Experiment 6: Heat Effects
Margarita Andrea S. de Guzman, Celine Mae H. Duran,
Celina Angeline P. Garcia, Anna Patricia V. Gerong
Department of Biological Sciences
College of Science, University of Santo Tomas
España, Manila
Abstract
Experiment 6 talks about heat effects. The following activities are focused on the concept of The Laws of Conservation of Energy. A specific heat of 1.2949 J/gC° is obtained from a metal sample in activity 1. A 303. 29 J/g latent heat of fusion of water was observed in activity 2. Lastly, a 4.46 ×〖10〗^(-6)/ C° of coefficient of thermal expansion of a rod was obtained in activity 3.
Introduction
Natural and physical objects are governed by the Laws of Conservation of Energy. This law specifically states that: “energy can never be created nor destroyed; but it can only be transformed into another form”.
Specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius. The relationship between heat and temperature change is usually expressed in the form shown 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.
The energy required to change a gram of a substance from the solid to the liquid state without changing its temperature is commonly called its "heat of fusion". This energy breaks down the solid bonds, but leaves a significant amount of energy associated with the intermolecular forces of the liquid state.
At the end of the experiment, the objectives to be accomplished are: (1) To determine the specific heat of a solid by method of mixtures. (2) To determine the latent heat of fusion and the latent heat of vaporization of water (3) To determine the coefficient of linear thermal expansion of a solid.
Theory
Eq. 1: Specific heat of Metal
〖-(Q〗_metal)=Q_water+Q_Al
Expanding this equation will give:
Where: m= mass (metal, water, aluminum)
S= specific heat (metal, water, aluminum)
Tf= final temperature
Ti= initial temperature
*SH2O= 4.184J/g°C
*SAl= 0.910J/g°C
Eq. 2: Heat of Fusion Water
〖-(Q〗_(H_2 O)+ Q_Al)=Q_melting+Q_(meltedH_2 O)
Expanding this equation will give:
Where: m= mass (water, aluminum, melted ice)
S= specific heat (water, aluminum, melted ice)
Tf= final temperature
Ti= initial temperature
Eq. 3: Thermal expansion of solids
α=∆L/(L_O (T_f-T_i))
where α=coefficient of thermal expansion
Tf= final temperature
Ti= initial temperature
LO= original length
Methodology
For this experiment, the following materials were used: calorimeter, hot plate, thermometer, ice, metal object, thread, metal jacket, beaker, linear expansion apparatus, boiler, and a meter stick.
The experiment is divided into three activities. For the first activity, the specific heat of a metal was determined. First, the metal object was weighed. Then, a piece of thread 30 cm long was attached to the metal object and the metal object was slipped inside the metal jacket. The metal jacket was then placed inside the beaker of water, which was heated until the temperature of the object is 80°C. Meanwhile, the inner vessel of the calorimeter was weighed. Then, water was poured into the vessel about 2/3 full. Then, the inner vessel with water was weighed. When the object reached 80°C, it was transferred from the beaker to the calorimeter, taking care not to splash any water. A thermometer was inserted through the cover, and the water was stirred. The temperature at equilibrium was recorded. Finally, computations were made. Using Eq. 1, the specific heat of the object was computed. With the accepted value of the metal, the percent error committed during the experiment was also computed.
Figure 1. Measuring the temperature at equilibrium of sample, water, and inner vessel of calorimeter (Activity 1)
The second activity deals with the heat fusion of water, and it utilizes some of the data obtained from the first activity. The same weight of the inner vessel of the calorimeter used in the first activity was used in the second activity. This time, the water was filled up to one-half, and the weight was measured. The inner vessel was then placed into its insulating jacket, and the initial temperature inside the calorimeter was recorded. Several pieces of ice were added to the water inside the calorimeter, and it was covered. The water was stirred until all the ice cubes have melted. At this point, thermal equilibrium has been established, thus, the temperature was recorded. The inner vessel with water and melted ice was recorded. Lastly, using Eq. 2, the heat of fusion of ice was computed, and applying the accepted value for the latent heat of fusion, the percent error was computed as well.
Figure 2. Mass of inner vessel of calorimeter with water only (left) and with water and melted ice (right)
The last activity is about the thermal expansion of solids. First, the initial length of the rod was measured. Then, the metal rod was placed inside the steam jacket, which was closed tightly at both ends with a stopper, leaving only a portion of each ends. Next, the steam jacket was mounded in the metal frame which has a micrometer disc at one end. Both ends were left free to expand. The steam jacket has two outlets: one for introducing the steam into the jacket and the other for letting steam out of the jacket. The first outlet was connected to the boiler by using rubber tubing. The initial temperature of the rod was measured by inserting a thermometer through the central hole of the jacket, making sure that the thermometer does not touch the rod. Then, the metal frame was connected to the galvanometer. The micrometer screw was moved so that it touches only the end of the rod as indicated by the sudden movement of the galvanometer needle. The initial reading of the micrometer disc was recorded. The disc was unwound, leaving the rod to expand freely. The rod was heated for twenty minutes through the steam coming from the broiler. The final temperature of the rod was recorded. The disc was moved until it is contact again with the rod. The final reading of the disc was recorded. Using Eq. 3, the coefficient of linear thermal expansion of the rod was computed. The percent error was computed using the accepted value of the linear thermal expansion of the rod.
