Experiment 6: Heat EffectsLaboratory Report

Kim Quiogue, Alexandra Reyes, Sameen Rizwan, Anne Sacayan, Bea Santiago

Department of Math and PhysicsCollege of Science, University of Santo Tomas

España, Manila Philippines

Abstract

The experiment deals with heat effects. It involves three activities, which are Specific Heat of Metal, Latent Heat of Fusion of Water and Thermal Expansions of Solids. The experimental results obtained were 383 J/kgC°, 313,285 J/kg and 3.4 x 10-

04/C°, respectively.

1. Introduction

Once thought to be a substance called Phlogiston, heat is now known to be one form of energy. According to the Kinetic Theory of Matter, heat is a result of the continuous motion and vibration of the atoms and molecules that constitute all matter. The transfer of heat between objects of different temperature involves a reduction in the average motion of the particles of the hotter object and an increase in the average motion of the particles of the cooler object. The SI unit of heat is Joules (J). [2]

The objectives of this experiment are: a.) To determine the specific heat of a solid by method of mixtures; b.) To determine the latent heat of fusion and latent heat of vaporization of water; c.) To determine the coefficient of linear thermal expansion of solid.

2. Theory

It is said that the length of an object changes when the temperature changes.

∆ L=α Lo ∆ TThe equation shows the change

in length (ΔL) which is the result of the coefficient of thermal expansion (α) multiplied with the initial length (Lo) and the change in temperature (ΔT). The common unit for the coefficient of linear expansion:

1C °

=(C ° )−1

The figure below is used to determine the different coefficient of Thermal expansion for solids and liquids. [1]

Figure 1. Coefficients of thermal expansions

In the volume thermal expansion, the volume of an object changes when the temperature changes.

∆ V =β V o ∆TThe equation shows the change

in volume (ΔV) is the result of the coefficient of volume expansion (β) multiplied with the initial volume (Vo) and change in temperature (ΔT). The common unit for the coefficient of volume expansion:

1C °

=(C ° )−1

The heat must be supplied or remove to change the temperature of an object. That is,

Q=mc ∆ TThe equation shows that heat

(Q) is the product of specific heat capacity of an object (mc) and change in temperature (ΔT). The common unit for specific heat capacity is J/(kg.C°).

The figure below is used to determine the different specific heat capacities of different solids and liquids. [1]

Figure 2. Specific heat capacity of solids and liquids

The heat that must be supplied or removed to change the phase of a mass m of a substance is: Q = mL.Where heat is the product of mass (m) and the latent heat (L). The SI unit of latent heat is J/kg. [1]

The figure below shows the latent heats of fusion and vaporization.

Figure 3. Latent heats of fusion and vaporization

3. Methodology

In activity 1, the steel was weighed to determine the specific heat then a 30cm long thread was attached to the steel and slip into the metal jacket. The metal jacket was placed in a beaker with water and the heated until it reaches 80°C. The inner vessel of calorimeter was weighed and then water was placed in the vessel until it is 2/3 full. The inner vessel with water was also weighed. The inner vessel was placed in its insulating jacket and the temperature was measured. The object was quickly transferred from the beaker to the calorimeter after the object was heated to 80°C. The calorimeter was covered and a thermometer was inserted through the cover and the equilibrium temperature was recorded. The specific heat of the object was computed using Energy Conservation and also the % error was computed.

Figure 4. Experimental set-up for Activity 1, the Specific Heat of Metal

In activity 2, the inner vessel of calorimeter was weighed and was filled with half full of water and was weighed again. The inner vessel was weighed inside the insulating jacket and the initial temperature of water inside the calorimeter was recorded. Pieces of dry ice were added to the water inside the calorimeter and were covered. After the ice melted and the thermal equilibrium was established, the thermal equilibrium was recorded. The inner vessel with water and melted ice was weighed. The heat of fusion of ice was computed by using Conservation of Heat Energy. The % error was also computed.

In activity 3, the initial length of the rod was measured and was placed inside the steam jacket and with both ends tightly closed with stopper. The steam jacket was mounted in the metal frame which has a micrometer disc at one end. Both ends were free to expand. The steam jacket has two outlets: one for introducing steam into the jacket and the other for the steam to come out of the jacket. The first outlet was connected to the broiler by means of rubber tubing. The initial temperature of the rod was measured by inserting a thermometer through the central hole of the jacket. The metal frame was connected to the galvanometer. The micrometer screw was moved so that it just touches 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 unwinded so that the rod can expand freely. The rod was heated for twenty minutes by means of steam coming from the broiler. The final temperature of the rod was recorded. The disc was move until it is contact again with the rod. The final reading of the disc was recorded again. The coefficient of linear thermal expansion of the rod was computed and also the % error.

