Experiment 6: Heat Effects
Laboratory Report
Abraham S.P. Francisco, Joselito A. Gardoce,
Marvin Lorenzo J. Gonzales, Maria Therese V. Ibarra, and Stephanie Lazo
Department of Math and Physics
College of Science, University of Santo Tomas España, Manila Philippines 1015
Abstract
The experiment is about the different effects of heat among objects specifically on the specific heat, heat fusion, and thermal expansion of solids. The students computed the coefficient linear thermal of the rod by using the formula , the formula Q=mc∆T for the specific heat capacity, and Q= mL for the heat fusion of water where L is the latent heat. All of the needed data were supplied by the experiment and were all imputed to these equations to solve for the unknown.
I. Introduction
Heat is energy produced or transferred from one body, region, set of components, or thermodynamic system to another in any way other than as work. The specific heat (also called specific heat capacity) is the amount of heat required to change a unit mass (or unit quantity, such as mole) of a substance by one degree in temperature. Therefore, unlike the extensive variable heat capacity, which depends on the quantity of material, specific heat is an
intensive variable and has units of energy per mass per degree (or energy per number of moles per degree). The heat capacity of a substance can differ depending on what extensive variables are held constant, with the quantity being held constant usually being denoted with a subscript.
The method of mixture based on the fact that when a hot substance is mixed with a cold substance, the hot body loses heat and the cold body absorbs heat until thermal equilibrium is attained. At equilibrium, final temperature of mixture is measured. The specific heat of the substance is calculated with the help of the law of heat exchange. During a phase transition of a given medium certain properties of the medium change, often discontinuously, as a result of some external condition, such as temperature, pressure, and others. For example, a liquid may become gas upon heating to the boiling
point, resulting in an abrupt change in volume.
The measurement of the external conditions at which the transformation occurs is termed the phase transition point. The term is most commonly used to describe transitions between solid, liquid and gaseous states of matter, in rare cases including plasma.
Thermal expansion is the tendency of matter to change in volume in response to a change in temperature. When a substance is heated, its particles begin moving more and thus usually maintain a greater average separation. Materials which contract with increasing temperature are rare; this effect is limited in size, and only occur within limited temperature ranges (see examples below). The degree of expansion divided by the change in temperature is called the material's coefficient of thermal expansion and generally varies with temperature.
The objectives of the experiment are as follows: to determine the specific heat of a solid by method of mixtures, to determine the latent heat of fusion and latent heat of vaporization of water and to determine the coefficient of linear thermal expansion of a solid.
II. Theory
The specific heat is the amount of heat per unit mass required to raise the temperature by one (1) degree Celsius. The relationship between heat and temperature change is usually expressed in the form shown in the equation below where c is the specific heat. The relationship does not apply if a phase change is encountered, because the heat added or removed during a phase change does not change the temperature.
Q is the heat added, c is the specific heat, m is the mass, and ∆T is the change in temperature. The next formula is the one that is used in the activity in finding the specific heat of a metal.
[ ]
Where Co is the specific heat of the
object, MC is the mass of empty calorimeter,
Cc is the specific heat of the calorimeter, MW
is the mass of the water, ∆TC is the
difference between the final and initial temperature of the water and calorimeter, MO is the mass of metal cylinder, and ∆TO is
the difference between the initial temperature of metal cylinder and the final temperature of the system.
In finding the activity which includes the thermal expansion of solids, the formula below could be used.
Where ∆L is the difference between the final and initial reading of the micrometer disc, LO is the initial length of
the rod, and ∆T is the change in temperature. The most easily observed examples of thermal expansion are size changes of materials as they are heated or cooled. Almost all materials (solids, liquids, and gases) expand when they are heated, and contract when they are cooled. Increased temperature increases the frequency and magnitude of the molecular motion of the material and produces more energetic collisions. Increasing the energy of the collisions forces the molecules further apart and causes the material to expand.
In the heat fusion of water activity, the following formula is being used.
