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EAS1600
Spring 2014
Lab 05 “Heat Transfer”
Objectives
In this lab we will investigate the ways heat can be transferred between bodies. Using the
calorimetric approach, we will verify the energy conservation principle and determine the value of
the latent heat of fusion for water. We also will determine the relative humidity of the air in the lab
using temperature measurements with “dry” and “wet” thermometers.
At the end of this lab, you should be able to:





understand the difference between heat and temperature;
understand the ways heat can be transferred between bodies;
know the energy and heat conservation principles;
be able to perform thermodynamic calculations related to heat transfer and phase change.
understand the concept of relative humidity and be able to determine the relative humidity in
the room using a thermometer.
Theoretical background
The laws of thermodynamics state that heat is transferred from hot objects to cold objects
(unless work is done by external forces on the system). A common unit for measuring heat is calorie
(cal). This is not the same as the calories in a Varsity Chili Dog! In the food industry, what they call
a calorie is actually 1 Kcal = 1000 cal. One calorie of heat will increase the temperature of 1 gram
of water by one degree Celsius (or Kelvin) at constant pressure. Another (standard) unit for heat is
also the unit for energy itself, the Joule. These units can be easily converted back and forth using
the following ratio: 1 cal = 4.184 J.
There are several ways by which heat can be transferred (also illustrated in figure 1):
1)
2)
3)
4)
via conduction;
via radiation (this process is studied in the Radiation Lab);
via convection;
via phase change (vaporization, melting, etc.)
If your coffee is too hot, you might add a little cold milk…or wait. In the first case, the
coffee transfers thermal energy to the cold milk by conduction until they are both at the same
temperature. If you wait, hot coffee radiates in infrared, and the heat is transferred to the
surroundings. At the same time air in the room circulates by the coffee mug’s walls and top, which
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removes heat through convection. Some liquid evaporates (recall steam coming from the freshbrewed cup) and removes heat through the latent heat of vaporization.
Figure 1. Ways to transfer heat: (a) radiation; (b) conduction; (c) convection.
All four ways of heat transfer are utilized by the human body to get rid of excessive heat on a
summer day: we perspire, we stay in front of the fan, we lean against a cool wall, and we radiate in
infrared!
When an object absorbs heat, its temperature increases. The degree to which the temperature
increases depends on the specific heat capacity, Cs, of the object. Specific heat capacity of a solid or
liquid is defined as the amount of heat required to raise the temperature of a unit mass of a substance
by one degree Kelvin. As heat capacity increases, more heat is necessary to reach a specified
temperature increase. For vapors and gases there are two definitions of heat capacity:
Cp = specific heat capacity at constant pressure, and
Cv = specific heat capacity at constant volume.
Specific heat capacity at constant pressure Cp is defined as the amount of heat required to
raise the temperature of 1 g of the substance by 1 K, at a constant pressure. Specific heat capacity Cv
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is defined as the amount of heat required to raise the temperature of 1 g of the substance by 1K while
keeping the volume of the substance constant.
As it was mentioned above, when a warm object and a cold object come into contact, heat is
transferred from the warm object to the cold one. The First law of thermodynamics states that total
energy of the closed system, q, is conserved. This can be written as (note the sign!):
qgained + qlost = 0,
or
qgained = - qlost
(1)
(This equation is actually a simplification of the First Law of thermodynamics, neglecting work and
internal energy variation). The amount of heat absorbed or transferred can be expressed as
qlost, gained = mCs T
(2)
Where m - mass (g); Cs - specific heat capacity (JK-1g-1), for water Cs is 4.184 JK-1g-1 ; T temperature change : T = Tequilibrium - Tinitial . (Note that T is the same whether you are working
in C or K).
In general, the specific latent heat, or heat of transformation, is designated by the letter L and
represents the heat absorbed or released per unit mass during a phase change. The amount of heat
q, absorbed or released by a mass m, is equal to:
q = L m
(3)
Therefore, for a solid mass m at its melting point, the amount of energy needed to change phase from
solid into liquid is:
q = Lf  m,
(4)
where Lf is the latent heat of fusion for the substance. It is called latent heat because it is associated
with the phase state of the object, but not with its sensible temperature. The energy to change the
phase of the object comes from its surroundings.