4. Results and Discussion

Calorimetry was used in activities 1 and 2 which use a calorimeter to insulate an unknown 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 MetalMass of sample (mm) 0.04894 kgMass of inner vessel of calorimeter (mc)

0.04391 kg

Mass of inner vessel of calorimeter with water

0.1846 kg

Mass of water inside inner vessel of calorimeter (mw)

0.14067 kg

Initial temperature of water and inner vessel of calorimeter (T1)

30 °C

Temperature of sample (T2) 99 °CEquilibrium temperature of sample, water and inner vessel of calorimeter (T3)

32 °C

Calculated Specific heat of sample (cm)

383 J/kgC°

Accepted value of specific heat (Csteel)

452 J/kgC°

% error 18 %

The experimental specific heat of sample was computed by using the formula, Qlost=Qgained which then gives the formula,

mmcm(T2-T3)=mwcw(T2-T1)+mccc(T3-T1)

Where cw is the specific heat capacity of water and cc is the specific heat capacity of calorimeter.

To compute for the cm, the formula was derived giving,

cm = mwcw(T3-T1)+mccc(T3-T1)mm(T2-T3)

substitute the values,

Cm=(0.14067 ) (4186 ) (32−30 )+(0.04391 ) (900 )(32−30)

0.04894 (99−32)

cm = 383 J/kgC°

The % error was computed by using the formula,

% error= t h eoretical−experimantalexperimental

×100

% error=452−383383

×100

% error=18 %

The calculated % error which is only 18 % indicates that the calculated experimental value 383 J/kgC° is near the theoretical or accepted value of the specific heat capacity of steel 452 J/kgC°.

Activity 2 states the heat and phase change of matter.

Table 2. Heat of Fusion of WaterMass of inner vessel of calorimeter (mc)

0.04391 kg

Mass of inner vessel of calorimeter with water

0.18458 kg

Mass of water inside inner vessel of calorimeter (mw)

0.14067 kg

Mass of inner vessel of calorimeter, water and melted ice

0.21536 kg

Mass of melted ice (mice) 0.03076 kgInitial temperature of water and inner vessel of calorimeter (T2)

31 °C

Equilibrium Temperature of inner vessel of calorimeter, water and melted ice (T3)

13 °C

Calculated latent heat of fusion (Lf) 313,285 J/kg

Accepted value of latent fusion of heat

335,000 J/kg

% error 7 %

The experimental latent heat of fusion was computed using the formula, Qgained = Qlost which gives,

mice Lf =mice cw ( T3−T1 )=mw cw (T 2−T3 )+mc cc (T2−T3)

Where cw is the specific heat capacity of water and cc is the specific heat capacity of calorimeter.

To compute for the experimental Lf, the formula was derived giving,

Lf =mw cw (T 2−T3 )+mc cc (T 2−T 3 )−mice cw(T 3−T 1)

mice

Substitute the values,

Lf =(0.14067 ) ( 4186 ) (31−13 )+(0.04391 ) (900 ) (31−13 )−(0.03076 ) 4186¿(13−0) ¿0.03076

Lf =313,285 J /kg

The % error was computed by using the formula,

% error= t h eoretical−experimantalexperimental

×100

% error=335000−313285313285

×100

% error=7%

The calculated % error which is 7% indicates that the calculated experimental value 313,285 J/kg is near the theoretical or accepted value

of latent heat of fusion which is 335,000 J/kg.

Table 3. Thermal Expansion of SolidsInitial length of rod (L˳) 0.51 mInitial reading of micrometer disc 0.0433 mmFinal reading of micrometer disc 0.0311 mmElongation of rod (ΔL) 0.0122 mmInitial temperature of rod (T˳) 27 °CFinal Temperature of rod (Tf) 90 °CExperimental value of coefficient of thermal expansion

3.8x10-04/C°

Accepted value of coefficient of thermal expansion

2.3x10-5/C°

% error 9%

The experimental value of coefficient of thermal expansion was calculated by deriving the formula

ΔL=αL ˳∆ T giving αL= ∆ LL ˳∆ T

.

Substitue the values:

αL=0.0433−0.0311(0.51 )(90−27)

αL=3.8 ×10−04 /C °

The % error was computed by using the formula,

% error= t h eoretical−experimantalexperimental

×100

% error=2.3 ×10−5−3.4 ×10−04

3.4 ×10−04 ×100

% error=9%

Since the calculated % error which is 9% indicates that the calculated experimental value of coefficient of thermal expansion 3.8x10-04/C° is near the theoretical or accepted value of 2.3x10-5/C°.

