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.