Formal Report Exp6.docx

Experiment 6: Heat Effects Andrea Benavides, Rachel Cajiles, Ralph Vincent Canivel, Kimiko Beltran Department of Math and Physics College of Science, University of Santo Tomas España, Manila Philippines Abstract

Theory

This experiment presents the determination of the effects of heat in different substances using almost similar processes. There were three activities done in this experiment. The first one was about determining the specific heat of a metal, specifically aluminium, and proving that its specific heat is about 910 J/kgC°. The second one is determining the latent heat of fusion of water. This activity aims to prove that 0.334 J/kg is the latent heat of fusion of water. In the third experiment, a different process was used. The effect of heat in the expansion of a metal rod was determined. The accepted value of the coefficient of thermal expansion, 0.00024/°C, was proven and determined.

This experiment used the method of mixture to determine the specific heat of a particular solid. Specific heat can be defined as the amount of heat required to change a unit mass of a substance by one degree in temperature. Taking the conservation of energy into account, the following equations were used to determine specific heat:

Introduction Specific heat is a physical property of a substance defined as the amount of energy or heat energy that is needed to change the temperature of a substance by 1 degree. It has a unit of Joules per kilogram*Kelvin. It may also be expressed as cal/gC°. Heat capacity is defined as ratio of the amount of transferred energy and its change in temperature with the formula of C = Q / ΔT with C being the heat capacity, Q being the energy usually expressed in Joules and ΔT pertaining to the change in temperature. Specific heat and heat capacity are related by mass with the formula of C = m x S. This experiment aims 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

𝑄𝑔𝑎𝑖𝑛𝑒𝑑 = 𝑄𝑙𝑜𝑠𝑡 or 𝑐1 𝑚1 𝛥𝑇1 = 𝑐2 𝑚2 𝛥𝑇2

in which Q is heat added or lost, c is specific heat, m is mass, and 𝛥𝑇is change in temperature.

The experiment also tackled the concept of heat of fusion. Heat of fusion can be conceptualized as the heat absorbed by a unit mass of a given solid at its melting point that completely converts the solid to a liquid at the same temperature. It can be computed through the use of the following equations:

𝐻𝑒𝑎𝑡𝑔𝑎𝑖𝑛𝑒𝑑𝑏𝑦𝑖𝑐𝑒 = 𝐻𝑒𝑎𝑡𝑙𝑜𝑠𝑡𝑏𝑦𝑤𝑎𝑡𝑒𝑟 or

Experiment 6: Heat Effects Andrea Benavides, Rachel Cajiles, Ralph Vincent Canivel, Kimiko Beltran Department of Math and Physics College of Science, University of Santo Tomas España, Manila Philippines 𝑚𝑖𝑐𝑒 𝐿𝑓 + 𝑐𝑖𝑐𝑒 𝑚𝑖𝑐𝑒 (𝑇𝑓 − 0°𝐶) = 𝑐𝑤𝑎𝑡𝑒𝑟 𝑚𝑤𝑎𝑡𝑒𝑟 (𝑇0 − 𝑇𝑓 )

     

where m is mass, Lf is the latent heat of fusion, c is specific heat, T0 is initial temperature, and Tf is final temperature.

Thermal expansion can be observed when an amount of heat is applied to certain substances. The change in length that can be observed in an object is directly proportional to the object’s original length and change in temperature, and can be computed for by using a proportionality constant 𝛼, which is the coefficient of linear expansion. This coefficient, in turn, can be retrieved using the following formula:

𝛼=

    

beaker, linear expansion apparatus, boiler, and meter stick. In activity 1, the metal object whose specific heat must be determined was weighed. A piece of thread which about 30 cm long was attached to the metal object and was slipped into the metal jacket. The metal jacket was placed inside the beaker of water. The water was heated until the temperature of the object has reached 80°C. As the object was being heated, the inner vessel of calorimeter was weighed. The water was placed into the vessel until it is 2/3 full. The inner vessel with water was weighed. After that, the inner vessel was placed in its insulating jacket and its temperature was measured. Soon the object has been heated to 80°C, it was quickly transferred from the beaker to the calorimeter without splashing any water. The calorimeter was covered. Another thermometer was inserted through the cover and was used to stir the water. The equilibrium temperature was recorded. The specific heat of the object, using Energy Conservation and percent error were computed.