5. Conclusion

In thermodynamics, 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 a closely related concept: it is the amount of heat necessary to raise a unit mass (such as one gram) of matter by one degree of temperature. The specific heats of most materials remain essentially constant over the common range of temperatures. At extremely low temperatures, however, specific heats become considerably smaller.

Conduction heat transfer is the flow of thermal energy in matter as a result of molecular collisions. For example in activity 3, if one metal rod is heated, heat is conducted along the rod. The conduction is initiated by the excitation or increased vibration of metal molecules of the rod. The excited molecules then collide with other molecules allowing them to excite also. This process passes thermal energy along the length of the rod and continues as long as the temperature is maintained at the metal rod.

6. Applications

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

You can add heat without raising temperature during the phase change of a pure substance. When a pure substance (water for example) changes from one state of matter to another, it must absorb energy. This energy goes into melting or evaporating the substance rather than causing a rise in temperature. Because this additional heat does not cause a change in temperature, it is called latent heat.

2.) Explain why steam burns are more painful than boiling water burns.

Steam and boiling water are both the same temperature (100 degrees). However, since steam is a gas the particles are further apart and move active. They therefore contain more energy. It is this potential energy that makes steam so much more dangerous than water. Another thing is the fact that steam condenses very quickly because when it is cooled it first has to lose its potential energy. The steam has more thermal energy than the boiling water and that`s why it is more dangerous.

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 and the water is still warm. Why?

During the day, it takes a long time for the water to absorb the heat which is why it is cold and it also takes a long time to release heat at night. The reason why it’s hot because its specific heat capacity is very high, unlike the sand which can absorb and release heat easily because of its low specific heat capacity.

4.) Explain why alcohol rub is effective in reducing fever.

Rubbing alcohol cools the skin by convection, as the alcohol evaporates it carries the heat away from the body with it, just like perspiration, only faster. 

5.) Cite instances where thermal expansion is beneficial to man. Cite also instances where thermal expansion is a nuisance.

One advantage is a bimetallic strip consists of two different metals such as brass and iron joined together. At normal temperature the bimetallic strip is straight. As it is heated the brass

expands more than the iron. So the brass forms the outside of a curve with the iron on the inside. A bimetallic strip can be used in a thermostat to break an electrical circuit. A thermostat is used to maintain a steady temperature in a system. 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 of shape and dimensions of objects such as doors, wall collapsing due to bulging, cracking of glass tumbler due to heating and bursting of metal pipes carrying hot water or steam are some of the disadvantages of thermal expansion of matter.

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

It is probably due to the fact that water has a no linear thermal expansion (Its thermal expansion coefficient at 20°C is not the same as at 90°C). Also, at atmospheric pressure, water is only liquid over a narrow temperature range of 100°C which limits its usefulness. Further it has massive problems at phase transitions. 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.

7.) The density of aluminum is 2700 kg/m3 at 20°C. What is its density at 100°C?

Linear thermal expansion coefficient of Aluminum is 24e-6 /K.

∆L/L = α∆T, α is linear thermal expansion coefficient.Take a cube 1 meter on a side, which at 20º weighs 2700 kg.∆L/L = α∆T∆L = Lα∆T = (1)(24e-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 calculated by dividing 2700 kg with 1.00577 m³, resulting to 2685 kg/m³.

8.) How much heat is needed to change 1g of ice at 0°C to steam at 100°C?

According to the previous concept of heat gain is equal to heat loss, it can be applied that 1g of ice at 0°C requires 100°C convert into steam. It is because there is only a change in physical state in water. The heat loss in ice turns into heat gained after heating, turning the physical state into steam.

9.) An aluminum calorimeter has a mass of 150g and contains 250g of water at 30°C. Find the resulting temperature when 60 g of copper at 100°C is placed inside the calorimeter.

First, combine Al with H2O. Using specific heat with c1 as 900 J/kg-K for Al and c2 as 4186 J/kg-K for H2O.The calorimeter and water together have: m3 = m1+m2 = 0.4 kg.To get c3: (m1c1+m2c2)/m3 = 2953.75 J/kg-K. Therefore, T3 = 30° C.Second, combine Cu with Al-H2O comboby putting m3, c3, T3 values from step 1 into m1, c1, T1. Using specific heat c2 = 386 J/kg-K for Cu.Solving for new T3 using basic equation c1m1(T3-T1) =-c2m2(T3-T2).T3=(m1c1T1+m2c2T2)/(m1c1+m2c2) = 31.3458° C.

7. References

[1] Silverio, Angelina A., Selected Physics Experiments for BS Biology Students, Revised ed., University of Santo Tomas, 2010.[2] Serway, Raymond A., and Vuille, C., Physics Fundamentals 1, Philippine ed., Cengage Learning Asia Pte. Ltd. Phil., 2011.