Names: _________________ and
_______________
GEOS 110 Winter 2014
INTRODUCTORY EXPERIMENTS: Sensible and Latent Heat
Changes for Water, Sand and Ice
The lectures, notes and lab and field activities in GEOS 110 are all aimed at an increased
understanding of the nature of the processes within, and the interactions among, the solid
Earth, its ocean and its atmosphere. In order to attain this objective, we will do “handson” lab experiments and field activities, as well as “paper” exercises and web-based labs.
In these introductory labs we will investigate properties of water and of rocks. Our
experiments will be simple, yet quantitative. Later we will see that subtle differences in
material properties determine large-scale Earth processes. What is important is that we
will become familiar with some properties, of water for example, and how it responds to
energy, such as heat or mechanical energy (On a small scale we will be simulating Earth
scale processes). This will give us a foundation on which to better understand the heat
engines in the ocean and atmosphere which produce ocean circulation, waves and
weather systems. Similarly some general knowledge of rocks will aid in understanding
solid Earth mechanisms.
DEFINITIONS
Here are a few definitions for matter and energy:
For our purposes, matter (stuff) exists in 3 main states, or phases:
Solid: matter having a definite shape and a definite volume, whose particles cohere
rigidly to one another, as in a rock.
Liquid: matter in which the particles move about freely relative to one another while the
substance retains a definite volume; liquids flow and take the shape of the container.
Gas: matter in which the particles are not in continuous contact, the substance has no
definite shape or volume.
Other properties of interest:
Mass: m the quantity of matter in an object.
Volume: v the amount of space occupied by matter.
Density: d the ratio of an object’s mass to its volume, d = m/v.
Energy: the capacity of matter to do work.
Work: the movement of a mass through a distance.
Kinetic energy: the energy that matter has due to its motion. K.E. = ½ m(velocity)2
Potential energy: the stored energy has because of its position or its chemical
composition
Energy conservation (using ‘conservation’ in its scientific sense): energy is neither
created nor destroyed
Energy conversion: energy can be converted from one form to another.
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Specific heat capacity: the amount of heat required to raise the temperature of one gram
of a substance by 1oC.
Heat: a form of energy associated with the motion of particles of the substance.
Temperature: a measure of the intensity of heat in a system, i.e. how hot or cold it is,
measured in degrees Celsius, oC or in degrees Kelvin, K.
Latent heat: heat involved in the change of state of a substance, e.g. melting, without
change in temperature.
Experiments
1) Density of (a) water, (b) rocks
Initially this may seem trivial, but subtle differences in density can initiate and maintain
movement of large masses of matter (rock, ocean water, air parcels).
Density is usually expressed in grams/centimeter3 = g/cm3 = g.cm-3
(a) Density of Water. (This varies with temperature and amount of solutes.)
Equipment:
Graduated cylinders
Distilled water, % saline solution
It will be useful to be aware of the density of “fresh” water and seawater later in the
course. In reality all water in nature contains dissolved and suspended substances.
Here we will use (i) distilled water, and (ii) a saline solution of known concentration.
(i) Distilled water
Measure the mass in grams of a graduated cylinder using a balance.
Mass(cyl) = ____g (1)
Partially fill the graduated cylinder with distilled water and record the water level in
milliliters, mL, (1mL = 1cm3) and record the value.
Volume(water) = ____ mL (1)
Determine the mass of cylinder plus water, in grams. Mass(cyl + water) = ____g (1)
Determine the difference in g, the mass of the water.
Mass(water) = ___ g (1)
Calculate the density of the distilled water. The greek letter Rho is used to denote
density.
ρ = _________ g/mL (2)
(ii) Saline solution Repeat the procedure using the salt water.
Partially fill the graduated cylinder with salt water and record the water level in
milliliters, mL, and record the value.