𝑒 𝐿0 (𝑡𝑓 − 𝑡0 )

where 𝛼 is the coefficient of linear expansion, e is the elongation or change in length, L0 is the initial length, tf is the final temperature, and t0 is the inital temperature.

Methodology Activity 1 In this experiment, the materials used are: calorimeter, hot plate, thermometer, ice blocks, metal object, metal jacket, thread,

Activity 2 In activity 2, the inner vessel of the calorimeter was weighed. The calorimeter was filled with half full of water and was weighed again. The inner vessel was placed inside its insulating jacket. The initial temperature of water inside the calorimeter was recorded. Some ice blocks were added to the water inside the calorimeter and were covered. The mixture of ice and water was stirred until all the ice has been melted and the thermal equilibrium was established. The equilibrium temperature was recorded. The inner vessel with melted ice and water was weighed. The heat of fusion of ice,

Experiment 6: Heat Effects Andrea Benavides, Rachel Cajiles, Ralph Vincent Canivel, Kimiko Beltran Department of Math and Physics College of Science, University of Santo Tomas España, Manila Philippines using Conservation of Heat energy and percent error were computed.

Activity 3 In activity 3, the initial length of the rod to be tested was measured. It was placed inside the steam jacket and it’s both ends were close tightly with stopper, leaving out a small portion of each end of the rod for necessary contacts. The steam jacket was mounted in the metal frame as this frame has a micrometer disc at one end. The steam jacket has two outlets: one for introducing steam into the jacket and the other one was for the steam to come out of the jacket. The first outlet was connected to the boiler by the means of rubber tubings. The initial temperature of the rod was measured by inserting a thermometer through the central hole of the jacket (Noted that the thermometer is touching the rod). The metal frame was connected to the galvanometer. The micrometer screw was moved so that it touched the end of the rod as indicated by the sudden movement of galvanometer needle, in other case of other models of linear expansion, faint light of bulb could be used. The initial reading of the micrometer disc was recorded. The disc was unwind in order the rod could expand freely. The rod was heated for twenty minutes by the means of steam coming out from the boiler. The final temperature of the rod was recorded. The disc was moved until it came into contact with the rod. The final reading of the disc was recorded. The expansion of the rod is the difference between the two readings. The coeffiecient of the linear expansion of the rod was computed by the following equation:

𝛼=

𝑒 𝐿0 − (𝑡𝑓 − 𝑡𝑜 )

The accepted value of the coefficient of linear expansion was given to be 0.000024/°C. The percent error was computed. Results and Discussion Table 1. Results to Specific Heat of Metal Mass of sample Mass of Inner vessel of calorimeter Mass of inner vessel of calorimeter with water Mass of water inside inner vessel of calorimeter Initial temperature of water and inner vessel of calorimeter Temperature of Sample Equilibrium temp of sample, water, and inner vessel of calorimeter. Calculated specific heat of sample Accepted Value of Specific Heat Percent error

16.61g 43.42g 228.63g

185.21g 27 o C 82 o C 28 o C 908.83 J/kg*Co 910 J/kg*Co 0.13 %

Table 2. Results to Heat of Fusion of Water Mass of inner vessel of calorimeter Mass of inner vessel of calorimeter with water Mass of water inside inner vessel of

43.42 g 228.63 g 185.21 g

Experiment 6: Heat Effects Andrea Benavides, Rachel Cajiles, Ralph Vincent Canivel, Kimiko Beltran Department of Math and Physics College of Science, University of Santo Tomas España, Manila Philippines calorimeter Mass of inner vessel of calorimeter, water, and melted ice. Mass of melted ice Initial temperature of water and inner vessel of calorimeter. Equilibrium temperature of inner vessel of calorimeter, water, and melted ice. Calculated latent heat of fusion Accepted value of latent heat of fusion Percent error

talks about when heat is fused and when heat is vaporized. 241.98 g 13.35 g 27o C

21 o C 257.88x103 J/Kg Co 334x103 J/Kg Co 22.81 %

Table 3. Results to Thermal Expansion of Solids. Initial Length of Rod 535mm Initial Reading of 0.45mm Micrometer Disk Final Reading of 1.09mm Micrometer Disk Elongation of Rod 0.64mm Initial Temperature 23o C of Rod Final Temperature of 99 o C Rod Experimental Value of Coefficient of 1.57x10-5 1/Co Thermal Expansion Accepted Value of Coefficient of 2.4 X10-5 1/Co Thermal Expansion Percent Error 34.58% The three experiments done above involve the concept of the effects change of temperature does on metals, water, and solids. It

