Physics Formal Report- Expt 6

Experiment 6: Heat Effects Laboratory Report Angelo G. Tolentino, Eloisa Francesca T. Umali, Janine Pearl G. Villaflor, Kim Nicole M. Yap Department of Biology College of Science, University of Santo Tomas Espana, Manila Philippines

Abstract

the coefficient of linear thermal expansion of a solid.

This report shows how to determine specific heat of the object (aluminum), heat of fusion of ice and the thermal expansion of solid (rod). The equilibrium temperature was carefully observed to obtain approximately correct reading. Recording the equilibrium temperatures plays a great role in calculation of the specific heat and latent heat because this data is very fast to occur and improper reading may lead to big percent errors. Errors are kept within 6% 7% range.

2. Theory Second law of thermodynamics: Energy will disperse from a concentrated form to a dilute form if it is not hindered from doing so. [1] Q=cmΔT where, Q = heat m = mass c = specific heat ΔT = change in temperature

1. Introduction Heat is the transfer of energy from a hot material to a cold material. Specific heat is the amount needed by heat per unit mass to raise a temperature by 1 degree Celsius. Latent heat is the energy that can be linked to a phase change of a substance. Thermal expansion is a characteristic of an object to increase in volume with respect to change in temperature.

Q=mL

where, Q = heat m = mass L= specific latent heat

The objectives of the experiment is to determine the specific heat of a solid by method of mixtures, the latent heat of fusion and latent heat of vaporization of water and

ΔL=L ₀ αΔT

where, Page 1

ΔL = increase in length L ₀ = initial length

inside the calorimeter was recorded. A piece of ice was dried and was added to the water inside the calorimeter and was covered. The mixture of water and ice was stirred until the ice has melted and the thermal equilibrium has been established. The equilibrium temperature was recorded. The inner vessel with water and melted ice was weighed. The heat of fusion of ice by Conservation of Heat energy was computed. The % Error was computed.

α = coefficient of linear expansion ΔT = change in temperature

3. Methodology Activity 1: Specific Heat of Metal A metal object, whose specific heat has to be determined, was weighed. The metal object was attach to a 30cm thread and was slipped into the metal jacket. The metal jacket was placed into a beaker of water then the water was heated until the temperature of the object is 80°C. While heating the water, the inner vessel of the calorimeter was weighed. Water was placed into the vessel until it is 2/3 full; the inner vessel with water was weighed. The inner vessel was placed in its insulating jacket and the temperature was measured. Once the object has been heated to 80°C, the object was transferred quickly to the calorimeter. The calorimeter was covered. A thermometer was inserted through the cover and stirred the water. The equilibrium temperature was recorded. The specific heat of the object using Energy Conservation was computed. The % error was computed.

Activity 2: Heat of Fusion of Water The inner vessel of the calorimeter was weighed. The calorimeter was filled with half full of water and was weighed. The inner vessel was placed into its insulating jacket. The initial temperature of water

Activity 3: Thermal Expansion of Solids The initial length of the rod to be tested was measured. A boiler was connected to the first outlet by means of a rubber tubing. A thermometer was inserted through the central hole of the jacket and the initial temperature was measured. The metal frame was connected to the galvanometer. The micrometer screw was moved that it just touched the end of the rod. The initial reading of the micrometer disc was recorded. The disc was unwinding so that the rod can freely expand. The rod was heated for twenty minutes by means of steam coming from the boiler. The final temperature of the rod was recorded. The disc was moved again until it touched the rod. The final reading of the disc was recorded. The expansion of the rod was determined by subtracting the two readings of the disc. The coefficient of linear thermal expansion of the rod was computed. The accepted value of the coefficient of linear thermal expansion was obtained. % Error was computed.