Volume(salt water) = ______ mL (1)
Determine the mass of cylinder plus salt water. Mass(cyl + salt water) = _____ g (1)
Determine the mass of the salt water
Mass(salt water)
2
= _______ g (1)
Calculate the density of the salt water =
ρ
= ______ g/mL (2)
Significance: This difference in density which is dependent on salinity is one of the
factors which aid in the generation of large-scale ocean currents. For example, the
thermohaline current, a world-wide deep water current is largely caused by density
differences. Look up the current in the text.
At a smaller scale, river runoff and ice melt into the ocean leads to a stratification
(layering) determined by difference in density.
(2) Equipment: plastic tube, water, salt water
Obtain a long plastic tube and with the help of a lab partner, carefully add tap water
to one end, making sure that there are no air pockets left in the tube. Hold the 2 ends
adjacent to one another. Fill to about 30 cm from the top of each end. Notice that the
water level is the same at each end. This is because the water is of uniform density
and air pressure acts equally on the system. Move the tube gently, then allow the
water to equilibrate; note that it always returns to the same level.
Next obtain some saline water (% solution). Add enough food colouring to identify
the water. Hold the tube at a nonvertical angle and GENTLY add the saline water to
one end of the tube without causing mixing. Carefully move the tube in order to
compare the water levels. Which level is higher, the fresh or salt water level? Why?
Sketch tube and levels:
(3)
Significance: The difference in water levels at the scale of large water bodies can
induce gravitational currents. These currents transport heat and nutrients; or equally
significantly, density stratification can mean the absence of currents, and an absence
of such transport.
.
(b) Density of Rocks
Equipment:
Basalt samples, granite samples, Graduated cylinders
Basalt is the fine-textured igneous rock which makes up the ocean crust on top of
which sediments accumulate and most sedimentary rocks form. Granite is a coarsegrained igneous rock. It is convenient to use granite to typify the average density of
the continental crust. Thus we can use these 2 rocks to distinguish between some
characteristics of oceanic and continental crusts.
Determine the mass in grams of a sample of basalt and of a sample of granite using a
metric balalance.
Mass(basalt) = ______ g (1)
Mass(granite) = _____ g (1)
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Get a graduated cylinder that can accommodate the rock sample, and fill the cylinder
~2/3 full of water. Record the level of water in milliliters,
Water ______mL.(1)
Carefully place the basalt in the cylinder and note the new water level. ______mL (1)
Determine the difference in volume, which is the volume of the rock.
∆Volume(basalt) = _____mL (1)
Repeat the procedure, using the granite
Water ______mL.(1)
Carefully place the granite in the cylinder and note the new water level. _____mL (1)
∆Volume(granite) = ________ mL (1)
Knowing the mass and volume, determine the density of both the basalt and the
granite.
Basalt density ρ =
Granite density ρ =
g/mL =
g/mL =
g/cm3
(should be about 3.1 g/cm3)
g/cm3
(should be about 2.8 g/cm3) (2)
(2)
Significance: This small difference in density will determine whether a lithospheric
plate will subduct (as in the case of the Juan de Fuca plate moving under the North
American plate) or not, which will determine whether volcanoes or other mountains
will form or not, which in turn influences weather, climate, weathering and
sedimentary processes…
2) Thermal Properties
So far we have investigated materials at room temperature. In other words, we have held
temperature constant. In the real world energy may be added to or removed from
materials. Different materials respond to energy input differently, so we need to
investigate thermal properties, in particular specific heat and latent heat, of some
common Earth substances.
(a) Specific Heat Capacity
When energy in the form of heat is added to a material, the temperature of the material
rises. Note that temperature, in units of degrees Celsius (°C) or Kelvin (K) is a measure
of how hot or cold a substance is, while heat, in units of joules (J) or calories (cal), is a
measure of its thermal energy. 1 calorie = 4.19 joules.
A measure of the efficiency with which a substance can store heat energy is known as
specific heat capacity, or simply the specific heat, c. The greater the material's specific
heat, the more energy that must be added or removed to change its temperature (in its
current state, e.g. as a liquid, without phase change).