In Table 1, it explains the concept of Specific Heat. It is the physical property of a substance to increase in Temperature by one degree. The results show that 908.83 J/kgCo is the calculated amount of energy needed to increase the temperature of the given metal which in the experiment, Aluminum was used. The calculated number is very close to the accepted value which is 910 J/kgCo. A good amount of water and a larger difference in the temperatures are factors that would help achieve a better result in the experiment. In Table 2, it explains the Heat of Fusion of water. The researchers solved how much energy is needed to melt water in solid form (ice) into water in liquid form (water). This time we are discussing the effect of heat during a change in Phase, in this case from Solid to Liquid. In this experiment, it also discusses the concept of latent heat of fusion. The accepted value is said to be 334x103 J/Kg Co. Based on the calculation of the researchers the solved latent heat of fusion of water from solid to liquid is 257.88x103 J/Kg Co. It is a very near value to the said accepted with a percentage error of 22.80 %. That is the required latent of fusion to change water in solid form into liquid form. In Table 3, it discusses what happens to certain solids once heat is applied to them. Thermal expansion is the name of the experiment and in this experiment the researchers solved for the Coefficient of Thermal Expansion. The metal used in this experiment is Aluminum and its accepted value is 2.4 X10-5 /Co. During the experiment, the researches made use of a large difference in

Experiment 6: Heat Effects Andrea Benavides, Rachel Cajiles, Ralph Vincent Canivel, Kimiko Beltran Department of Math and Physics College of Science, University of Santo Tomas España, Manila Philippines temperatures and waited for the metal to expand a short distance of 1.5mm. Using the proper calculations and formula, the calculated coefficient of the experiment was 1.57 X10-5 /Co, which is relatively close to the original value with the percentage error of 34.58%. Through all three experiments a great factor that helps the calculations better is when the initial and final temperatures have a large difference. A good amount of water is also another factor that comes into consideration in all three experiments. Conclusion The first activity was done to determine the specific heat of a solid by method of mixtures. The group was given an object made from aluminum and they had to determine its specific heat. The result of the activity that the group obtained was that 908.83 J/kgCo. It was compared to the true specific heat of aluminium which is 910J/KgC ֯. With that, the group had a % error of 0.13%. In the second activity, the latent heat of fusion and latent heat of vaporization of water was determined by heating water and taking its initial temperature and placing ice inside the calorimeter. The result calculated by the group was 257.88x103 J/Kg Co and the accepted value of latent heat of fusion of water is 334x103 J/Kg Co. The %error obtained by the group was 22.80%. The last activity was done to determine the coefficient of linear thermal expansion of a solid. The elongation of the rod was measured after it was heated. The result of the experiment was not as good as the group expected because the % error obtained was 34.58%. The calculated value was 1.57 X10-5 /Co, while the accepted value is 0.000024/°𝐶.

Application 1. Is it possible to add heat to a body without changing its temperature? Adding heat does not always increase the temperature. For example, when water is boiling, adding heat does not increase its temperature. This happens at the boiling temperature of every substance that can vaporize. Adding heat to a boiling liquid is an important exception to general rule that more heat makes a higher temperature. When energy is added to a liquid at the boiling temperature, its converts the liquid into a gas at the same temperature. In this case, the energy added to the liquid goes into breaking the bonds between the liquid molecules without causing the temperature to change. 2. Explain why steam burns are more painful than boiling water burns. When water goes from gas to liquid it is undergoing what is called a phase change. Phase changes require a lot more energy than just a temperature change. The energy required for water to go from a liquid to a gas is called the heat of vaporization. When steam hits the skin, a lot of energy will be released as it condenses into a liquid, undergoing a phase change. This energy release causes a much worse burn than if the same amount of boiling water were to hit the skin where it would decrease in temperature (to your skins temperature) but would not have to go through a phase change. 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

