(cold water)
= C
water
x ΔT
Cold
x
mass of water
4.18J/(°C)g x 60(°C) x 50g =
12,540 J ∆q
(cold water)
= C
water
x ΔT
Cold
x mass of water
4.18J/(°C)g x 60(°C) x 50g =
12,540 J
5
Change in q Hot Water (J)
∆q
(hot water)
= C
water
x ΔT
Hot
x mass of water
∆q
(hot water)
= C
water
x ΔT
Hot
x
mass of water
4.18J/(°C)g x -5(°C) x 50g =
-1,045 J
∆q
(hot water)
= C
water
x ΔT
Hot
x mass
of water
4.18J/(°C)g x -5(°C) x 50g =
-1,045 J
6
Heat (Energy) gained by the Calorimeter (final unit should be J) ∆q
cal
= |
∆q
(cold water)
+
∆q
(hot water)
|
**∆q
cal
should be a positive value** ∆q
cal
= |
∆q
(cold water)
+
∆q
(hot water)
|
|12,540J + -1,045J| = 11,495 J
∆q
cal
= |
∆q
(cold water)
+
∆q
(hot water)
|
|12,540J + -1,045J| = 11,495 J
7
Heat capacity of calorimeter
C
calorimeter
(final unit 11,495J/60(°C) = 11,495J/60(°C) = ©2016 2 Carolina Biological Supply Company
should be J/°C)
∆q
cal
= C
calorimeter x |ΔT
Cold
|
191.583 J/(°C)
191.583 J/(°C)
8
Average Heat Capacity of Calorimeter, (
C
AVE
) in J/°C
191.583 J/(°C)
Data Table 2a You must include your work, with units, in the Data Table
. For C
w
use 4.18 J/g∙
℃
Calculation
#
5g CaCl
2
10g CaCl
2
15g CaCl
2
Mass of water (g)
m
w
100.0 g
100.0 g
100.0 g
Mass of salt (g) m
s
5 g
10 g
15 g 9
Moles of salt solution
(
m
s
x mol/g) 5g / 110.98g/mol =
0.0451 moles of
CaCl
2
10g / 110.98g/mol =
0.0901 moles of CaCl
2
15g / 110.98g/mol =
0.1352 moles of CaCl
2
Initial Temperature, T
i
(°C) 20 (°C)
20 (°C)
20 (°C)
Final Temperature, T
f
(°C) 27 (°C)
33 (°C)
42 (°C)
10
Change in Temperature (°C) ∆T = T
f
- T
i
∆T = T
f
- T
i
27(°C) - 20(°C) =
7 (°C)
∆T = T
f
- T
i
33(°C) - 20(°C) =
13 (°C)
∆T = T
f
- T
i
42(°C) - 20(°C) =
22 (°C)
11
Heat absorbed by the solution (J) q
w
= -
[c
w
x m
w
x ∆T
] q
w
= -
[c
w
x m
w
x ∆T
]
-[4.18J/g(°C) x 100g
x 7(°C) =
-2,926 J
q
w
= -
[c
w
x m
w
x ∆T
]
-[4.18J/g(°C) x 100g x
13(°C) =
q
w
= -
[c
w
x m
w
x ∆T
]
-[4.18J/g(°C) x 100g x
22(°C) =
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-5,434 J
-9,196 J
8
Average Heat Capacity of the Calorimeter,
C
AVE
. From Data Table 1B 191.583
J/(°C)
191.583
J/(°C)
191.583
J/(°C)
12
Heat absorbed by the calorimeter (J) q
c
=
-
[ C
AVE
x ∆T
] q
c
=
-
[ C
AVE
x ∆T
] -[191.583J/(°C) x
7(°C)] =
-1,341.081 J
q
c
=
-
[ C
AVE
x ∆T
] -[191.583J/(°C) x
13(°C)] =
-2,490.579 J
q
c
=
-
[ C
AVE
x ∆T
] -[191.583J/(°C) x 22(°C)] =
-4,214.826 J
13
Enthalpy of solution (J) ∆H = q
w
+ q
c
∆H = q
w
+ q
c
-2,926J+ -1,341.081J
=
-4,267.081 J
∆H = q
w
+ q
c
-5,434J + -2,490.579J
=
-7,924.579 J
∆H = q
w
+ q
c
-9,196J + -4,214.826J
=
-13,410.826 J
14
∆H (
Enthalpy of solution) in kJ *Note: 1 kJ = 1000 J -4,267.081J / 1000kJ
= -4.267081 kJ
-7,924.579J / 1000kJ
= -7.924579 kJ
-13,410.826J / 1000kJ
= -13.410826 kJ
15
∆H/moles of salt Enthalpy per mole of solution in kJ/mol
5g CaCl
2
x 1mol
CaCl
2
/ 110.98g CaCl
2
= 0.0450mol
-4.267081kJ/
0.0450mol =
-94.8 kJ/mol
10g CaCl
2
x 1mol
CaCl
2
/ 110.98g CaCl
2
= 0.0901mol
-7.924579kJ/
0.0901mol =
-87.9 kJ/mol
15g CaCl
2
x 1mol CaCl
2
/
110.98g CaCl
2
= 0.1352mol
-13.410826kJ/
0.1352mol =
-99.2 kJ/mol
Data Table 2b You must include your work, with units, in the Data Table.