When water evaporates, it changes state from liquid to gas and the energy is absorbed from
its surroundings. Similar to the melting, the energy required for the evaporation is proportional to the
mass m:
q = Lv  m ,
[5]
where Lv is the specific latent heat of vaporization for the substance. For water at 100 oC, Lv = 540
cal/g or 2260 kJ/kg.
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Figure 2. Phase transitions. The energy should be added to the system for melting, vaporization and
sublimation to occur (red arrows), while the energy is released by the system during deposition,
freezing and condensation (blue arrows).The numbers shown are the approximate number of calories
either absorbed or released during the phase change per 1 g of water.
Water can (and does) evaporate at temperatures less than 100 oC and, by evaporating, absorbs
heat from its surroundings. Therefore, evaporation has a net cooling effect on its surroundings. This
is why we sweat when we are hot, so that our body can cool. The degree of cooling, which is
directly related to the rate of evaporation, can be measured directly with a thermometer. The rate
of evaporation depends on the temperature and humidity of the surrounding air. Humidity is a
measure of water vapor content in the air; relative humidity is defined as the concentration of water
vapor in an air parcel divided by the concentration that would be present if the air parcel were 100 %
saturated with water vapor. If the ambient air has a high relative humidity, then it contains a lot of
moisture and, therefore, evaporation is less favorable (the rate of evaporation is lower).
Online exercises/activities
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/heatra.html - on this page you can find
interactive review of different aspects of thermodynamics, including basic principles of heat transfer,
heat engines and examples of thermodynamic processes in everyday life.
http://www.wisc-online.com/objects/index_tj.asp?objID=SCE304 - animation of basic heat
transfer principles.
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Safety precautions
-
Caution: water used for this experiment is hot and can cause burns.
Clean all spills immediately.
To fill the calorimeter with hot water, put the calorimeter on the lab table, hold it, and slowly
pour in the water from the electric kettle.
Check the calorimeters before the experiment – look if they are in a good condition with
no cracks or holes. If your calorimeter is in poor condition, report to the lab instructor immediately.
Be careful with thermometers – they break easily if dropped on the table.
Please read through all the following steps before starting.
Equipment and procedures in the lab
Part I. Energy Transfer
In the first part of this lab exercise, you will investigate how energy is conserved during the
transfer of heat. Two quantities of water, each of known temperature and mass, will be mixed
together and the final equilibrium temperature will be measured.
The amount of energy gained by the water that became warmer should equal the amount of
energy lost by the water that cooled down.
Procedures:
1. Label two large 500-ml Styrofoam cups as “#1” and “#2”. These cups will act as your
calorimeters. Record the mass of each empty calorimeter in your data sheet (at the end of this
manual). Prepare two small Styrofoam cups, one about 1/2 full with hot water and the other
with crushed ice.
2. Fill calorimeter #1 with approximately 100…110 ml of cold tap water. Use electronic scales
to determine the exact weight of the water.
3. Measure the temperature of the water in the calorimeter. Then, by adding either hot or cold
water/ice, bring the temperature of the water exactly to 20 C. Mix your sample carefully to
ensure uniform temperature distribution.
4. Put the calorimeter on the electronic scales and remove the excess water with the plastic
pipette, bringing the weight of the water in the calorimeter to 100 g. Record the mass and the
temperature of your sample in Table 1.
5. Fill calorimeter #2 with a little more than 200 ml of hot water. Be careful not to spill the
water and not to burn yourself.
6. Measure the temperature of hot water and make sure it’s at or little above 70 C. You will
have to hold the calorimeter to prevent it from tipping over when measuring the temperature
or mixing the water.
7. Put the calorimeter on the electronic scales and remove the excess water with a plastic
pipette, bringing the weight of the water in the calorimeter to 200 g.
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8. Now wait (if necessary) until the water cools down to 70 C. While measuring the
temperature, mix your sample carefully to ensure the uniform temperature distribution within
your calorimeter. Hold the calorimeter to prevent it from tipping over.
9. Record the mass and the temperature of your sample into the Table 1.
10. Immediately pour the content of calorimeter #1 (cold water) into calorimeter #2 (not vice
versa!) and stir with the thermometer until the temperature stops changing. Record the
equilibrium temperature of the mixture.
