Expert answer:Heat effects and calorimetry pre lab questions
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CHEM 403
Exp 9
Experiment 9:
HEAT EFFECTS AND CALORIMETRY*
Heat is a form of energy, sometimes called thermal energy, that can pass spontaneously
from an object at a high temperature to an object at a lower temperature. If the two objects are in
contact, they will, given sufficient time, both reach the same temperature.
Heat flow can be measured in a device called a calorimeter. A calorimeter is simply a
container with insulating walls, made so that essentially no heat is exchanged between the contents
of the calorimeter and the surroundings. Within the calorimeter chemical reactions may occur or
heat may pass from one part of the contents to another, but, ideally, no heat flows into or out of
the calorimeter from or to the surroundings.
Specific Heat
When heat flows into a substance, the temperature of that substance will increase. The
quantity of heat (q) required to cause a temperature change (ΔT) of any substance is proportional
to the mass (m) of the substance and the temperature change, as shown in Equation 1. The
proportionality constant is called the specific heat, c, of that substance.
q = specific heat * mass of substance * temperature change = c∙m∙ΔT
(1)
The specific heat is the amount of heat required to raise the temperature of one gram of the
substance by 1 °C. Amounts of heat are measured in either joules or calories. To raise the
temperature of 1 g of water by 1 °C, 4.18 joules of heat must be added to the water. The specific
heat of water is therefore 4.18 joules/g∙°C. Since 4.18 joules equals 1 calorie, the specific heat of
water is 1 calorie/g∙°C.
Heat flow into or out of a substance is often determined by the effect that that flow has on
a known amount of water. Because water plays such an important role in these measurements, the
calorie, which was the unit of heat most commonly used until recently, was defined to be equal to
the specific heat of water.
The specific heat of a metal can readily be measured in a calorimeter. A weighed amount
of metal is heated to some known temperature and is then quickly poured into a calorimeter that
contains a measured amount of water at a known temperature. Heat flows from the metal to the
water, and the two equilibrate at some temperature between the initial temperatures of the metal
and the water. Assuming no heat is lost from the calorimeter to the surroundings, and that a
negligible amount of heat is absorbed by the calorimeter walls, the amount of heat that flows from
the metal as it cools is equal to the amount of heat absorbed by the water.
In thermodynamic terms, the heat flow for the metal is equal in magnitude but opposite in
direction, and hence in sign, to that for the water. For the heat flow q,
qwater = –qmetal
*Adapted from Slowinski, E. J., Wolsey, W. C. Chemical Principles in the Laboratory 9th ed.
(2)
CHEM 403
Exp 9
If heat flow is expressed in terms of Equation 1 for both the water and the metal, the result is:
qwater = cwater ∙ mwater ∙ ΔTwater = –[cmetal ∙ mmetal ∙ ΔTmetal]
(3)
In this experiment, you will measure the masses of water and metal and their initial and
final temperatures. Note that ΔTmetal < 0 and ΔTwater > 0, since ΔT = Tfinal – Tinitial. Given the specific
heat of water, you can find the positive specific heat of the metal by Equation 3.
The specific heat of a metal is related in a simple way to its molar mass. Dulong and Petit
discovered many years ago that about 25 joules were required to raise the temperature of one mole
of many metals by 1 °C. This relation, shown in Equation 4, is known as the Law of Dulong and
Petit:
25
MM ≈
(4)
c
where MM is the molar mass of the metal. Once the specific heat of the metal is known, the
approximate molar mass can be calculated by Equation 4. The Law of Dulong and Petit was one
of the few rules available to early chemists in their studies of molar masses.
Heat of Reaction
When a chemical reaction occurs in water, the situation is similar to that which is present
when a hot metal sample is put into water. With such a reaction, there is an exchange of heat
between the reaction mixture and the solvent, water. As with measuring specific heat, the heat flow
for the reaction mixture is equal in magnitude but opposite in sign to that for the water. The heat
flow associated with the reaction mixture is also equal to the enthalpy change, ΔH, for the reaction,
so the equation obtained is:
qreaction = ΔHreaction = –qwater
(5)
By measuring the mass of the water used as solvent, and by observing the temperature
change that the water undergoes, you can find qwater by Equation 1 and ΔH by Equation 5. If the
temperature of the water goes up, heat has been given off by the reaction mixture, so the reaction
is exothermic; qwater is positive and ΔH is negative. If the temperature of the water goes down, the
reaction mixture has absorbed heat from the water and the reaction is endothermic. In this case
qwater is negative and ΔH is positive.