Calculation #
5g NH
4
Cl 10g NH
4
Cl 15g NH
4
Cl Mass of water (g)
m
w
100 g
100 g
100 g
Mass of salt (g) m
s
5 g
10 g
15 g
©2016 2 Carolina Biological Supply Company
16
Moles of salt solution
(
m
s
x mol/g) 5g / 53.49 mol/g =
0.093 moles of NH
4
Cl
10g / 53.49 mol/g =
0.187 moles of NH
4
Cl
15g / 53.49 mol/g =
0.280 moles of NH
4
Cl
Initial Temperature, T
i
(°C) 20 (°C) 20 (°C) 20 (°C) Final Temperature, T
f
(°C) 17 (°C)
12 (°C)
7 (°C)
17
Change in Temperature (°C) ∆T = T
f
- T
i
∆T = T
f
- T
i
17
(°C)-20(°C) =
-3 (°C)
∆T = T
f
- T
i
12
(°C)-20(°C) =
-8 (°C)
∆T = T
f
- T
i
7
(°C)-20(°C) =
-13 (°C)
18
Heat absorbed by the solution (J) q
w
= -
[c
w
x m
w
x ∆T
] q
w
= -
[c
w
x m
w
x ∆T
] - [4.18J/g
(°C) x 100g x
-3
(°C)] =
1,254 J
q
w
= -
[c
w
x m
w
x ∆T
] - [4.18J/g
(°C) x 100g x
-8
(°C)] =
3,344 J
q
w
= -
[c
w
x m
w
x ∆T
] - [4.18J/g
(°C) x 100g x
-13
(°C)] =
5,434 J
8
Average Heat Capacity of the Calorimeter, C
AVE
. From Data Table 1B 191.583 J/(°C)
191.583 J/(°C)
191.583 J/(°C)
19
Heat absorbed by the calorimeter (J) q
c
=
-
[ C
AVE
x ∆T
] q
c
=
-
[ C
AVE
x ∆T
] - [
191.583J/(°C) x
-3(°C)]
= 574.749 J
q
c
=
-
[ C
AVE
x ∆T
] - [
191.583J/(°C) x
-8(°C)]
= 1,532.664 J
q
c
=
-
[ C
AVE
x ∆T
] - [
191.583J/(°C) x
-13(°C)]
= 2,490.579 J
20
Enthalpy of solution (J) ∆H = q
w
+ q
c
∆H = q
w
+ q
c
1,254J + 574.749 J
= 1,828.749 J
∆H = q
w
+ q
c
3,344J + 1,532.664J
= 4,876.664 J
∆H = q
w
+ q
c
5,434J + 2,490.579J
= 7,924.579 J
21
∆H (
Enthalpy of solution) in kJ *Note: 1 kJ = 1000 J 1,828.749J / 1000kJ =
1.828749 kJ
4,876.664J / 1000kJ =
4.876664 kJ
7,924.579J / 1000kJ =
7.924579 kJ
22
∆H/moles of salt Enthalpy per mole of solution in kJ/mol
5g NH
4
Cl x 1mol
NH
4
Cl / 53.49g NH
4
Cl
= 0.0935mol
1.828749kJ/ 0.0935mol
=
19.6 kJ/mol
10g NH
4
Cl x 1mol
NH
4
Cl / 53.49g NH
4
Cl
= 0.18695mol
4.876664kJ/
0.18695mol =
26.1 kJ/mol
15g NH
4
Cl x 1mol
NH
4
Cl / 53.49g NH
4
Cl
= 0.2804mol
7.924579kJ/ 0.2804mol
=
28.3 kJ/mol
Question 2: Create a spreadsheet and graph for CaCl
2
and NH
4
Cl, using the data from Data Table 2a and b. Plot mass on the X axis and change in temperature on the Y axis for both graphs. The slope will be the change in temperature per gram of salt dissolved. Insert a trendline and display the equation
on the graph. Include your name and date in the titles.