[ ] [ ] Where Lf is the heat fusion of ice,
MW is the mass of water, MC is the mass of
empty calorimeter, Cc is the specific heat of
the calorimeter, TO and T is the initial and
final temperature of the system, respectively. Mi is the mass of the ice.
Melting and freezing behaviour are some of the characteristic properties that give one’s substance its unique identification. Pure solid water or ice at 0°C changes to liquid water at also 0°C when energy is added.
To compute for the percent error:
III. Methodology
Activity 1: Specific Heat of Metal
The metal object was weighed. A 30-cm long thread was attached to the metal object which is immediately put into the metal jacket. The metal jacket was placed inside a beaker with water. The beaker was subjected to heat of 80°C. The inner vessel of the calorimeter was weighed. After weighing, 2/3 of it was filled with water and weighed again. The inner vessel was placed in its insulating jacket and the temperature was measured. When the object in the beaker reached 80°C it is quickly transferred into the calorimeter. The water was stirred with the thermometer inside it. The equilibrium temperature was recorded. The specific heat of the object and percent error were computed.
Activity 2: Heat Fusion of Water
The empty inner vessel of the calorimeter was weighed. It is then filled with water and weighed again. The inner vessel was placed into its insulating jacket. The initial temperature of the water was recorded. Dried pieces of ice were added into the calorimeter. The equilibrium temperature was recorded after the ice melted. The inner vessel was again weighed together with the water and melted ice inside it. The heat of fusion of ice was computed by Conservation of Heat Energy. The percent error was also computed.
Activity 3: Thermal Expansion of Solids
The initial length of the rod was measured. It is then placed inside the steam jacket. The steam jacket was mounted in the metal flame. The first outlet of the jacket was connected to the boiler by rubber tubing. The initial temperature of the rod was recorded. The metal frame was then connected to the galvanometer. The micrometer screw was moved so that it will touch the end of the rod. The initial reading of the micrometer disc was recorded. The disc was unwound so that the rod can expand freely. Using a steam coming from the boiler, the rod was heated for twenty
times. The final temperature of the rod was recorded. The disc was then moved until it is in contact again with the rod. The final reading of the disc was recorded. The coefficient of linear thermal expansion of the rod and the percent error were computed.
IV. Results and Discussion Activity 1. Specific Heat of Metal
In activity 1, the specific heat of a sample metal was calculated (Table 1). The ability of a substance to absorb or release energy is known as specific heat. The specific heat of a substance is defined as the amount of heat energy required to change the temperature of one gram of a substance one degree Celsius. If a substance absorbs energy easily, it is said to have a low specific heat capacity. Most metals have a low specific heat capacity Which means they will absorb energy easily. In the experiment, Heat energy flows from the sample to the water and its container
(the calorimeter), causing the temperature of the water and container to rise and the sample's temperature to fall. It is assumed that the heat lost by the sample is absorbed by both the water and the calorimeter thus we can calculate the specific heat of a sample metal. We calculated the specific
heat of a sample metal to be 0.097 cal/g . 0C and yielded 18% error. Possible sources of error include the error in reading the thermometer and temperature changes due to heat transferring to the environment.
Table 1. Data on Specific Heat of a metal
Activity 2. Heat Fusion of Water
In Activity 2, The latent heat of Fusion was determined (table 2). During the process of melting, the solid and liquid phases of a pure substance are in equilibrium with each other. The amount of heat required to convert one unit amount of substance from the solid phase to the liquid phase — leaving the temperature of the system unaltered — is known as the latent heat of fusion (Wakeham, 2011). The
calculated Latent heat was 117.6 cal/g yielding a high % error of 47 %. Possible causes of error includes Improper Stirring that causes the final temperature to be too warm and gives an experimental value of the Latent Heat of Fusion that is too low. Another one is not drying the ice, If the ice is not dried there will be water at 0ᵒC on the ice. The added water will contribute to the final mass of liquid but it will not gain the amount of heat that an equivalent amount of ice would gain. The initial temperature of the water in the calorimeter will not have to drop as far. Hence the final temperature will be too high. The result will be an experimental value of the Latent Heat of Fusion that is too low.