Page 2

Figure 3 Thermal Expansion Setup

Figure 1 Calorimeter Setup

4. Results and Discussion Tables 1, 2 and 3 show the measured mass of objects and temperatures; and the calculated specific heat, latent heat and thermal expansion for activities 1, 2 and 3 respectively. % error was also tabulated. Activity 1 Specific Heat of Metal Table 1 Results for Specific Heat of Metal

Figure 2 Heating of the metal at 90°

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 temperature of sample, water, and Page 3

16.63g 44.61g 158.52g

113.91g

24°C 90°C 26°C

inner vessel of calorimeter Calculated specific heat of sample Accepted value of specific heat % error

heat is 0.9723J/gC° yielding 6.85% error. Errors may be due to a slight difference in reading of the temperature from its true value.

0.9723J/gC° 0.910J/gC°

Activity 2 Heat Fusion of Water

6.85%

Table 2 Results for Heat Fusion of Water

Mass of inner vessel 44.61g of calorimeter Mass of inner vessel of calorimeter with 158.32g −[mw C w ΔT w + mcal C cal ΔT cal ] C s= water ms ΔT s Mass of water inside inner vessel of 113.71g Knowing the values for each part of the calorimeter equation, direct substitution was done Mass of inner vessel of calorimeter, water 177.74g 4.186 J 113.91 gw ∙C ° ( 26−24 ) ° w and melted ice g w Mass of melted ice 19.42g −[¿+ 44.61 g cal 0.910 J / g ∙ C ° cal (26−24)° cal ] Initial temperature of C s= 16.63 g s (26−90)° s water and inner 26°C vessel of calorimeter Equilibrium C s=0.9723 J / g ∙C ° temperature of inner vessel of 12°C T ow=T ocal Remember that calorimeter, water and melted ice The metal that is used in this activity Calculated latent 322.18J/g is aluminum. Specific heat is the amount heat of fusion needed by heat per unit mass to raise Accepted value of 344J/g temperature by a degree. Metals were latent heat of fusion % error 3.54J/g known to have low specific heat. Aluminum as a metal has low specific heat which means it can absorb energy easily. The flow m w C w ΔT w +mcal C cal ΔT cal of energy is from the sample to the water −[¿ +m M .i . C w ΔT M .i . ] and calorimeter. Increase in temperature of Lf = mI water and calorimeter and decrease in temperature of the sample is used as basis in Same with specific heat, direct substitution calculating the specific heat of aluminum. was done. Specific heat of the aluminum is known to be 0.910J/gC° and the calculated specific This activity uses one equation presented below.

Page 4

4.186 J ∙C ° (coefficient 12−26 ) ° w of g w thermal expansion ¿ % error 6.96% −[+ 44.61 gcal 0.910 J / g ∙C ° cal ( 12−26)° cal ¿ +19.42 gm .i . 4.186 J /g ∙ C ° w (12−0)° m .i . ] Lf = 19.42 g I ΔL α= L o ΔT Lf =322.1775 J /g 113.71 gw

−4

Remember

that

mI =mass of ice

α=

and

7.4 × 10 0.517 m× 67

mM .i . =mass of melted ice −5

α =2.14 ×10 The substance that is used in this activity is dry ice. Latent heat is the energy that is linked to a phase change. This is also the heat amount needed to convert solid phase of ice to its liquid phase without changing the system’s temperature. Latent hear of fusion of ice is known to be 344J/g and the calculated latent heat of fusion is 322.18J/g yielding 6.34% error. Errors may be due to not drying the ice completely or as temperature was not read correctly. Activity 3 Thermal Expansion of Solids Table 3 Results for Thermal Expansion of Solids

Initial length of rod Initial reading of micrometer disc Final reading of micrometer disc Elongation of rod Initial temperature of rod Final temperature of rod Experimental value of coefficient of thermal expansion Accepted value of

0.517m

Thermal expansion presented in the experiment is mainly linear expansion. Errors may be due to in incorrect reading of the micrometer disc or the length of the rod using meter stick. 5. Conclusion In the experiment the determination of specific heat of aluminum, the latent heat of fusion of water and the thermal expansion of solids was done. The computed values are: 0.9723J/gC° for specific heat, 322.18J/g for latent heat of fusion and the coefficient for linear expansion as 2.14x10-5/C°.