Insert graphs below the Data Tables 3 and 4. Data Table 3
Mass CaCl
2
∆T 5 g
7
10 g
13
15 g
22
Data Table 4 ©2016 2 Carolina Biological Supply Company
Mass NH
4
Cl ∆T 5 g
-3
10 g
-8
15 g
-13
Copy and Paste Question 2 graph(s) here
: 4
6
8
10
12
14
16
0
5
10
15
20
25
7
13
22
f(x) = 1.5 x − 1
Sodium Chloride
Kristen Brown 11/19/2023
Temp (Celsius)
Linear (Temp (Celsius))
Mass (g)
Temperature (Celsius)
4
6
8
10
12
14
16
-14
-12
-10
-8
-6
-4
-2
0
-3
-8
-13
f(x) = − x + 2
Ammonium Chloride
Kristen Brown 11/19/2023
Temp (Celsius)
Linear (Temp (Celsius))
Linear (Temp (Celsius))
Mass (g)
Temperature (celsius)
Question 3: What sort of relationship exists between the temperature change and the mass of the solid? Explain
why that relationship exists. Hint: See your lab manual Background.
The relationship this question is referring to is when the salts are dumped into the water and dissolved. The heat released or absorbed by the dissolving salts during the reaction is what is calculated by measuring the temperature change.
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Question 4: How do the calculated molar heats (∆H/moles of salt) of solution for calcium chloride compare to one another? How do the calculated molar heats of solution for ammonium chloride compare to one another? Hint: Compare the values
of the determined Enthalpy/mole of solution for each compound at the different masses. Are they similar? How similar?
Did the values you determine differ a lot? A little? Include numerical information in your response. 1.
Calcium Chloride
∆H/moles of salt
Enthalpy per mole of solution in kJ/mol
5g CaCl
2
x 1mol CaCl
2
/
110.98g CaCl
2
=
0.0450mol
-4.267081kJ/ 0.0450mol
=
-94.8 kJ/mol
10g CaCl
2
x 1mol CaCl
2
/
110.98g CaCl
2
=
0.0901mol
-7.924579kJ/ 0.0901mol =
-87.9 kJ/mol
15g CaCl
2
x 1mol CaCl
2
/
110.98g CaCl
2
= 0.1352mol
-13.410826kJ/
0.1352mol =
-99.2 kJ/mol
I think it is very interesting how the numbers appeared they were gradually decreasing with each mass at
first glance, but the 10g mass dipped as compared to the smaller mass (5g), and the biggest mass (15g). I
redid my math for it several times and it seems to be correct, especially when you look at the kJ values, which show a significant decrease. That is what I was expecting out of the kJ/mol values as well--a steady, but gradual decrease. The largest decrease between numbers in the kJ/mol values is 12.
2.
Ammonium Chloride
∆H/moles of salt
Enthalpy per mole of solution in
kJ/mol
5g NH
4
Cl x 1mol NH
4
Cl / 53.49g NH
4
Cl =
0.0935mol
1.828749kJ/ 0.0935mol =
19.6 kJ/mol
10g NH
4
Cl x 1mol NH
4
Cl / 53.49g NH
4
Cl =
0.18695mol
4.876664kJ/ 0.18695mol =
26.1 kJ/mol
15g NH
4
Cl x 1mol NH
4
Cl / 53.49g NH
4
Cl =
0.2804mol
7.924579kJ/ 0.2804mol =
28.3 kJ/mol
These values came out differently than I expected as well. The 10g and 15g kJ/mol values were similar, and the smallest mass (5g) has a pretty decent gap between the other two kJ/mol values. I was expecting these to have a steady incline, but they differ greatly. I thought you’d be able to estimate based on the kJ values since they are evenly spread out with no gap difference. Each value was off by 2 in the kJ values, but the kJ/mol value is off by 7 between 5g and 10g, while there is only a difference of 2 between 10g and 15g with the kJ/mol values.. Question 5: The actual molar enthalpy of solution (∆H/moles of salt) for calcium chloride is -81.3 kJ/mol, whereas the molar enthalpy of solution of ammonium chloride is 14.8 kJ/mol. Calculate the average molar enthalpy of solution for each compound based on your data, and then calculate the percentage error for calcium chloride and ammonium chloride using your calculated average. Do not calculate the percent error for each mass. Percent error is (|Actual – Experimental|/Actual) * 100% Calcium Chloride average:
Average = -94.8 + -87.9 + -99.2 / 3
Average = -93.96
Ammonium Chloride Average:
©2016 2 Carolina Biological Supply Company
Average = 19.6 + 26.1 + 28.3 / 3
Average = 51.13
©2016 2 Carolina Biological Supply Company
Percent error:
Percent error = |-33.25 – (-21.415) / -33.25| x 100%
Percent error = 35.6%
Activity 4 Question 6: Based on the data and graphs from question 2 for calcium chloride and ammonium chloride, determine which compound to use and what quantity of each compound will be needed to make a chemical hot pack and cold pack. Hint: You need to use the equations of the lines. Both packs should be calculated based on using 100 g (100 mL) of water. The hot pack should reach
60 °C, and the cold pack should go down to
3.0 °C from a room temperature of 25 °C. This is a 4-part question! You must show work to earn credit for this question. Hot Pack: Compound needed to reach 60 °C? __
Calcium Chloride
____________________
How many grams are needed? Show your work for this calculation. The equation of the line is y = 1.5x – 1
y is the final temperature and x is the mass of salt
60 = 1.5x - 1
x = (60 + 1) / 1.5
x = 40.7
We need 40.7 g
of calcium chloride per 100 g of water.