One of the simplest reactions that can be studied in solution occurs when a solid is dissolved
in water. As an example of such a reaction note the solution of NaOH in water:
NaOH(s) → Na+(aq) + -OH(aq)
ΔH = ΔHsolution
(6)
When this reaction occurs, the temperature of the solution becomes much higher than that
of the NaOH and water that were used. If a known amount of NaOH is dissolved in a measured
amount of water in a calorimeter, and the temperature change that occurs is measured, Equation 1
can be used to find qwater for the reaction and Equation 5 used to obtain ΔH. Noting that ΔH is
CHEM 403
Exp 9
directly proportional to the amount of NaOH used, ΔHsolution can be easily calculated for either a
gram or a mole of NaOH. In the second part of this experiment, you will measure ΔHsolution for an
unknown ionic solid.
Chemical reactions often occur when solutions are mixed. A precipitate may form, in a
reaction opposite in direction to that in Equation 6. A very common reaction is that of
neutralization, which occurs when an acidic solution is mixed with one that is basic. In the last
part of this experiment, you will measure the heat effect when a solution of HCl, hydrochloric acid,
is mixed with one containing NaOH, sodium hydroxide, which is basic. The heat effect is quite
large, and is the result of the reaction between H+ ions in the HCl solution with -OH ions in the
NaOH solution:
H+(aq) + -OH(aq) → H2O(l)
ΔH = ΔHneutralization
(7)
Procedure
Your calorimeter for this experiment consists of two nested expanded polystyrene coffee
cups fitted with a styrofoam cover. There are two holes in the cover for a thermometer and an
optional stirring rod with a loop bend on one end. Assemble the experimental setup as shown in
Figure 1.
Figure 1: Diagram of a “Coffee-cup” Calorimeter
CHEM 403
Exp 9
A. Specific Heat
Obtain a sample of metal in a large stoppered test tube from your instructor. Fill a 600-mL
beaker two-thirds full of water and begin heating it to boiling. While the water is heating, weigh
your sample of unknown metal to the nearest 0.1 g. Replace the metal in the test tube and put the
loosely stoppered tube into the hot water in the beaker. The water level in the beaker should be
high enough so that the top of the metal in the test tube is below the water surface. Continue heating
the metal in the water for at least 10 minutes after the water begins to boil to ensure that the metal
attains the temperature of the boiling water. Add water as necessary to maintain the water level.
While the water is boiling, weigh the calorimeter to 0.1 g. Place about 40 mL of water in
the calorimeter and weigh it again. Insert the stirrer and thermometer into the cover and put it on
the calorimeter. The thermometer bulb should be positioned completely under the water.
Measure the temperature of the water in the calorimeter to 0.1 °C. Carefully take the test
tube out of the beaker of boiling water, remove the stopper, and pour the metal into the water in
the calorimeter. Hot hands or mitts may be used here as the test tube will be hot. Be careful that
no water adhering to the outside of the test tube runs into the calorimeter when you are pouring
the metal. Replace the calorimeter cover and swirl the calorimeter to mix the metal and water.
Record to 0.1 °C the maximum temperature reached by the water.
Repeat the experiment, using about 50 mL of water in the calorimeter. Be sure to dry your
metal before reusing it; this can be done by heating the metal briefly in the unstoppered test tube
in boiling water and then pouring the metal onto a paper towel to drain.
B. Heat of Solution
Place 50 mL of distilled water in the calorimeter and weigh it. Measure the temperature of
the water to 0.1 °C. The temperature should be within a degree or two of room temperature. Obtain
a sample of solid from the instructor, weigh out about 5 g of the solid compound to 0.1 g. Break
up any large clumps of the solid sample with a spatula, then add the compound to the calorimeter.
