calorimetry (formal)
TRANSCRIPT
Bettina De Mesa
Chemistry 111 T/Th (1:10-4:45pm)
11/01/2013
Experiment 25: Calorimetry
Conclusion:
The specific heat for unknown metal #21 was .97 J/g˙˚C
The average enthalpy of neutralization for Part 1 (HCl and NaOH) was 57.24 kJ/mol.
The average enthalpy of neutralization for (HNO3 and NaOH) was 57.037 kJ/mol
Abstract:
The purpose of this lab was to determine the average specific heat of unknown metal #21
by utilizing a calorimeter to measure the amount of heat released by the metal. In order to
calculate the specific heat of the metal and the heat of neutralization of the acids and bases, the
mass, temperature change, volume and molarity were measured. Due to the fact that a
calorimeter prevents heat loss to the surroundings, the assumption is that all of the exothermic
energy released from the metal was absorbed by the water in the insulator. In essence, part A of
the experiment consisted of heating the unknown metal to a constant temperature and then placed
in a calorimeter containing water whose initial temperature was recorded. The changes in
temperature indicated the amount of heat being released in the coffee cup, which can then be
used to calculate the specific heat of the metal. The specific heat for my metal was 0.9710
J/g˙˚C. The data to support the validity of my results can be seen through the volume and mass
of the water and the temperature changes in both the metal and its solution. In part B of the
experiment, the enthalpy of neutralization in an acid-base reaction was needed. The
neutralization reaction between acids and bases occur when equal molar amounts of each
substance are present to form a salt and water. By assuming that the density and the specific heat
of the acid and base solutions are equal to water, the heat of reaction can be determined. The
average ΔH in the reaction between HCl and NaOH was 57.24 kJ/mol while the average for
HNO3 and NaOH was 57.037kJ/mol.
Introduction:
The purpose of experiment 25, calorimetry, was to determine the specific heat of
unknown metal #21 and the enthalpy of neutralization for an acid-base reaction. Any changes
that occur physically or chemically are sure to release heat. This experiment highlights the heat
transfer between physical changes between water and the metal as well as two independent
neutralization reactions.
In this experiment there were two parts to be facilitated. The first part consisted of
observing and recording the heat transfer in a physical reaction between an unknown metal and
water. The second part consisted of analyzing and recording the enthalpy of neutralization of an
acid-base reaction. These procedures are delineated in our laboratory manual.
Knowledge about the calorimeter is needed before starting the experiment. A calorimeter
is a laboratory tool which simulates a perfect insulated system. Due to the lack of funds, our
calorimeter was fashioned from a Styrofoam coffee cup. The heat exchange found in the
calorimeter is then used to calculate the amount of heat transfer that takes place between the
physical and chemical reactions. The transfer of heat is usually flowing in one direction,
generally from a section of high temperature to a section of low temperature in order to achieve
equilibrium. Part A of this experiment called for the transfer of heat from a high heat, constant
temperature sample of unknown metal to a sample of colder water. Part B parallels this theory of
heat transfer but with respect to acid-base neutralizations. The equation,
Q=Cs * mass * ∆T (Equation 1)
illustrates the amount of energy released or absorbed by a sample by multiplying the specific
heat of the substance by the mass and the change in temperature.
In part A of this experiment, the objective of determining the specific heat of an object is
given. The methods used of achieving that task is recording the changes in temperature,
analyzing the flow of heat between substances and heating the metal to a constant temperature.
The first step in the procedure of part A is to keep the metal at a constant temperature by
heating the substance in a test tube submerged in boiling water. The metal must be kept dry
during the process of heating due to the fact that any moisture on the solid will cause unequal
heating. The metal is then transferred to a water filled calorimeter whose initial temperature has
been recorded. Once the metal is placed inside, the changes in temperature are recorded. This
increase could be explained by the amount heat being transferred from the metal to its nearby
surroundings. According to the law of conservation of energy, the heat delivered by the metal
(system) will then be absorbed by the water (surroundings). The basic equation is:
-qmetal = qwater (equation 2)
If the q has a negative sign, the heat is released by the system while inversely if the q is positive
the heat is absorbed. In order to solve for the specific heat of the metal. Arranged algebraically to
solve for specific heat of metal:
Csmetal = Cs water∗mass of water∗ΔT of water
mass metal∗ΔT metal (equation 3)
All of these equations assume that there is a perfect system and there is no heat loss by the
calorimeter.