As an example, the specific heat of water is given as
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(1)
c = 1.00 cal/g oC which means that 1.00 calorie of heat is necessary to raise the
temperature of one gram of water one degree Celsius, or
(2)
c = 4190 Joules/kg K, which means that 4190 = 4.190x103 Joules of heat are
necessary to raise the temperature of one kilogram of water one Kelvin.
The heat Q necessary to cause a temperature change, ΔT, in a material of mass, m, is
given by the equation
Q = mc ΔT
(Equation #1)
where c is the specific heat capacity of that material (how much energy is needed to raise
a mass of material one degree);
Specific and Latent Heat Values for common materials.
Specific Heat
Latent Heat of Fusion
Material
(cal/g °C)
(J/kg K)
(cal/g)
(J/kg)
Aluminum
0.215
900
94.5
3.96x105
Copper
0.092
385
49.0
2.05x105
Iron
0.107
448
63.7
2.67x105
Lead
0.031
130
5.5
0.23x105
Brass
0.092
385
Unknown Unknown
Magnesium
0.245
1030
88.0
3.7x105
Zinc
0.093
390
27.0
1.1x105
Styrofoam
0.27
1131
Unknown Unknown
Air
0.240
1006
N/A
N/A
Water
1.000
4190
N/A
N/A
Ice
0.500
2095
79.7
3.34x105
ΔT, is the difference between its final and initial temperatures, Tf , Ti , ΔT = Tf - Ti
If Temperature increased the difference is positive, if it decreased it is negative.
The table gives values of specific heat, plus latent heat, to be discussed later.
Look at the table of values. For water at 28oC the value of c = 4190 J kg-1 K-1. For dry air
it is 1006 J kg-1 K-1; thus the value of c for air is about 1/4 that of water. With its density
some 800 times less, the heat capacity of the entire column of atmosphere is
equivalent to that of just 5 m of ocean beneath. For this reason the ocean is an
active heat source and sink for the atmosphere, and generally more exchange occurs
as latent heat of evaporation than as ‘sensible’ conducted heat by water touching air
or vice versa.
Try to have an intuitive quantitative sense of heating. One visualization is the heating of
a tea-kettle, where a 500W stove-top heating unit (500 J sec-1) takes about 300,000 J to
boil 1 kg of water initially at 250C, and 10 minutes to do it. (As a liquid is heated, it
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develops an increasing vapor pressure. At a given temperature and pressure, a vessel of
liquid has an equilibrium vapor pressure which describes the escape of water molecules
from the surface. At the boiling point, the vapor pressure has risen to equal the
atmospheric pressure. Raising the water to the higher energy level requires a great deal of
heat. For this reason, it takes longer to boil a tea kettle dry, than the time initially to bring
it to a boil.
When two bodies having different temperatures come into contact with each other, heat
energy is transferred between the bodies. Take, for example, placing a piece of hot metal
into a container of cool water. From experience we know that over time the metal sample
will become cooler, while the water and its container will become warmer, until an
equilibrium temperature is reached. Put another way, according to the law of energy
conservation the total heat energy lost by the metal is the total heat energy gained by the
water and container.
(b) Latent Heat of Fusion
Water changes state, or phase, from liquid to vapour, liquid to solid, solid to vapour, etc
in the atmosphere and ocean and on land.
If the materials undergo a phase change, for example by melting or evaporating, the
internal energy of the material changes, but its temperature does not change as the phase
transformation occurs, that is at the temperature of phase change. For example, water
changes from solid to liquid or liquid to solid at 0oC, not at any other temperature (at
Earth’s surface, at 1 atmosphere pressure). Because this energy change does not alter the
temperature of the material, we refer to this heat as latent, or "hidden", heat L. In the
case where a solid changes into a liquid (melting) we refer to the latent heat of fusion,
Lf, which is defined as the amount of heat necessary to change one gram of a solid to a
liquid without a temperature change. Note that latent heat of fusion literally means
"hidden heat which causes melting".
