MAPUA UNIVERSITY

School of Chemical, Biological, Material Engineering and

Sciences (CBMES)

3
Experiment No. : _____
Calorimetry
Title: ___________________________________________________

5
Group No. : ______
Libo-on, Eldon Ryan
Name: ______________________________________________

Pre/Post Lab Quiz

Evaluation
Prelim Data Sheet
Laboratory Attire
Title Page
Report Form
Conclusion

MARILYN A. MIRANDA
Instructor
A4 Libo-on, Eldon Ryan

10.0 g

98.0 OC

5.47 g
24.41 g

18.94 g
36.0 OC

39OC

1.0 OC

79.17 J
-61.0 OC

0.1298 J g-1 OC-1

0.1298 J g-1 OC-1
 50 mL

30 OC
50 mL

36 OC
1.0M

43 OC

33 OC

5.0 OC
100 mL

100 g

-2.090 kJ
0.05 mol

0.05 mol

41.80 kJ mol-1
 5.0 g

0.093 mol

5.47 g

25.41 g

19.94 g
30 OC

18 OC

-12.0 OC

-1000.19 J

-89.4 J

-1089.59 J
-11716.02 J
mol-1 salt
-11716.02 J
mol-1 salt
A. Calculations for Specific Heat of a Metal

1. Temperature change of water, ΔT (OC)

ΔT = Tf - Ti
ΔT = 37 OC -36 OC
ΔT = 1 OC

2. Heat gained by water (J)

Q = mcΔT
Q = (18.94 g)(4.18 J g-1 OC-1)(1 OC)
Q = 79.17 J

3. Temperature Change of Metal, ΔT (OC)

ΔT = Tf - Ti
ΔT = 37 OC -98 OC
ΔT = -61 OC

4. Specific Heat of Metal

Specific HeatMetal = -(specific HeatWater)(masswater)( ΔTwater)
= -(4.18 J g-1 OC-1)(18.94 g)( 1 OC)
(10.0 g)(-61 OC)
= 0.129786 J g-1 OC-1
= 0.1298 J g-1 OC-1

5. Average Specific Heat of Metal = 0.1298 J g-1 OC-1

B. Calculations for Enthalpy (Heat) of Neutralization for an Acid-Base reaction

1. Average Tf of acid and NaOH

(30 OC+36 OC)/2 = 33 OC

2. Temperature Change, ΔT (OC)

ΔT = Tf - Ti
ΔT = 41 OC -36 OC
ΔT = 5 OC

3. Volume of Final Mixture

50 mL + 50 mL = 100 mL

4. Mass of Final Mixture

100 mL ( 1g/ 1.0 mL ) = 100 g

5. Specific Heat of Mixture = 4.184 g-1 OC-1

6. Heat Evolved (J)

Q = mcΔT
Q= -(100 g)(4.18 J g-1 OC-1)(5 OC)
Q= -2090 J = -2.090 kJ

7. Moles of OH-1 reacted, the limiting reactant (mol)

nOH= M(V) = (1.0 M)(50 mL) = 0.05 mol OH
1000 1000

8. Mole of H2O formed (mol)

nH2O = 0.05mol OH (1 mol H2O) = 0.05 mol H2O
1 mol OH
9. ΔHB (kJ/mol H2O)
ΔHB= -(specific heatH20)(combined massesacid+base)(ΔT)
mol of H2O
ΔHB= -(4.184 J g C )(100 g)(5 OC)
-1 O -1

0.05mol
ΔHB= 41800 J mol-1 = 41.80 kJ mol-1

10. Average ΔHB (kJ/mol H2O) = 41.80 kJ mol-1

C. Calculations for Enthalpy (Heat) of Solution for the Dissolution of the Salt
1. Temperature Chang of Solution, ΔT (OC)
ΔT = Tf - Ti
ΔT = 18 OC – 30 OC
ΔT = -12 OC

2. Heat change of water (J)

Q = mcΔT
Q= (19.94 g)(4.18 J g-1 OC-1)(-12 OC)
Q= -1000.19 J

3. Heat change of Salt (J)

Q = mcΔT
Q= (5 g)(1.49 J g-1 OC-1)(-12 OC)
Q= -89.4J

Total Enthalpy change (J) = (-1000.19 J )+( -89.4J)= -1089.59 J

4. ΔHS (J/mol salt)

ΔHS = -1089.59J
0.093 mol salt
ΔHS = -11716.02J mol-1 salt

Average ΔHS (J/mol salt) = -11716.02J mol-1 salt

1. The temperature change in the calorimeter will be lower. Aside from the metal,
the test tube contains some water which will add to the mass of water in the
calorimeter. A greater mass means that higher heat is required to heat it
leading to lower change in temperature. In addition, given a higher specific heat
for the water means that more energy is required for the water to increase its
temperature, resulting in a lower calorimeter temperature change.

2. If Josh used the highest measured temperature for calculating the specific heat
of the metal, his decision will result in a lower specific heat value for the metal.
By using the observed maximum temperature, the increase in change in
temperature of the metal will be greater while the change in water temperature
will be less. By relating this into the formula, c=Q/mΔT , a greater change in
temperature leads to a greater denominator thus the resulting specific heat will
be lower.

3. In strong-acid-base reactions, both the acid and the base are ionized during the
neutralization process thus the same amount of energy was used up. On the
contrary, this is not the same for the weak-acid-strong-base reactions because
the enthalpy of neutralization is dependent on the acid used in the reaction.