When the change is temperature is recorded, the maximum temperature reached must be
measured by means of a graph. This graph consists of comparing the time and temperature of
the data and observing the peak points. When the peak temperature has been determined
equation 1 can be utilized to find the amount of energy transferred which can then be input into
equation 3 in order to find the specific heat of the metal.
For part B of the experiment, the enthalpy of change in a strong acid/base reaction was
determined. Four combinations of two acids and bases were mixed together with the intent of
observing the amount of heat exchange and calculating the heat of neutralization.
The reactions formed by these strong acids and bases are both characterized by their
reactant. Due to the fact that both substances are strong, this factor implies that they will both
dissociate in the solution. Their exothermic nature will ultimately release which will allow the
reactants to be in a more stable state. These two molecular equations exemplify the reaction.
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) + heat (equation 4)
HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l) + heat (equation 5)
The initial temperature of the NaOH and the acid has to be recorded. Once the two are
mixed together, the changes in temperature are recorded. Unlike Part A, this section consists of a
chemical reaction stemming from two reactants. The heat of neutralization can be exemplified by
the equation
ΔHn = Cs(water) • mass of both reactants • ΔT (Equation 6)
Similar to part A, the max temperature has to be recorded by means of a time versus
temperature graph and inputted into equation 6 in order to determine the amount of heat evolved.
Results:
The data found in Part A can be summarized by observing the Time v. Temperature
graph done. At zero the initial temperature was 24.9˚C and immediately increased to 55˚C once
the metal was added. The heated metal was acquired by heating the substance in a test tube
submerged in a beaker of hot water. Once the metal was at constant temperature it was added the
water.
0 50 100 150 200 250 300 3500
10
20
30
40
50
60
Temp Vs. Time Part A
Time (Seconds)
Tem
pera
ture
(Cel
sius)
Title ResultsMass of Metal (g) 38.4881Temp of Metal- Initial (˚C) 97.5Temp of Metal –Final (˚C) 34.2∆Temperature (˚C) 63.3Mass of Water (g) 20.0Temperature of Water in Calorimeter (˚C) 24.9Maximum temperature of metal and H2O (˚C) 55.0Temperature Change of Water, ΔT (˚C) 30.1Heat Gained by Water (J) 2518.768Temperature Change of Metal, ΔT (˚C) -67.4Specific Heat of Metal (J/g˙˚C) .9710
The data found in Trial I of Part B is illustrated in the graph below. The first initial
temperature of the 1.015M NaOH was 21.1˚C. Once the 21.1˚C HCl was added the exchange of
energy occurred resulting in a sharp spike in temperature to 28.0˚C and ultimately plateaus.
0 50 100 150 200 250 300 3500
5
10
15
20
25
30Temperature vs. Time (Trial 1 HCl/NaOH)
Time (Seconds)
Tem
pera
ture
(Cel
sius)
The data found in Trial II of Part B is illustrated in the graph below. The first initial
temperature of the 1.015M NaOH was 21.1˚C. Once the 20.9˚C HCl was added the exchange of
energy occurred resulting in a sharp spike in temperature to 27.1˚C and ultimately plateaus.
0 50 100 150 200 250 300 3500
5
10
15
20
25
30
Temperature Vs. Time (Trial 2 HCl/NaOH)
Time (Seconds)
Tem
pera
ture
(Cel
sius)
The data found in Trial II of Part B is illustrated in the graph below. The first initial
temperature of the 1.015M NaOH was 20.1˚C. Once the 21.0˚C HNO3 was added the exchange
of energy occurred resulting in a sharp spike in temperature to 27.3˚C and ultimately plateaus.
0 50 100 150 200 250 300 3500
5
10
15
20
25
30
Temperature vs. Time Trial 2 (HNO3/NaOH)
Time (Seconds)
Tem
pera
ture
(Cel
sius)
The data found in Trial II of Part B is illustrated in the graph below. The first initial
temperature of the 1.015M NaOH was 21.1˚C. Once the 20.9˚C HNO3 was added the exchange
of energy occurred resulting in a sharp spike in temperature to 28.1˚C and ultimately plateaus.