The heat required to completely melt a given mass, m, of a substance is given as
Q = m Lf
(Equation 3)
To determine the latent heat of fusion of ice, L, you could simply add some ice cubes to a
water bath at room temperature. Heat is removed from the water bath to both melt the ice
and warm the melted ice water to an equilibrium temperature. Again, from energy
conservation arguments we can say the heat lost from the hot reservoir is equal to the heat
gained by the colder reservoir). We can write this mathematically as:
-Q(hot) = Q(cold) ;
-Q(to cool container) – Q(to cool water) = Q(to melt ice) + Q (to warm melted ice)
(Equation 4)
- mc cc Δ Tc - cw ΔTw = mi Li + mi cw Δ T i
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Note that while in each experiment the initial temperatures of each material is different,
they end up at the same the final temperature once they equilibrate.
Experiments
We will use the above relationships to estimate the specific heat of sand, which
we will take to be broadly representative of the specific heat of “land” on earth.
Measuring the Specific Heat of Sand
Given that the specific heat of water is 1.000 calorie/gm/°C (which we will take to be
representative of sea water and hence “ocean”), we can measure the specific heat of sand
(which we will take as representative of “land”) using the experimental procedure
described below. This will allow us to compare the “ocean” and the “land” in terms of
their response to input of energy, e.g. from solar radiation.
Equipment
styrofoam calorimeter
beaker
thermometer
weighing scale
water
hot plate
sand
Procedure
(1) Record the temperature of the room in degrees Celsius. This should also be the initial
temperature of the sand, Ts(to) where T s is temp of sand and to is time zero assuming that
the sand has been sitting around in the room sufficiently long.
(2) Place the calorimeter on the scale and record its mass in grams.
(3) Add water to the beaker (no more than half full), place the beaker on the hot plate,
and heat the water to a temperature lower than the maximum temperature measurable by
the thermometer.
(4) Pour heated water into the calorimeter (no more than half full) and record its mass.
Subtract the result of (2) from the result of (4) to determine the mass of water in the
calorimeter, mw.
(5) Record the initial temperature of the water, Tw(t0), when the temperature has largely
stabilized (or is changing only very slowly).
(6) Add sand to the water in the calorimeter (less than one quarter of the total volume of
the thermos). Record the mass of the partly filled calorimeter and subtract the mass
measured in (4) to determine the mass of sand, ms, added to the calorimeter.
(7) Cover the top and swish the sand around in the water briefly. Repeatedly record the
temperature of the water/sand mixture and swish it around, until the temperature has
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largely stabilized. We’ll call this the “final” temperature, T(tf ) . It should be the
temperature of both the sand and the water.
The Calculation
The calculation of the specific heat of the sand depends on several assumptions,
some better than others:
 When the approximately room-temperature sand is added to the heated
water, the sand will warm and the water will cool by conduction, and the amount of
heat lost by the water will equal the heat gained by the sand. No significant amount of
heat is lost from the heated water to the calorimeter, to the thermometer, or to the air
(by conduction or evaporation because we do this at a low enough temperature and
quickly).
 The sand and water end up at the same temperature.
 The sand consists of material that all has the same specific heat.
 (And is it reasonable to extrapolate from this experiment to the world
scale?)
