Experiment 6 (Formal Report)

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

intensive variable and has units of energy

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.

the quantity being held constant usually

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

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

I. Introduction Heat

is

the cold body absorbs heat until thermal energy

produced

or

equilibrium is attained. At equilibrium, final

transferred from one body, region, set of

temperature of mixture is measured. The

components, or thermodynamic system to

specific heat of the substance is calculated

another in any way other than as work. The

with the help of the law of heat exchange.

specific heat (also called specific heat

During a phase transition of a given medium

capacity) is the amount of heat required to

certain properties of the medium change,

change a unit mass (or unit quantity, such as

often discontinuously, as a result of some

mole) of a substance by one degree in

external condition, such as temperature,

temperature. Therefore, unlike the extensive

pressure, and others. For example, a liquid

variable heat capacity, which depends on the

may become gas upon heating to the boiling

quantity of material, specific heat is an

point, resulting in an abrupt change in

II. Theory

volume.

The specific heat is the amount of

The measurement of the external

heat per unit mass required to raise the

conditions at which the transformation

temperature by one (1) degree Celsius. The

occurs is termed the phase transition point.

relationship between heat and temperature

The term is most commonly used to describe

change is usually expressed in the form

transitions between solid, liquid and gaseous

shown in the equation below where c is the

states of matter, in rare cases including

specific heat. The relationship does not

plasma.

apply if a phase change is encountered, because the heat added or removed during a

Thermal expansion is the tendency of matter to change in volume in response to

phase

change

does

not

change

the

temperature.

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

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.

temperature ranges (see examples below).

[

]

The degree of expansion divided by the the

Where Co is the specific heat of the

material's coefficient of thermal expansion

object, MC is the mass of empty calorimeter,

and generally varies with temperature.

Cc is the specific heat of the calorimeter, MW

change

in

temperature

is

called

is the mass of the water, ∆TC is the The objectives of the experiment are

difference between the final and initial

as follows: to determine the specific heat of

temperature of the water and calorimeter,

a solid by method of mixtures, to determine

MO is the mass of metal cylinder, and ∆TO is

the latent heat of fusion and latent heat of

the

vaporization of water and to determine the

temperature of metal cylinder and the final

coefficient of linear thermal expansion of a

temperature of the system.

solid.

difference

between

the

initial

In finding the activity which includes the

Melting and freezing behaviour are

thermal expansion of solids, the formula

some of the characteristic properties that

below could be used.

give

one’s

substance

its

unique

identification. Pure solid water or ice at 0°C changes to liquid water at also 0°C when Where ∆L is the difference between the

final

and

initial

reading

of

the

energy is added. To compute for the percent error:

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

III. Methodology

materials as they are heated or cooled. Almost all materials (solids, liquids, and

Activity 1: Specific Heat of Metal

gases) expand when they are heated, and

The metal object was weighed. A 30-

contract when they are cooled. Increased

cm long thread was attached to the metal

temperature increases the frequency and

object which is immediately put into the

magnitude of the molecular motion of the

metal jacket. The metal jacket was placed

material and produces more energetic

inside a beaker with water. The beaker was

collisions. Increasing the energy of the

subjected to heat of 80°C. The inner vessel

collisions forces the molecules further apart

of the calorimeter was weighed. After

and causes the material to expand.

weighing, 2/3 of it was filled with water and

In the heat fusion of water activity, the

weighed again. The inner vessel was placed

following formula is being used.

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

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.

with

the

thermometer

inside

it.

The

equilibrium temperature was recorded. The specific heat of the object and percent error were computed.

times. The final temperature of the rod was

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. 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.

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.

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

Activity 3: Thermal Expansion of Solids

the temperature of one gram of a substance one degree Celsius. If a substance absorbs

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

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

.0

C

calculated Latent heat was 117.6 cal/g

and yielded 18% error. Possible sources of

yielding a high % error of 47 %. Possible

error include the error in reading the

causes of error includes Improper Stirring

thermometer and temperature changes due

that causes the final temperature to be too

to heat transferring to the environment.

warm and gives an experimental value of the Latent Heat of Fusion that is too low.

Table 1. Data on Specific Heat of a metal Mass of empty Calorimeter (Mc) Mass of Calorimeter with Water Mass of Metal Cylinder (Mo) Initial Temperature of water & calorimeter Initial Temperature of the metal cylinder Final temperature of the system Mass of Water Calculated specific heat of sample Accepted value of specific heat % error

Another one is not drying the ice, If the ice is not dried there will be water at 0ᵒC on the

44.04 g

ice. The added water will contribute to the

142.61g

final mass of liquid but it will not gain the amount of heat that an equivalent amount of

49.43 g

ice would gain. The initial temperature of 25ᵒ C

the water in the calorimeter will not have to

96 ᵒC

drop as far. Hence the final temperature will be too high. The result will be an

28 ᵒC

experimental value of the Latent Heat of

95.57 g

Fusion that is too low.

0.097 cal/g . 0C

Table 2. Data on Heat fusion of water

0.118 cal/g . 0C 18%

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

Mass of Calorimeter Mass of Calorimeter w/ Water Mass of water Initial temp of water and calorimeter Mass of ice, water and calorimeter Mass of ice Final temp of the system Calculated latent heat of fusion Accepted value of latent heat of fusion % error

43.82 g 158.10g 114.28g 25 ᵒC 175.42 g 17.32 g 7.5 ᵒC 117.6 cal/g 80 cal/g 47%

Percent error can be calculated by dividing

Activity 3. Thermal Expansion of Solids

the difference of the accepted value of

Activity 3 Thermal Expansion of Solids

thermal coefficient and experimental value Initial length of the rod Initial temperature of the rod Initial reading of micrometer disc Final temperature of the rod Final reading of micrometer disc Experimental value of coefficient of thermal expansion Accepted value of thermal coefficient of thermal expansion % error

54.80 cm 25⁰C

of coefficient of thermal expansion to the accepted value of thermal coefficient of

0.245 cm

thermal expansion and multiply it by 100.

95⁰C 0.271 cm 6.82 x10-6/⁰C

The increase in any one dimension of a solid is called linear expansion, linear in a

2.5x10-5/⁰C

sense that the expansion occurs along a line. When the temperature of the rod increases to T0 +∆T, the length becomes Lo + ∆L where

73%

∆T and ∆L are the changes in temperature In activity 3, the length L0 of an

and length respectively. Conversely, when

object changes by an amount of ∆L when its

temperature decreases to To – ∆T, the length

temperature changes an amount of ∆T.

decreases to Lo - ∆L. The experiment

Where α is the coefficient of linear

showed that the change in length is directly proportional to the change in temperature.

expansion.

∆L is proportional to both L0 and ∆T by

We can obtain the coefficient of thermal

using a proportionality constant α, which is

expansion through dividing both sides by

the coefficient linear expansion.

L0∆T V. Conclusion α In the experiment we were able to To get the elongation of the rod ∆L, the final reading

of

subtracted

the to

micrometer the

initial

micrometer disc. α

disc reading

was of

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 6.82 x10-6/⁰C

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.

3

7. The density of Aluminum is 2700kg/m 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?

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 0.40 kg Water Mass of Copper 0.60 kg Mass of Calorimeter, 1 kg water, and copper Initial Temperature of 30 water in Calorimeter We can combine first the calorimeter with water:

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:

)

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: AZ Guide to Thermodynamics, Heat & Mass transfer and Fluids engineering: http://www.thermopedia.com/content/915/?ti d=110&sn=16

[4] http://hyperphysics.phyastr.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