Table 2. Data on Heat fusion of water Mass of Calorimeter 43.82 g
Mass of Calorimeter w/ Water
158.10g Mass of water 114.28g Initial temp of water
and calorimeter
25 ᵒC Mass of ice, water and
calorimeter
175.42 g Mass of ice 17.32 g Final temp of the
system
7.5 ᵒC Calculated latent heat
of fusion
117.6 cal/g Accepted value of
latent heat of fusion
80 cal/g % error 47% Mass of empty Calorimeter (Mc) 44.04 g Mass of Calorimeter with Water 142.61g Mass of Metal Cylinder (Mo) 49.43 g Initial Temperature of
water & calorimeter 25ᵒ C Initial Temperature of
the metal cylinder 96 ᵒC Final temperature of
the system 28 ᵒC
Mass of Water 95.57 g Calculated specific
heat of sample 0.097 cal/g . 0
C Accepted value of
specific heat 0.118 cal/g .0
C
Activity 3. Thermal Expansion of Solids Activity 3 Thermal Expansion of Solids
Initial length of the rod 54.80 cm Initial temperature of the
rod
25⁰C Initial reading of micrometer
disc
0.245 cm Final temperature of the rod 95⁰C Final reading of micrometer
disc 0.271 cm Experimental value of coefficient of thermal expansion 6.82 x10-6/⁰C
Accepted value of thermal coefficient of thermal expansion
2.5x10-5/⁰C
% error 73%
In activity 3, the length L0 of an
object changes by an amount of ∆L when its temperature changes an amount of ∆T. Where α is the coefficient of linear expansion.
We can obtain the coefficient of thermal expansion through dividing both sides by L0∆T
α
To get the elongation of the rod ∆L, the final reading of the micrometer disc was subtracted to the initial reading of micrometer disc.
α 6.82 x10-6
/⁰C
Percent error can be calculated by dividing the difference of the accepted value of thermal coefficient and experimental value of coefficient of thermal expansion to the accepted value of thermal coefficient of thermal expansion and multiply it by 100.
The increase in any one dimension of a solid is called linear expansion, linear in a sense that the expansion occurs along a line. When the temperature of the rod increases to T0 +∆T, the length becomes Lo + ∆L where
∆T and ∆L are the changes in temperature and length respectively. Conversely, when temperature decreases to To – ∆T, the length
decreases to Lo - ∆L. The experiment
showed that the change in length is directly proportional to the change in temperature. ∆L is proportional to both L0 and ∆T by
using a proportionality constant α, which is the coefficient linear expansion.
V. Conclusion
In the experiment we were able to determine the specific heat of a metal by method of mixtures and the computed specific heat of the metal is 0.097 cal/g0C. The latent heat and of vaporization of water was also computed based from the results and it is 117.6 cal/g. Lastly, we were also
able to determine the coefficient of linear thermal expansion of a solid and the computed value is 6.82 x10-6/⁰C. We are tasked to complete all the values needed for the computations and it is shown in Tables 1, 2 and 3.
VI. Application
1. Is it possible to add heat to a body without changing its temperature?
Answer: No, it is not possible to add heat without changing its temperature because heat is defined as energy in transit from a high temperature object to a lower temperature object. Therefore to transfer heat, there must be movement of heat energy from a high temperature object to a low one thus affecting the temperature of the object itself. But it should be taken noted that the heat energy of the system is conserved all throughout.
2. Explain why steam burns are more painful than boiling water burns.
Answer: A steam burn is worse than a hot water burn because the steam is in a different phase. When the steam comes in contact with your body, the steam must turn into water before it can cool down to body temperature. This releases more energy into the skin due to the phase change, thus causing a worse burn.
3. Early in the morning when the sand in the beach is already hot, the water is still cold. But at night, the sand is cold while the water is still warm. Why?
Answer: Sand has the property of getting an environmental temperature very quickly that’s why it is warm in the morning and cold at night.