The relations between temperature change and heat lost by the water and gained
by the sand can be written separately as:
(1)
Qw  c Hw m w Tw  c Hw m w Tw t f  Tw t 0 
Qs  c Hs m sTs  c Hs m s Ts t f  Ts t 0 
where cHw and cHs are the specific heats of water and sand, respectively. By assumption,
the heat gained by the sand equals the heat lost by the water:
(2) 

Qs  Qw
Substituting Eqs.(1) into Eq.(2) to eliminate the heat gain/loss terms gives:
(3) 
cHsms Ts t f  Ts t 0  cHw mw Tw t f  Tw t 0 
The assumption that the sand and water end up at the same temperature can be
written as:
Tf  Tw t f  Ts t f
(4) 
 
 
Substituting Eq.(4) into Eq.(3) gives:





(5)  cHs ms Tf  Ts t 0   cHw mw Tf  Tw t 0   cHw m w Tw t 0   Tf

We have measured or otherwise know everything in Eq.(5) except cHs, so we can solve
for it:

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(Equation 6)
m Tw t 0   T f 
c Hs  c Hw  w 
m s T f  Ts t 0 
substitute and calculate this.
What is the specific heat of sand (based on your measurements and Eq. 6)? _______ (7)
How does the
 specific heat of “land” (sand) compare to the specific heat of “ocean”
(water)?
(2)
(The actual specific heat of quartz sand is 0.19 cal/gm/°C, though for sandy clay,
which is more like soil or damp sand, it is 0.33 cal/gm/°C.)
3) Some other investigations of water properties and their significance
In this section you will perform simple experiments to learn about some properties of
water and ice. Caution: You will be using hot plates so be careful not to burn your
hands. Also, take care in handling the glassware.
Equipment: 500 mL beaker, 250 mL beaker, thermometer, ice, watch, hot plate, and
gloves (or tongs) to hold hot beakers.
Although the experiments you will be conducting are relatively simple, you must take
great care in your measurements, observations and data in order to get the correct answer!
Experiment 1
To the 500 mL beaker add precisely 400 mL of tap water. Be careful not to under- or
overfill. Measure the room temperature in oC. Be sure to allow the thermometer to come
to equilibrium. Turn the heater to the highest setting and allow it to heat for 2 minutes.
Place the beaker of water on the hot plate and immediately start timing. Fill in the table
below making sure you take the data every minute consistently. If for some reason you
miss a reading by more than 5 seconds then don't guess on what the temperature might
have been but simply skip that data point. It's more important to have fewer data points
that are accurate than more data that are not. The goal here is to measure the temperature
change until the water begins to boil. At that point continue to boil off water until the
table is filled out and the water has boiled for at least 5 minutes. You will need to
accurately determine: 1) the volume of water when the temperature reaches 95oC
(Question 1); 2) the temperature at which the water began to boil (see question 3); and 3)
the volume of water after it has boiled for 5 minutes and subsequently cooled back down
to 95oC (see question 3). Plot the temperature change (y axis) with time (x axis) on a
sketched graph.
Exact H2O mass = __________ g Room Ti ________°C (2)
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Time, minutes Temp oC
1
2
3
4
5
6
7
8
9
10
Time, minutes Temp oC
11
12
13
14
15
16
17
18
19
20
Time, minutes Temp oC (30)
21
22
23
24
25
26
27
28
29
30
1. At 95oC note any changes in the water level from your initial volume and estimate the
volume of water at 95oC in cm3 (Note: 1 mL = 1 cm3)
Water95°C = ______mL (1)
What is the change in volume? (V = volume at 95oC – initial volume) = _______mL (1)
What is the difference in temperature (T) from initial to 95oC? T = ___ - ___ = ___ (1)
The change in water level is due to the thermal expansion of the liquid. The equation for
thermal expansion of any substance is given by
(Equation 7)
B = 1/V (V/ T)
where B = thermal expansion coefficient, V = Volume, V = change in Volume, and T
= change in temperature. Substitute the appropriate numbers from your data and
calculate the thermal expansion coefficient for water . Be sure to include units.
Bmeasured =
(4)
Compare it to the accepted value of 2.1x10-4oC-1. Calculate the percent difference .
% difference = [( Baccepted – Bmeasured ) x 100 ] / Bmeasured
(4)
2. What is your measured boiling point? Let the boiling point be defined as that
temperature at which the temperature ceases to increase. Find the percent difference
between your result and the true boiling point of water at atmospheric pressure, 100oC.