0 50 100 150 200 250 300 3500
5
10
15
20
25
30
Temperature vs. Time Trial 2 (HNO3/NaOH)
Time (Seconds)
Tem
pera
ture
(Cel
sius)
Table 1:
In order to find the moles of the KHP, the molar mass of KHP was divided by the
molecular weight of KHC8H4O4 (204.44g/mol). The molarity of NaOH was calculated by
dividing moles of NaOH by the volume dispensed of the NaOH dispensed. The amount of NaOH
dispensed was calculated by taking the difference of the initial and final volumes of the recorded
data. Molarity of the NaOH was calculated by taking the moles of KHP and dividing it by the
amount of NaOH dispensed.
Part A: Specific Heat of Unknown Metal #21
Title ResultsMass of Metal (g) 38.4881Temp of Metal (˚C) 97.5Mass of Water (g) 20.0Temperature of Water in Calorimeter (˚C) 24.9Maximum temperature of metal and H2O (˚C) 55.0Temperature Change of Water, ΔT (˚C) 30.1Heat Gained by Water (J) 2518.768Temperature Change of Metal, ΔT (˚C) -67.4Specific Heat of Metal (J/g˙˚C) .9710
Table 2:
Title HCl/NaOH
Trial 1
HCl/NaOH
Trial 2
HNO3/NaOH
Trial 1
HNO3/NaOH
Trial 2
Volume of Acid (mL) 50 50 50 50
Temperature of Acid
(˚C)
21.1 20.9 21 20.9
Volume NaOH (mL) 50 50 50 50
Temperature of NaOH
(˚C)
21.1 21.1 20.1 21.1
Exact Molar
Concentration of
NaOH (mol/L)
1.015 1.015
Exact Molar
Concentration of HCl
and HNO3 (mol/L)
1.1 1.1
Maxiumum 28.0 27.1 27.3 28.1
Temperature of Graph
(˚C)
Average initial
temperature of
acid/base (˚C)
21.1 21 20.55 21
Temperature Change
(˚C)
6.9 6 6.75 7.1
Volume of Final
Mixture (mL)
100 100 100 100
Mass of Final Mixture
(g)
100 100 100 100
Specific Heat of
Mixture (J/g˙˚C)
4.18
Heat Evolved (J) 2884.2 2926 2821.5 2967.8
Moles of OH- reacted
(mol)
0.05075 0.05075 0.05075 0.05075
Moles of H20 formed
(mol)
0.05075 0.05075 0.05075 0.05075
ΔHn (kJ/mol H20) 56.83 57.655 55.59 58.47
Average ΔHn (kJ/mol
H20)
57.24 57.04
Calculations
Part A: Find Specific Heat of Metal
1. Find the energy absorbed by water.
q = Cs • mass • ΔT (Equation 1)
Example:
4.184 J/g•°C •20.0g •21.1°C = 2518.76 Joules
2. Manipulate equation algebraically to find the specific heat of metal.
Cs = q/(mass • ΔT) (Equation 1)
Example:
2518.76 J/(38.48g • 30.1°C) = 0.9710 J/g•°C
3. Find the Joules released with equation 6
ΔHn = Cs(water) • mass of both reactants • ΔT (Equation 6)
Example:
4.184 J/g•°C • (50.0g + 50.0g) • 6.9°C = -2884.2 Joules
4. Find moles of the water by finding the moles of the reactants with molarity and then use
equation 4 and 5 to aid in stoichiometry.
Molarity = moles/L (Equation 7)
Moles = Molarity • L (Equation 7)
Example:
1.1M • 0.05L = 0.05075 moles
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) + heat (Equation 4)
HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l) + heat (Equation 5)
5. Heat of Nuetralization by dividing the found kJ of the reactants by the moles of water
formed.
Example:
-2.8842kJ/.05075mol = -58.83kJ/mole of water.