4. By using the formula Q=mcΔT, we can solve for the heat loss of the inner
Styrofoam cup.

Given:
m= 2.35 g
c= 1.35 J g-1 OC-1
ΔT= 6.22 OC

Solution:
Q= mcΔT
Q= (2.35 g)( 1.35 J g-1 OC-1)(6.22 OC)
Q= 19.6 J
5. From the balanced equation, the mole ratio of NaOH and HCL is 1:1. Given a 50
mL of 1.0 M NaOH, if Jacob carelessly added only 40.0 mL in the reaction instead
of the 50 mL then the change in enthalpy(delta H) will decrease (if compared to
the actual) because some of NaOH did not react.

6. Even though there is a miscalibration of 2 OC in the thermometer used in the

experiment, the change in temperature will still be the same compared a
correctly calibrated thermometer. Therefore, the reported energy of
neutralization remains unaffected.

7. Since some of the salt was not transferred to the calorimeter but rather
adhered to the weighing paper then the enthalpy of the solution will be reported
as too low because the solution being measured has lower mass compared to
the actual or original mass.

8. It is stated that the calorimeter is not a perfect insulator, therefore the reported
enthalpy of the solution will be too high if the heat change is ignored. This is due
to the incapability of the calorimeter to absorb heat coming from the
surroundings thus more heat would then be put in the system.
 Part A Part B Part C
Time (second) Temperature (Celsius) Temperature (Celsius) Temperature (Celsius)
0 36 36 29
2 36 36 30
4 36 36 30
6 36 36 30
8 36 36 30
10 36 36 30
10.08 36 42 19
10.16 37 42 19
10.25 37 43 19
10.33 38 43 18
10.42 38 43 18
10.5 38 43 18
10.58 38 43 18
10.67 39 43 18
10.75 38 43 18
10.83 38 43 18
10.91 38 42 18
11.5 38 42 18
12 37 42 18
12.5 37 42 18
13 37 42 18
13.5 37 42 18
14 37 42 18
14.5 37 42 18
15 37 42 18
15.5 37 41 18
16 37 41 18
Table 1 Time and Temperature correlation – Table 1 shows the temperature recorded for the three
experiments.
The gathered temperature (°C ) in the experiment is
plotted against the time (min) The initial temperature is at
36°C at the 10th min. After 0.08 min, the temperature
increased by 1°C. It continues to rise reaching the highest
temperature of 39 °C at 10.67 min. Then, the temperature
gradually decreases and reached the final temperature of
37 °C at 16 mins.
The maximum temperature was determined through the
intersection of the line from the cooling portion of the
graph and the line drawn perpendicular to the x-axis at the
10th minute which is about 50-55 °C.
The gathered temperature (°C ) in the experiment is plotted against the time
(min) The initial temperature is at 36°C at the 10 th min. then it suddenly
increased to 42°C. It continues to rise reaching the highest temperature of 48 °C
where it stayed constant for aout one minute. Then, the temperature gradually
decreases and reached the final temperature of 41 °C at 16 mins.
The maximum temperature was determined through the intersection of the line
from the cooling portion of the graph and the line drawn perpendicular to the x-
axis at the 10th minute which is also about 50-55 °C.
The gathered temperature (°C ) in the experiment is plotted against the time
(min) The initial temperature is at 30°C at the 10th. then it suddenly decreased
to 19°C. It continues to fall and reached the final temperature of 18 °C where it
stayed constant for until the end.
The maximum temperature was determined through the intersection of the line
from the cooling portion of the graph and the line drawn perpendicular to the x-
axis at the 10th minute which is also about 23 °C.
 Our group were able to perform the experiment and we were able to determine
the specific heat of a metal, enthalpy of neutralization for a strong acid- strong base
reaction, and the enthalpy of solution for the dissolution of the salt through Coffee cup
Calorimetry.

A coffee cup is considered to be a constant pressure calorimeter. It is commonly

used for solution-based chemistry that involves little or no change in volume. Since the
work applied in the system would be very small, heat which is equal to the enthalpy
change is also approximately equal to the change in internal energy. The experiment is
simply done in a Styrofoam cup which makes it a good adiabatic wall hall that helps keep
the heat absorbed or release inside the cup.

Based on the data collected, for the first part of the experiment, the temperature
change of water when the unknown heated metal was deposited inside the cup is 1°C.
Meanwhile, the change in temperature of the metal is -61°C. The heat absorbed by the
water is 79.17 J, therefore -79.17 J is the heat released by the metal. The specific heat
was then calculated giving us an answer of 0.1298 J g-1 OC-1. For the second part of the
experiment, the average temperature of the acid and base used for the neutralization
was 33°C. The heat evolved in the reaction is -2.090 kJ. Using the combined masses of
the acid and base, and the calculated moles of the water, the enthalpy (heat) of
neutralization of acid and base was determined giving us 41.80 kJ mol-1 H20. Finally, for
the last part of the experiment, the total enthalpy of change was determined by adding
the heat change of water (-985.14 J ) and heat change of salt (-116.4J) which gives us an
answer of -1101.54 J. The enthalpy (Heat) of solution for the dissolution of the salt was
then calculated by dividing the total change of enthalpy (-1101.54 J) and the mole of the
salt (0.093 mol) leading to an answer of -11844.52J mol-1 salt.

In addition, in-order to obtain accurate results in doing the coffee cup calorimetry,
all substances to be used in the experiment should be utilized or put inside the
calorimeter. The set-up should also be done correctly because in doing this experiment,
it has a plenty of sources of error that can affect your result. For instance, in reading
the temperature, measuring the exact amount of NaOH and HCL, as well as making sure
that all the salt should’ve been transferred to the calorimeter. Therefore, the group must
know what to do in order to prevent such errors.