11. Determine the mass of the combined water sample (use the electronic scale with the larger
weight limit, if necessary). Record the mass in the Table 1.
Part II. Latent Heat of Fusion
In the second part of this lab exercise you will melt a known amount of ice in the warm water and
calculate the heat of fusion of the ice using a heat balance equation.
Procedures:
1. Prepare a calorimeter with about 200 g of hot water, as you did in Part I. Record the mass of the
water and its temperature in Table 2.
2. Place approximately 100 g of ice (note that ice should be at 0 C temperature) into the calorimeter
and carefully stir with the thermometer until all the ice melts. Do not allow water to overflow
from the calorimeter.
3. Record the final temperature and mass of the water in the calorimeter (in Table 2).
4. Calculate the mass of water added by the ice (i.e. obtain the exact mass of the melted ice). Record
it in Table 2.
Part III. Evaporative Cooling: Using the Wet-Bulb Temperature
to Determine Relative Humidity
In this section you will use the thermometer to determine the relative humidity of the air in the lab.
1. Record the room temperature in Table 3. This temperature is called the dry-bulb temperature.
2. Wrap a small piece of paper towel around the bulb of the thermometer and secure it with a rubber
band. Place this thermometer in a Styrofoam cup and mix warm and cool water with the
thermometer until you have a room temperature water.
3. Now remove the thermometer from the water and hold it in front of the fan for several minutes,
until the temperature reaches equilibrium (this shouldn't be more than 5 minutes). Be careful not
to drop or break the thermometer. Read the temperature every 30 seconds until it no longer
changes. Record this equilibrium temperature. This temperature is called the wet-bulb
temperature.
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Section _______________
Measurement results
Table 1. Energy Transfer (14 pts)
Calorimeter #1
Mass of calorimeter (g)
Mass of water (g)
Initial temperature of water (C)
Calorimeter #2
Mass of calorimeter (g)
Mass of water (g)
Initial temperature of water (C)
Mass of combined water sample (g)
Final equilibrium temperature of water (C)
Table 2. Latent Heat (10 pts)
Mass of water (g)
Mass of ice (g)
Initial temperature of water (C)
Initial temperature of ice (C)
Assumed ~ 0C
Final mass of water (including melted ice) (g)
Final temperature of water (C)
Table 3. Evaporative Cooling (2 pts)
Dry bulb temperature (C)
Wet bulb temperature (C)
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Analysis of the measurement results
Questions 1-6 refer to the experiment in Part 1.
Question 1. Why where the calorimeters in your experiments made from Styrofoam? Choose one
from the following and enter the letter that corresponds to your answer into Clicker (2 pts).
A. Because calorimeters should be made from material with high specific heat capacity.
B. Because calorimeters should be made from material with high thermal conductivity .
C. Because calorimeters should be made from material with low specific heat capacity.
D. Because calorimeters should be made from material with low thermal expansion.
E. Because calorimeters should be made from material with high thermal expansion.
Question 2. Using equation (2) and data from your measurements (mass, initial and equilibrium
temperatures), calculate the energy lost (qlost) by the hot water originally in calorimeter #2 after it was
mixed with cold water from calorimeter #1. Enter the number into your Clicker in Joules (3 points)
Hint: to get the correct sign of the energy, make sure you use the proper Tequilibrium and Tinitial.
Question 3. Using equation (2) and data from your measurements (mass, initial and equilibrium
temperatures), calculate the energy gained (qgained) by the cold water originally in calorimeter #1 after
it was mixed with hot water from calorimeter #2. Enter the number into your Clicker in Joules (2
points)
Question 4. Calculate the difference between absolute values of gained and lost energies. Enter the
number into your Clicker in Joules (2 points)
Question 5. Is this difference significant? Choose one from the following and enter the letter that
corresponds to your observations into Clicker (2 pts).
A. The difference is small comparing to the energies lost and gained, and represents the errors in
measuring the mass and temperature.