Discussion:
The basic concept that shadows every part of this experiment is to determine how much
heat is reacted in physical or chemical changes. In part A, when the heated metal is added to the
cooler water, there was a temperature rise that occurred indicating the release of energy. This
release of energy continues until both substances are in equilibrium. A way to measure if a
system is at equilibrium is to test if the water warmed up (gained energy) or if the metal cooled
down (lost energy). In an ideal situation, the water should have absorbed all of the lost heat from
the unknown metal. However, due to the fact that the calorimeter used was only a coffee cup,
energy could have been lost.
Part A consisted up measuring the mass of the metal and ultimately heating the sample to
a constant temperature. It was vital to ensure that the metal was not wet, in order to prevent
unequal heating. Since 1mL of water equates to 1gram of water, it could be assumed there was
20g of water in the calorimeter. A relevant source of error could be the fact that the instruments
provided were inaccurate and out of date. Due to the lack of precise instruments, the highest
constant temperature achieved was 97°C as opposed to the desired 100°C.
Once the metal was added into the calorimeter, the temperature was to be taken at five
second intervals for one minute and then fifteen second intervals from minute two to minute five.
Due to the compact nature of the calorimeter, the thermometer kept in constant contact with the
heated metal as opposed to the surrounding water. Our experiment called for analyzing the
change in temperature within the water, but instead could have been skewed by the sudden spike
in temperature caused by the prolonged interaction of the thermometer with the metal.
It is essential that the data be graphed in order to determine the maximum temperature at
which the substances reached. Due to the poor construction of the calorimeter, energy could have
been lost during the time period where the metal was added and when the thermometer was
inserted in order to measure the temperature. In order to acquire a more accurate reading, the
thermometer and the metal should have been added simultaneously to the calorimeter.
The specific heat that I had calculated was 0.97 J/g˙˚C, which was derived from
the data gathered up during the experiment. With simple manipulation of equation 1, the specific
heat could be calculated by multiplying the temperature change, mass, and energy released in the
calorimeter.
Part B of the experiment called for the measure of enthalpy in acid/base reactions. This
chemical reaction released energy by the dissociation of the strong acid and strong base mixture.
Due to the fact that all of the reactions had the same net ionic equation (H+ + OH- H2O), they
should have followed in direct accordance with Hess’s law. All equations with the exactly the
same products and reactants should get at least 57.04 kJ/mol of water and at maximum 57.24
kJ/mol of water.
The procedure called for the mixture of the acid and base substances in the calorimeter,
before the addition of a thermometer. With proper stirring, an equal distribution of temperature
could be achieved. However, due to time wasted stirring the solution, some energy may have
escaped and remained unrecorded.
Similar to protocol in part A, the acid/base reactions were graphed with the intent of
determining the maximum temperature. Due to heat lost by means of inappropriate lab tools,
graphing maintains an accurate depiction of the temperature that occurred within the calorimeter.
The average heat of neutralization for acid/base pair #1 was 57.04kJ/mol. The second
acid/base pair possessed 57.24kJ/mol heat of neutralization. By calculating the amount of Joules
released in each reaction and dividing that number by the amount of moles, the enthalpy change
of the two reactions could be identified.
Conclusion: The specific heat for unknown metal #21 was .97 J/g˙˚C. The average enthalpy of
neutralization for HCl and NaOH was 57.24 kJ/mol. The average enthalpy of neutralization for
HNO3 and NaOH was 57.037 kJ/mol
Post Lab Questions
1.) The temperature change will be lower since the water will absorb more energy.
2.) The decrease in mass of water will cause the temperature of the water to increase. Due to the
fact that the change in temperature will increase, transitively the amount of heat of the metal will
be reportedly too high.
3.) The net ionic equation for any strong acid or base reaction would be H+ + OH- H2O. The
only reaction that is occurring is between the OH- and the H+ ions. Since the reaction is the
same, then the enthalpy will be the same as well. With weak acid and weak base reactions,
energy is needed to separate the hydrogen ion from the acid. Due to this dependence, the amount
of energy needed is proportional to the amount of weak acid added.
4.) Q = 2.35g * 6.22 ˚C * 1.34 J/g˙˚C. (Q = 19.6 J)
5.) The resulting solution will have an alkaline pH due to the fact that there will be excess
NaOH.
6.) The energy of neutralization will be unaffected since we are measuring the change in
temperature not absolute temperature.