1.53x10-3

6. Application

2.27x10-3

1. Is it possible to add heat to a body without changing a temperature?

7.4x10-4

Yes because most substances can exist in three states–solid, liquid, and gas. The change of state or phase changes usually involve a transfer of heat energy. During a phase change, the substance can absorb heat energy without changing its temperature until the phase change is complete. Because

23°C 90°C 2.14x10-5/C° 2.3x10-5/C° Page 5

this additional heat does not cause a change in temperature, it is called latent heat.

something removes heat from your skin, that feels cold to you. Evaporation is a process that requires a lot of energy, so it can feel quite cold in your skin absorbing heat when you have a fever.

2. Explain why steam burns are more painful than boiling water burns. Because change of phases requires a lot more energy than a temperature change. When steam (water in the gas phase) hits your 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 your skin where it would decrease in temperature (to your skins temperature) but would not have to go through a phase change.

5. Cite instances where thermal expansion is beneficial to man. Cite also instances where thermal expansion is a nuisance. Advantages 1).Fuel used is cheaper. 2).Smaller space is required to hydro power plant. 3).Economical in initial cost compared to hydro plants. 4).Thermal plants can be placed near load centres unlike hydro and nuclear plants. 5).plants can withstand for certain extent.

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?

Disadvantages 1).requires higher maintenance and operational costs. 2).pollution of atmosphere. 3).huge requirement of water. 4).Handling of coal and disposal of ash is quite difficult. 5).Efficiency of thermal plant is less (3035%).

Sand has the property of getting an environmental temperature very quickly. That is why it is warm in the morning and cold at night. 4. Explain why alcohol rub is effective in reducing fever. As the alcohol evaporates (or any liquid in fact, but alcohol evaporates easily, so it feels more cold than say, water), it goes from a liquid to a gas. The evaporation process takes energy, and that energy goes into breaking apart the liquid alcohol molecules to make isolated alcohol molecules in the gas phase. If it's on your skin, it takes energy from you skin, in the form of heat. If

6. Why is water not used in liquid in glass thermometer?

Page 6

Water has a no linear thermal expansion (Its thermal expansion coefficient at 20C is not the same as at 90C). Also, at atmospheric pressure, water is only liquidous 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. Mercury is a better choice since it doesn’t have any phase transitions in the temperature experience in most everyday situations

multiplying its mass by the latent heat needed to melt ice into liquid (80 cal/g 0C). The sum will be 720 cal. 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.

7. The density of aluminum s 2700 kg/m 3 at20°C. What is its density at 100°C?

Mass of Calorimeter Mass of Water Mass of Calorimeter and Water Mass of Copper Mass of Calorimeter, water, and copper Initial Temperature of water in the Calorimter

Linear thermal expansion coefficient of Aluminum: 24x10-6 /K Formula to be used: ∆L/L = ∆T, is linearα α 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 = ∆Tα ∆L = L ∆T = (1)(24x10α-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 C is 2685 ᵒkg/m³ or 2.69 g/cm 8. How much heat is needed to change 1 g of ice at 0°C to stem at 100°C?

6. References [1] Second Law of Thermodynamics, Retrieved on Nov 2, 2015, Retrieved from http://ceaccp.oxfordjournals.org/content/8/3/ 104.full.pdf [2] Definition of Thermal Expansion, Retrieved on Nov 7, 2015, Retrieved from http://hyperphysics.phy-astr.gsu.edu/hbase/ thermo/thexp.html#c3 [3] Definition of Latent heat of Fusion, Retrieved on Nov 7, 2015, Retrieved from http://study.com/academy/lesson/latent-heatdefinition-formula-examples.html

First compute for the amount of heat needed to turn ice into water by

Page 7