B. The difference is several Joules and therefore significant.
C. The difference is a result of human error.
D. The difference is only a few Joules, and therefore it’s not significant.
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Question 6. Using only the mass and initial temperature of the samples in your experiment (assume
they were exactly 100 g and 20 C for sample #1; 200 g and 70 C for sample #2) calculate the
(theoretical) temperature water would have after two samples are mixed together. Enter the number
into your Clicker in C. (4 points)
Questions 7-10 refer to the experiment in Part 2.
Question 7. In the experiment for Part 2, where does the energy to melt the ice come from?
Choose one from the following and enter the letter that corresponds to your answer into Clicker (2
pts).
A. The energy comes from hidden (latent) energy of the ice.
B. The energy comes from heating up the ice.
C. The energy comes from the calorimeter.
D. The energy comes from ambient air (the air temperature drops).
E. The energy comes from hot water.
Question 8. List all thermodynamical processes that took place in your experiment (Part 2). Choose
from the following and enter all numbers (in increasing order) that correspond to plausible processes
into Clicker. For example, if choices 1, 3 and 6 are plausible, enter “136” (3 pts).
1.
2.
3.
4.
5.
6.
The ice temperature decreased.
The ice gained some heat and melted.
The hot water lost some heat and cooled down to equilibrium temperature.
The total energy of the system increased.
The water, formed from melting ice, warmed up to equilibrium temperature.
The latent energy of the ice became latent energy of the water.
Based on this list of the processes in your experiment, write an energy balance
equation in the provided space below, which would include latent heat changes and
the transfer of the heat between components of your system. Use this equation
to calculate the latent heat of fusion, Lf, for water. (4 points)
Energy balance equation:
___________________________________________________________________________
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Question 9. Enter the value of Lf that you have found into Clicker in units of J/g. (2 pts)
Question 10. How does this found value compare to the known constant of 333.7 kJ/kg (333.7 J/g) ?
Calculate the difference between your value and reference value in %, and enter this number into your
Clicker (2 pts)
Questions 11 and 12 below refer to the Part 3 of the experiments.
Question 11. Which temperature is higher, dry-bulb or wet-bulb? Choose one from the following
and enter the letter that corresponds to your answer into Clicker (1 pt)
A. Dry-bulb temperature is higher.
B. Wet-bulb temperature is higher.
C. Temperatures are the same.
Question 12. Why is there a difference in temperature between the dry-bulb and the wet-bulb ?
Choose one from the following and enter the letter that corresponds to your answer into Clicker (2
pts)
A. Because the fan is blowing air at the wet-bulb thermometer, and it always feels colder when the
wind is blowing (i.e. because of wind chill factor).
B. Because the water is evaporating from the wet-bulb thermometer, and latent heat of evaporation is
cooling the thermometer down.
C. The wet-bulb thermometer is colder because it’s wet.
D. Because the water is condensing on the wet-bulb thermometer, and latent heat of evaporation is
warming the thermometer.
E. The thermometers were initially at different temperatures.
Problems for individual work
Question 13. You have a 250-gram cup of coffee at 90 C, too hot to drink. Assuming that there is no
other heat exchange with the surroundings, how much would you be able to cool the coffee by adding
50 g of milk at 3 C? Assume that milk has the same heat capacity as water. Answer by entering the
final temperature of coffee in C into your clicker (4 pts)
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Question 14. You want to make ice cubes for your drink. In order to do so, you have filled the ice
cube tray with 500 g of water at 10 C, and put it in the freezer compartment of your refrigerator that
is at -18 C. How much energy (in kilo-Joules) your refrigerator will need in order to make those
ice cubes? Enter your answer to clicker in kJ. The specific heat capacity of ice is Cice ~ 2.1 JK-1g-1,
the latent heat of fusion for ice is Lf ice = 334 J/g, and water heat capacity Cs is 4.184 JK-1g-1 (4 pts)
Question 15. Using the provided reference relative humidity data (Table 4 below), determine the
relative humidity of the air in the lab, based on your temperature readings. Enter the relative humidity
value (in %) into the Clicker (2 pts)
Question 16. When water condenses from gas to the liquid phase, what happens to the energy of its
surroundings? Choose one from the following and enter the letter that corresponds to your answer
into Clicker (2 pts)
A. Increases
B. Decreases
C. Remains the same
Question 17. Convection tends to develop when there is a temperature difference between air
parcels. Will changes in energy due to water condensation from gas to the liquid phase promote or
suppress convection? Answer “A” or “B” using your Clicker (2 points)
A. Suppress
B. Promote
Question 18. Based on your answers to questions 16-17, analyze how changes in energy (and
therefore, changes in temperature) associated with water condensation can affect the formation of
clouds. Choose one from the following and enter the letter that corresponds to your answer into
Clicker (2 pts)