Experiment 6: Heat Effects Andrea Benavides, Rachel Cajiles, Ralph Vincent Canivel, Kimiko Beltran Department of Math and Physics College of Science, University of Santo Tomas España, Manila Philippines cold while the water is still warm. Why? During the day, the sun heats up both the ocean surface and the land. Water is a good absorber of the energy from the sun. The land absorbs much of the sun’s energy as well. However, water heats up much more slowly than land and so the air above the land will be warmer compared to the air over the ocean. At night, the roles reverse. The air over the ocean is now warmer than the air over the land. The land loses heat quickly after the sun goes down and the air above it cools too. The ocean, however, is able to hold onto this heat after the sun sets and not lose it as easily. 4. Explain why alcohol rub is effective in reducing fever. The body uses evaporative cooling to maintain body temperature on hot days or during strenuous activity. The evaporating sweat cools the body. Same principle explains why alcohol rub is effective. When alcohol evaporates, it cools down the body temperature making it effective in reducing fever. 5. Cite instances where thermal expansion is beneficial to man. Cite also instances where thermal expansion is a nuisance. Thermal expansion can be beneficial when, for example, trying to open a glass jar with a metal cap. Heating the cap will make it expand. Making it easier to open. Thermal expansion can be a nuisance when engineers and architects design bridges because bridges are made of metals and metals expand when in it is heated. During the summer, bridges expand and needs

special joints to avoid the bridge to bend. 6. Why is water not used in liquid in glass thermometer? Water will not rise or fall at 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑐ℎ𝑎𝑛𝑔𝑒𝑠 𝑎𝑠 𝑚𝑒𝑟𝑐𝑢𝑟𝑦. water has a nonlinear thermal expansion. Also, at atmospheric pressure, water is only liquid over a narrow temperature range of 100C which limits its usefulness. 7. The density of aluminum is 𝟐𝟕𝟎𝟎 𝒌𝒈/ 𝒎𝟑 at 20°C. What is its density at 100°C? Assume initial volume is 1𝑚3 at 20°C 𝑚 = (2700𝑘𝑔/𝑚3 )(1𝑚3 ) = 2700 𝑘𝑔 𝑣𝑓 = 1 + 1(69𝑥10−6 )(100°𝐶 − 20°𝐶) = 1.00552𝑚3 𝜌=

2700𝑘𝑔 = 2685.18 𝑘𝑔/𝑚3 1.00552𝑚3

8. How much heat is needed to change 1 g of ice at 0°C to steam at 100°C? 1 g of ice at 0°𝐶 to 1g of water at 0°𝐶 Q= 1(80)= 80 cal 1g of water at 0°𝐶 to 1g of water at 100°𝐶 Q=1(1)(100-0)= 100 cal 1 g of water at 100°𝐶 to 1g of steam at 100°𝐶 Q= 1(540)= 540 cal Summation of Q = 80+100+540 = 720 cal 9. An aluminum calorimeter has a mass of 150g and contains 250g of water at

Experiment 6: Heat Effects Andrea Benavides, Rachel Cajiles, Ralph Vincent Canivel, Kimiko Beltran Department of Math and Physics College of Science, University of Santo Tomas España, Manila Philippines 30°C. Find the resulting temperature when 60g of copper at 100°C is placed inside the calorimeter. Mass of water + aluminum = 0.4kg C of water and aluminum 𝐽 4186𝐽 −𝐾)+(0.25𝑘𝑔)( ) 𝑘𝑔 𝑘𝑔

0.15𝑘𝑔(900

=

0.40𝑘𝑔

𝐽 =2953.75 𝑘𝑔

−𝐾

𝑇𝑓 𝐽

=

0.40𝑘𝑔(2953.75 𝑘𝑔 − 𝐾)(30°𝐶)(0.60𝑘𝑔) 0.40𝑘𝑔 (2953.75

𝐽 − 𝑘𝑔

𝐾) (0.60𝑘𝑔)

= 31.3458°𝐶

Reference Cutnell, J., Johnson, K., Young, D., & Stadler, S. (2015). Physics (10th ed.). Asia: John Wiley & Sons (Asia) Pte Ltd. Helmenstine, P. A. (n.d.). Here's How You define Specific Heat Capacity. Retrieved October 27, 2017, from https://www.thoughtco.com/definitionof-specific-heat-capacity-605672 Serway, R. & Vuille, C. (2015). College Physics (10th ed.). Singapore: Cengage Learning Asia Pte Ltd.