Figure 3. Setup for Activity 4
Results and Discussion
Calorimetry was used in activities 1 and 2 to insulate a hot and cold material. It states that if there is no heat loss to the surroundings, the heat lost by the hotter object equals the heat gained by the cooler ones.
Tables 1, 2 and 3 show the results obtained in Activities 1, 2 and 3 respectively.
Table 1. Specific Heat of Metal
The high temperature of the sample increased the temperature of the water by 1°C. The amount of heat that flows from the metal as it cools is equal to the amount of heat absorbed by the water and the calorimeter.
The experimental specific heat of sample was computed using the formula,
Q_lost = Q_gained.
To compute for the C_m, the formula was derived giving,
C_m = (M_w C_w (T_3-T_1 )+ M_c C_c (T_3-T_1 ) )/(M_m (T_2-T_3 ) )
Subsitute the values,
C_m = ((261.75)(4.184)(28°C-27°C)+ (43.35)(0.910)(28°C-27°C) )/((16.85)(80°C-28°C) )
C_m = 1.2949 J/gC°
Given an accepted value of 0.9100 J/gC°, a 42.31 % error was computed.
Activity 2 states the heat and phase change of matter from solid to liquid through heat of fusion.
Table 2. Heat of Fusion of Matter
The addition of ice caused the temperature to decrease by 13°C.
To compute for the experimental L_f, the formula was derived giving,
L_f= (M_w C_w (T_2-T_3 )+ M_c C_c (T_2-T_3 )-M_ice C_w (T_3-T_1 ) )/M_ice
Substitute the values,
L_f=█((222.56)(4.184)(27°C-14°C)+@(43.35)(0.910)(27°C-14°C)- @(34.87)(4.184)(14°C-0°C))/34.87
L_f=303.29 J/g
Given an accepted value of 334 J/g, a 9.19% error was computed. This indicates that that the calculated experimental value 303.29 J/g is near the theoretical or accepted value of latent heat of fusion which is 334J/g.
Lastly, thermal expansion of solids is observed in Activity 3. The data gathered are presented in Table 3.
Table 3. Thermal Expansion of Solids
The rod increased in length by 0.17 mm after being exposed to increasing temperature and heat is conducted along the rod.
The experimental value of coefficient of thermal expansion was calculated using the formula,
αL= ∆L/L∆T
Substitute the values,
αL= ((0.51-0.35))/((539)(93.0°C-26.5°C))
αL=4.46 ×〖10〗^(-6)/ C°
Given an accepted value of 1.20 ×〖10〗^(-5)/ C°, 62.83% error was obtained.
Conclusion
The calorimeter is the device used to measure heat flow. Chemical reactions occur within the calorimeter, heat passing from one part of the contents to the other, but no heat flows into or out of the calorimeter from or to the surroundings.
The heat capacity of an object or substance is the amount of heat energy required to raise the temperature of the object or substance by one degree. Specific heat is the amount of heat necessary to raise a unit mass (such as one gram) of matter by one degree of temperature. Heat flows from the metal and/or ice to the water, and the two equilibrate at some temperature between the initial temperatures of the metal and/or ice and water. Heat flow of both materials is equal in magnitude but opposite in direction to that of the water.
Conduction heat transfer is the flow of thermal energy in matter as a result of molecular collisions. Conduction is initiated by the excitation or increased vibration of metal molecules of the rod. The excitation of molecules passes thermal energy along the length of the rod and causes its elongation.
Applications
1. Is it possible to add heat to a body without changing its temperature?
Yes, it is possible to add heat to a body without changing its temperature by causing a change in phase. It is a change from one state to another without a change in chemical composition and occurs at constant temperature, pressure and it involves latent heat, for example when ice is heated, it takes energy to convert it from ice to water but it does not change the temperature of the body until all of the ice is converted.
2. Explain why steam burns are more painful than boiling water burns.
Steam burns are more painful than boiling water burns because the specific heat of steam is much higher than the specific heat of water. For the reason that steam contains latent (hidden) heat unlike boiling water, this pushes apart the water molecules and converts liquid water into hot water vapor (steam) which is the extra energy that steam contains.
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?