A. Cloud formation is suppressed.
B. Cloud formation is enhanced.
C. Formation of the clouds is not affected by the process of water condensation
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Name ___________________________________________________ Lab section ______________________
Question 19. Where would one expect to find the greatest difference between daily low and high
temperatures? (Think about water and its latent energy). Choose one from the following and enter the
letter that corresponds to your answer into Clicker (2 pts)
A.
B.
C.
D.
In the city
In the desert
In the rainforest
In the ocean
Question 20. Water condensation that appears on the outside of glass or can with a cold beverage is a
manifestation of … (Choose one from the following; enter the letter that corresponds to your answer
into Clicker, 2pts)
A.
B.
C.
D.
E.
F.
High humidity in the air
Low temperature of the drink
High heat conductivity of the glass/can
Low heat conductivity of the air
All of the above
None of the above
Conclusions
1. Discuss the difference between measured and calculated equilibrium temperatures obtained in
experiment Part 1. What are the possible reasons for these discrepancies? (5 pts)
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2. Think of what is happening when very cold wind (with a temperature significantly below freezing) is
blowing above the surface of the lake and ice is forming on the surface of the water. List several
processes that are happening to air, water and ice (5 pts).
A.
___________________________________________________________________
B.
__________________________________________________________________
C.
____________________________________________________________________
D.
_____________________________________________________________________
E.
_____________________________________________________________________
3. Give 2-3 examples of important heat transfer processes that you encounter in everyday life.
(3 pts)
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Table 4.
Relative Humidity Table
TDryBulb, C
Twb,C 15
16
17
18
7%
19
20
21
22
23
24
5%
2%
0%
5
12%
3%
0%
6
19% 14% 10%
6%
3%
0%
7
27% 22% 17% 13%
9%
5%
2%
0%
8
35% 29% 24% 19% 15% 11%
8%
25
26
27
28
29
30
9
44% 37% 31% 26% 22% 17% 14% 10%
7%
5%
2%
0%
10
52% 45% 39% 33% 28% 24% 19% 16% 12%
9%
7%
4%
2%
0%
11
61% 54% 47% 41% 35% 30% 26% 21% 18% 15% 12%
9%
6%
4%
2%
1%
12
70% 62% 55% 48% 42% 37% 32% 27% 23% 20% 16% 14% 11%
8%
6%
4%
13
80% 71% 63% 56% 50% 44% 38% 34% 29% 25% 22% 18% 15% 13% 10%
8%
14
90% 80% 72% 64% 57% 51% 45% 40% 35% 31% 27% 23% 20% 17% 14% 12%
15
100% 90% 81% 73% 65% 58% 52% 47% 41% 37% 32% 28% 25% 22% 19% 16%
16
100% 90% 81% 73% 66% 59% 53% 48% 43% 38% 34% 30% 26% 23% 20%
17
100% 91% 82% 74% 67% 60% 54% 49% 44% 39% 35% 31% 28% 25%
18
100% 91% 82% 75% 68% 61% 55% 50% 45% 41% 37% 33% 29%
19
100% 91% 83% 75% 68% 62% 56% 51% 46% 42% 38% 34%
20
100% 91% 83% 76% 69% 63% 57% 52% 47% 43% 39%
21
100% 91% 84% 76% 70% 64% 58% 53% 48% 44%
22
100% 92% 84% 77% 70% 65% 59% 54% 49%
23
100% 92% 84% 77% 71% 65% 60% 55%
24
100% 92% 85% 78% 72% 66% 61%
25
100% 92% 85% 78% 72% 67%
26
100% 92% 85% 79% 73%
27
100% 92% 86% 79%
28
100% 93% 86%
29
100% 93%
30
100%
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