Replace the calorimeter cover, and while swirling the calorimeter, determine to 0.1 °C the
maximum or minimum temperature reached as the solid dissolves. Check to make sure that all the
solid has dissolved at the time you take your temperature reading. A temperature change of
approximately 5 degrees should be obtained in this experiment. If necessary, with instructor
permission, repeat the experiment, increasing the amount of solid used.
C. Heat of Neutralization
Rinse out your calorimeter with distilled water, pouring the rinse into the sink. In a
graduated cylinder, measure 25 mL of 1.00 M HCl; pour that solution into the calorimeter. Rinse
out the cylinder with distilled water, and measure 25 mL of 1.00 M NaOH; pour that solution into
a dry 50-mL beaker. Measure the temperature of the acid and of the base to 0.1 °C, making sure
to rinse and dry your thermometer before immersing it in the solutions. Put the thermometer back
in the calorimeter cover. Pour the NaOH solution into the HCl solution and replace the cover of
the calorimeter. Swirl the reaction mixture, and record the maximum temperature that is reached
by the neutralized solution.
CHEM 403 – Exp 9
Name:______________________________ Section: _________
Data & Calculations
Mass of calorimeter: _____________
A. Specific Heat
Metal identity: _____________
Molar mass of metal: _____________ g/mol
Trial 1
Trial 2
Mass of metal
____________ g
____________ g
Mass of calorimeter + water
____________ g
____________ g
Mass of water
____________ g
____________ g
Initial temperature of water in calorimeter
____________ °C
____________ °C
Initial temperature of metal (assume 100 °C)
____________ °C
____________ °C
Equilibrium temperature of metal and water in
calorimeter
____________ °C
____________ °C
ΔTwater (Tfinal – Tinitial)
____________ °C
____________ °C
ΔTmetal
____________ °C
____________ °C
qwater (using Equation 1)
____________ J
____________ J
cmetal (using Equation 3)
____________ J/g∙°C ____________ J/g∙°C
Experimental MM of metal (using Equation 4)
____________ g/mol ____________ g/mol
How well does your experimental molar mass match the actual molar mass of your metal sample?
Briefly discuss your results.
B. Heat of Solution
Mass of calorimeter + water
____________ g
Mass of water
____________ g
Solid sample identity
____________
Mass of solid sample
____________ g
Initial temperature of water in calorimeter
____________ °C
Final temperature of water in calorimeter
____________ °C
ΔTwater
____________ °C
qwater (using Equation 1)
____________ J
ΔH for this reaction (using Equation 5)
____________ J
The quantity you have just calculated is approximately equal to the heat of solution of your sample.
Calculate the heat of solution per gram of your solid sample.
ΔHsolution = ____________ J/g
The process of dissolution for my sample is (circle one):
endothermic
exothermic
C. Heat of Neutralization
Initial temperature of HCl solution
____________ °C
Initial temperature of NaOH solution
____________ °C
Final temperature of neutralized mixture
____________ °C
ΔT (Tfinal – average Tinitial)
____________ °C
qwater (assume 50.0 g of water)
____________ J
ΔH for the neutralization reaction
____________ J
Calculate the heat of neutralization per one mole of this reaction. Hint: how many moles of HCl
or NaOH are reacting in your sample?
ΔHneutralization = ____________ J/mol
CHEM 403 – Exp 9
Name:______________________________ Section: _________
Prelab Assignment
1. In this experiment, the assumption is made that no heat is lost or gained from the surroundings
outside the calorimeter. However, the insulated system is not perfect. Consider the first part of
the experiment where you will determine the specific heat of the metal, and how poor insulation
could affect your calculations involving Equations 1 and 3.
a. If the calorimeter is not effectively insulated, and heat can escape to the surroundings, how
would this affect your recorded temperature change (Tfinal – Tinitial) for the water in the
calorimeter?
b. How would this error affect your calculation of qwater? Would your experimental value be too
high or too low?
c. How would this error affect your calculation of cmetal? Would your experimental value be too
high or too low?
2. When 2.0 g of LiBr were dissolved in 49.0 g water in a calorimeter at 24.0 °C, the final
temperature of the solution was 34.5 °C.
a. Is this reaction endothermic or exothermic? Explain your answer.
b. Given that LiBr exists in solution as lithium cations and bromide anions, write the equation
for the reaction that occurs when LiBr is dissolved in water.
…
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