Your Boiling point
TBP = _____°C
Thermometer error: % difference = [( Taccepted – Tmeasured ) x 100 ] / Tmeasured
10
(1)
(4)
3. After the water has boiled for at least 5 minutes (note exact time), use gloves and place
the beaker on the bench and measure the water level as the temperature cools to 95oC.
What is your observed volume? Compare it to your water level at 95oC in question 1.
This difference is the amount of water that was evaporated during the heating process.
Calculate this difference in mL or cm3. (It is better to estimate volumes than carry hot
water to scales!)
∆ V = (V 95 before – V 95 after )
∆ V = ( _____mL - _____ mL) = _________ mL
(3)
At this point. Before finishing the next few calculation problems jump ahead to
question 10 and do the combination experiment of ice plus boiling water.
4. The density of water at 95oC is ρ = 0.96 g/mL. The mass of water evaporated is then
the product of the density and the volume of the water evaporated or:
m = (ρ) x (v)
Calculate the mass of the water evaporated: m = 0.96 g/mL x _____ mL = _____g (3)
5. Calculate the number of calories it took to evaporate this mass of water. Remember
that the equation for this calculation is
Q = mLv where m = mass and Lv = latent heat of vaporization of water = 600 cal/g.
Q = _______g x 600 cal/g = _______ cal
(3)
6. Calculate the number of calories it took to raise the temperature of your total mass of
water from room temperature to the boiling point. Remember the equation for this
calculation is heat is mass times heat capacity times temperature change. Q = mcT
where m = mass of 400 mL of water, c = specific heat of water = 1 cal/goC, and
T = change in temperature.
Q = ~400 g x 1 cal/g x ( ____Tf – _____Ti) =
7. How many calories would it take to evaporate the entire 400 mL of water?
(see question 5 above: Q/g above x 400 g = Qtotal
Qtotal = (______ cal / _______ g) x 400 g = _______cal
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(4)
8. Assuming that all evaporation occurred at the boiling point, how long would it take to
boil the entire 400 mL of water?
(In ~ 5 minutes you evaporated ____ g so: (~ 5 min/ ____ g) x (~400 g) = ____ min (3)
9. From this experiment, what can you now say about the amount of energy it takes to
raise the temperature of a liquid a degree or even several degrees compared to the amount
of energy it takes to convert that same amount of liquid to vapor at its boiling
temperature? The energy released when water vapor condenses to form clouds (which are
composed of water droplets) is released back to the atmosphere in the form of latent heat.
As you can see it is very substantial. This is the source of energy that drives
thunderstorms, makes air masses expand, rise, convect and blows “weather” around. (3)
10. Take a clean dry beaker and weight out some ice at 0°C (e.g 200-300 g) and note the
weight. Weigh out an approximately equal volume of your nearly boiling water from the
last experiment, then pour it into the beaker with the ice, stir until all the ice melts with a
stirring bar (no heat) or stirring rod (not a thermometer!), then measure the equilibrium
temperature of your colled water plus melted ice. If the masses of ice and water were
equal and they were originally freezing and boiling, what would you expect the
equilibrium temperature to be?
_______ °C
(1)
How does your actual measurement compare to this and explain any differences in terms
of thermal properties (heat capacity or latent heat) of the substance H2O? Where did the
extra heat go? Use your data from this experiment to estimate the latent heat of fusion for
ice (cal/g°C).
Hint: You can assume: Qfusion ice + Qwarming ice water = -Qcooling hot water
Recall that Q’s are in calories or Joules while temperature changes are in °C so that the
cooling or heating orf water steps are done by sensible heat: cal/g°C
Calculate your heat of fusion for ice
(6)
Compare this to the tabulated value and calculate your % error.
(3)
Speculate as to how important glacial ice caps are for controlling Earth’s climate or for
buffering climate change? (Then and finish other calculations 4 through 10)
(2)
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