4. Explain why alcohol rub is effective in reducing fever.
Answer: Rubbing alcohol cools the skin by convection, as the alcohol evaporates it carries the heat away from the body with it.
5. Cite instances where the thermal expansion is beneficial to man. Cite also examples where thermal expansion is a nuisance.
Answer: An example of its advantage to man is bimetallic strip which is used in mechanical switch in thermostat. It is a nuisance it its use in roadway construction, if roadway were poured as one continuous slab (the cheapest way possible), when it expanded in the heat of the day, or contracted in the cool of night, it can fracture or crack and separate where the road meets the wall, or at some point on the road in between, causing surface defects and potholes.
6. Why is water not used in liquid in glass thermometer?
Answer: Water will not rise or fall at Temperature changes as mercury. water has a none linear thermal expansion (Its thermal expansion coefficient at 20C is not the same as at 90C). Also, at atmospheric pressure, water is only liquidus over a narrow temperature range of 100C which limits its usefulness. Further it has massive problems at phase transitions- for instance when it turns to a gas it consumes a lot of energy (latent heat). A thermometer should have a nice linear response to a rise in temperature.
7. The density of Aluminum is 2700kg/m3 at 200 C. What is its density at 1000 C?
Answer: Linear thermal expansion coefficient of Aluminum : 24x10-6 /K Forumla to be used: ∆L/L = α∆T, α is linear
thermal expansion coefficient
Take a cube 1 meter on a side, which at 20ºC weighs 2700 kg
What does the length change to at 100º ? ∆L/L = α∆T
∆L = Lα∆T = (1)(24x10-6
)(80) = 0.00192 meter
so the new cube is 1.00192 m on a side and the volume is that cubed or 1.00577 m³
Density is 2700 kg / 1.00577 m³ = 2685 kg/m³
The density of aluminum at 100 ᵒC is 2685 kg/m³ or 2.69 g/cm3
8. How much heat is needed to change 1g of ice at 0oC to steam at 100oC?
Answer: First compute for the amount of heat needed to turn ice into water by multiplying its mass by the latent heat needed to melt ice into liquid (80 cal/g 0C).
Liquid water must be allowed to boil to reach eva[oration. The heat needed to raise the temperature of liquid water from 0oC to its boiling point is given by:
After reaching its boiling point, water will begin to evaporate. The amount of heat needed to vaporize 1g of water is given by:
The sum of these is 720 cal.
9. An aluminum calorimeter has a mass of 150g and contains 250g of water at 30 . Find the resulting temperature when 60g of copper at 100 is placed inside the
calorimeter.
Answer:
Mass of Calorimeter 0.15 kg Mass of water 0.25 kg Mass of Calorimeter and
Water
0.40 kg Mass of Copper 0.60 kg Mass of Calorimeter,
water, and copper
1 kg Initial Temperature of
water in Calorimeter
30
We can combine first the calorimeter with water: ( )
We can then use m3, c3 and T3 to combine copper ( ) ( ) References:
[1] Austin Community College. (n.d.). Latent
Heat of Fusion. Retrieved February 25, 2012,
from
http://www.austincc.edu/mmcgraw/Labs_1401 /19c%20Latent%20Heat%20of%20Fusion.pdf [2] University of Brigham Young University - Idaho. (2005, October 14). Specidic Heats. Retrieved from Chemistry 105 : Experiment 2: http://www2.byui.edu/Chemistry/lab_manuals/ chem_105/chem_105_exp_2_specific_heat.pdf [3] Wakeham, W. (2011, February 11). Latent
heat of Fusion. Retrieved from Thermopedia:
A-Z Guide to Thermodynamics, Heat & Mass transfer and Fluids engineering:
http://www.thermopedia.com/content/915/?ti d=110&sn=16 [4] http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/spht.html [5]http://www.bookrags.com/research/ther mal-expansion-wop/ [6]http://www2.vernier.com/sample_labs/ CWV-04-COMP-heat_of_fusion.pdf