This is because of natural convection which is a heat transport that is not generated by any external source but only by the local changes in density due to temperature gradients. The driving force for natural convection is buoyancy, a result of differences in fluid density. The water has a greater heat capacity than sand and can therefore absorb more heat than sand, so the surface of the water warms up more slowly than the sand's surface. As the temperature of the sand rises it heats the air above it. The warm air is less dense therefore it rises. This is what brings the sea breeze, air cooled by the water, ashore in the day, and carries the land breeze, air cooled by contact with the ground, out to sea during the night.
4. Explain why alcohol rub is effective in reducing fever.
Rubbing alcohol is effective in reducing a fever since it cools the skin by means of convection, as the alcohol evaporates from the body it also carries the heat with it.
5. Cite instances where thermal expansion is beneficial to man. Cite also instances where thermal expansion is a nuisance.
Thermal expansion is the tendency of matter to change in volume in response to a change in temperature [1]. An example of a thermal expansion benefit to man is the bimetallic strip. This strip consists of two different metals, brass and iron joined together. At normal temperatures the bimetallic strip is straight. As it is heated the brass expands more than the iron, the brass forming the outside curve and the iron contained inside. It can be used in a thermostat to break an electrical circuit. As the temperature increases the strip bends and breaks electrical contact in the heater circuit. When the temperature decreases, the bimetallic strip returns to its original position and shape. Thus, contact is restored.
Changing the shape and dimension of an object, such as doors, walls collapsing due to bulging, cracking of glass due to heating, and bursting of metal pipes carrying steam are some of the disadvantages of thermal expansion of matter.
6. Why is water not used in liquid in glass thermometer?
Water will not rise or fall at temperature changes as mercury does. Water has no linear thermal expansion; its thermal expansion coefficient is at 20°C and not the same as mercury which is at 90°C. Moreover, at atmospheric pressure, water is only liquid over a narrow temperature range of 100°C, which limits its usefulness and accuracy. Further, it cannot be used for phase transitions. For instance when water turns to a gas it consumes a lot of energy (latent heat). Ultimately, a thermometer should have a steady linear response to rises in temperature.
7. The density of aluminum is 2700kg/m3 at 20°C. What is its density at 100°C?
Linear thermal expansion coefficient of Aluminum: 24×10-6/K.
Formula used: ∆L/L = α∆T,
Where, α is the linear thermal expansion coefficient.
Take a cube 1 meter on a side, which at 20ºC weighs 2700kg. What does the length change to at 100ºC?
∆L/L = α∆T
∆L = Lα∆T = (1)(24×10-6)(80) = 0.00192m
The new cube is 1.00192m on one of its sides and the volume is that cubed or 1.00577 m³.
Density is,
2700 kg/1.0057m³ = 2685kg/m³
Therefore, the density of aluminum at 100°C is 2685kg/m³ or 2.69 g/m3
8. How much heat is needed to change 1g of ice at 0°C to steam at 100°C?
First, compute for the amount of heat needed to turn ice into water by multiplying its mass by its latent heat to be able to melt the ice into liquid which is 80cal/g at 0°C.
1g x 80 cal/g = 80cal
Liquid water must then reach to a boil. The heat needed to raise the temperature of liquid water from 0°C to its boiling point is the following,
1g x 100 cal/g = 100cal
Upon reaching its boiling point, water then will begin to evaporate shown by,
1g x 540 cal/g = 540cal
Therefore, the sum of heat needed to change 1g of ice at 0°C to steam at 100°C is 720cal.
80cal + 100cal + 540cal = 720cal
9. An aluminum calorimeter has a mass of 150g and contains 250g of water at 30°C. Find the resulting temperature when 60g of copper at 100°C is placed inside the calorimeter.
Mass of Calorimeter 0.15kg
Mass of Water 0.25kg
Mass of Calorimeter and Water 0.40kg
Mass of Copper 0.60kg
Mass of Calorimeter, Water and Copper 1kg
Initial temperature of water in Calorimeter (T3) 30°C
First, combine aluminum with water by using their given specific heat (C), 900J/kg°K and 4186 J/kg°K for aluminum and water respectively.
Combine first the calorimeter with water:
m3 = 0.15kg + 0.25kg = 0.40kg
C3 = (0.15kg)( 900J/kg°K) + (0.25kg)( 4186 J/kg°K) / 0.40kg = 2,953.75J/kg°K
We can then combine copper with aluminum and water, placing m3, C3, T3 valuesfrom step 1 into m1, C1, T1. Using specific heat C2 = 386 J/kg°K for Cu. Therefore, solving for T3 using basic equation C1m1(T3-T1) = -C2m2(T3-T2)T3 = (m1C1T1+m2C2T2)/(m1C1+m2C2) = 31.35°C.