Cold Pack: Compound needed to reach 3.0 °C? ___
Ammonium Chloride
___________________
How many grams are needed? Show your work for this calculation.
The equation of the line is y = -x + 2, where y is the final temperature and x is the mass of salt
3 = -x + 2
x = 2 - 3
x = -1
We need -1 g
of ammonium chloride per 100 g of water.
Question 7: What were some potential sources of error in this investigation? Heat loss to the surroundings due to having a diy calorimeter, the salt not dissolving completely, and possible impurities in the water that could hinder the reaction.
Question 8: Suggest some ways in which the calorimeter or lab protocol could be improved to have lower percentage of errors. A lid could be helpful. After watching to calorimeter that bioengineers use, I noticed they had a heavy duty lid for theirs, and it looked to be made of stainless steel. A digital thermometer could be more precise too.
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25.0° C.
Under these conditions, calculate the reaction free energy AG for the following chemical reaction:
N,(g) + O,(g) = 2NO (g)
Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.
do
Data
HNO3 (1)
-174.1
HNO3 (g)
-133.9
Oxygen
OH (aq)
H20 (1)
-285.8
H20 (g)
-241.8
02 (9)
H202 (1)
-187.8
H202 (9)
-136.3
Phosphorus
Explanation
Check
P4 (s)
2021 McC
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Please help with 2-4
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Constants:
• c = 2.9979 x 108 m/s
h = 6.626 x 10-34 J s per one photon
• R = 8.314 J/(K mol) = 0.08206 L atm/(K mol)
NA = 6.022 x 1023 particles/mol
RH = 1.097 x 107 m1 = 2.178 x 10-18 J
.
.
You have an aqueous Glucose solution that is 12.0% Glucose by mass. What is the molality of
Glucose in the solution?
Molar Mass of Water = 18.015 g/mol
Molar Mass of Glucose 180.16 g/mol
=
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Please help me complete the chart with the information provided. The mass of water in the Calorimeter is 100 mL.
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23.4
KJ
STARTING AMOUNT
23.4 kJ
4184
ADD FACTOR
x( )
4.184
Convert 23.4
97.9
kcal
23.4
2.34 x 10¹
KJ
kJ to calories
1
ANSWER
5590
cal
9.79 x 10¹
J
RESET
5
1000
0.001
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I'm waiting for answer for this is image .
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Please help
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FP
HYDR
A bomb calorimeter, or a constant volume calorimeter, is a device often used to determine the heat of combustion of fuels and the energy content of foods.
Ignition
wires
heat
sample
esc
Water
Stirrer
Insulated Sample Burning Steel
outside dish
sample bomb
chamber
Combustion (bomb) calorimeter.
In an experiment, a 0.6087 g sample of 2-naphthylacetic acid (
C12H10 O2) is burned completely in a bomb calorimeter. The calorimeter is surrounded by
1.201 x 10³ g of water. During the combustion the temperature increases from 23.86 to 26.91 °C. The heat capacity of water is
4.184 J.g¹. °C-¹.
Thermometer
The heat capacity of the calorimeter was determined in a previous experiment to be 936.5 J. ˚C-¹.
Assuming that no energy is lost to the surroundings, calculate the molar heat of combustion of 2-naphthylacetic acid based on these data.
C12H10O2 (s) + (27/2)O₂(g) →→ 5H₂O(l) +12CO2 (g) + Energy
Molar Heat of Combustion =
FOT
2
W
F2
#3
3
F3
E
$
4
kJ/mol
DOO
F4
R
%
5
2
T
MacBook Pro
6
F6
&
7…
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* [ASSESSMENT IN PROGRESS] Su x
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* https://docs.google.
Portfolio Problems...
Jacoby Finley-Unti.
Jacoby Finley
Energy 1:Question 1
What is a difference between a 90°C cup of water and a 40°C cup of
water? Assume both cups of water are the same except for their
temperature.
Select one:
The 40°C water contains different kinds of molecules than the 90°C water.
A particle taken from the 90°C water will be larger than one from the 40°C
water.
The 90°C water has more thermal energy than the 40°C water.
The 40°C water contains fewer molecules than the 90°C water.
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Calculate the heat capacity of the calorimeter for which the sample data in Figure 2 was collected. Assume that mh=mc=50.0g.
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