CALORIMETER EXPERIMENT
I. PURPOSE
a. Can understand about heat, heat capacity of matter, and matter specific heat
b. Can determine the heat capacity of calorimeter and specific heat of solid matter.
c. Can understand the using of calorimeter set, thermometer, error theory, and equipment error.
II. BASIC THEORY
Heat is energy transferred from one object to another object which caused by the differences of temperature. Heat flow from the system with higher temperature to the system with lower temperature.
Every matter has different amount of heat to increase their temperature in certain amount of this matter. The comparison of the amount of heat needed by a matter to increase their temperature as DT is called C heat capacity of a matter.
If the heat capacity and calorimeter is known, calorimeter can use to determine the specific heat of a matter. The main factor which must be pay attention in this experiment of using calorimeter is take care the calorimeter system in isolated condition maximally.
III. EQUIPMENTS AND MATERIALS
1. Thermometer
2. Calorimeter set and stirring rod
3. Thermometer and protector vessel.
4. Beaker glass 100 ml
5. Bunsen lamp
6. Heating vessel
7. 2 solid matter
8. Water
9. Analytic weight
10. Matches
11. Clamper tong
12. Alcohol lamp
13. Tripod and gauze
IV. PROCEDURE
A. 1st experiment (Determine the heat capacity of calorimeter)
1. Prepare the equipment and materials that needed
2. Arrange the calorimeter set and prepare to heat the water
3. Weight + 50 gram water and record it as Map. Then heat the water until the temperature is 500 C
4. Weight + 50 gram water which has lower temperature than room temperature. Then put in the calorimeter and record the mass as Mad
5. Observe the thermometer then record the initial equilibrium of cold water and calorimeter as t.
6. Take the water which heated and put it quickly to calorimeter then record the temperature as t2.
7. The mixture of cold water and heat water stir slowly while observe the change of temperature which showed by temperature. After the temperature is stable and the temperature nearly decrease, record the temperature as t3.
8. Throw the water in calorimeter and dry the calorimeter.
9. Repeat the procedure number 3 until 10 for 3 times.
10. Record the data which get in experiment sheet which available.
B. 2nd experiment (Determine the specific heat of solid substance)
1. Take the solid matter which available, weight the mass + 100 gram then put it in heating vessel.
2. Heat the solid matter until the temperature is 800C
3. Repeat the procedure in experiment 1 and change the heat water by the solid substance.
4. Take the record of the data in experiment sheet which available.
V. EXPERIMENT DATA
1st Experiment
| No | Mad (gram) | Map (gram) | t1 0C | t20C | t30C |
| 1. 2. 3. | 50,52 50,18 50,07 | 38,71 38,51 38,26 | 21,5 23 21 | 47,5 46 44 | 31 35 39 |
Information :
Mad : The mass of cold water
Map : The mass of heat water
t1 : equilibrium temperature of calorimeter and cold water
t2 : The temperature of heat water when put in calorimeter
t3 : The equilibrium temperature of calorimeter, cold water, and heat water.
2nd Experiment
| No | Mad (gram) | Map (gram) | t1 0C | t20C | t30C |
| 1. 2. 3. | 50,08 50,04 50,63 | 100,41 100,41 100,41 | 16 18 15 | 80 78 75 | 34 35 29 |
Information :
Mad : The mass of cold water
Map : The mass of heat solid matter
t1 : equilibrium temperature of calorimeter and cold water
t2 : The temperature of heat solid matter when put in calorimeter
t3 : The equilibrium temperature of calorimeter, cold water, and heat solid matter.
VI. DATA ANALYSIS
1st Experiment
| No | Mad (gram) | Map (gram) | t1 (0C) | t2 (0C) | t3 (0C) | C cal (cal/g |
| 1. 2. 3. | 50,52 50,13 50,07 | 38,71 38,51 38,26 | 21,5 23 21 | 47,5 46 44 | 31 35 29 | 16,71 -14,87 21,67 |
Ø Determine the Calorimeter Capacity
ca = 1 cal/gram 0C
The measurement of CK is influenced by the measurement of Map, Mad, t1,t2,t3 so the propagation error is appeared. In this experiment all independent variable measure once (single measurement), so the each error number determine by the equipment error number.
Mad DMad = 0,01 gram
t1 Dt1= 0.5 x 10C =0,50C
t2 Dt2= 0.5 x 10C =0,50C
t3 Dt3= 0.5 x 10C =0,50C
From formula above, we get the differential as below:
=
= 1,74 cal/g0C
=7,08 cal g-1 0C-1
=
= 4,07 cal g-1 0C-1
U = Ma.Ca(t2-t3)
V = (t3-t1)
Then we used the formula:
= 0,14 cal g-1 0C
= 2,72 Cal g-1 0C-1
To determine the average value of CK can be calculated use the weight average formula:
Based on calculation above, we can know the result of Cu and DCu, so we can know the propagation error of calorimeter by indirect calculation in single measurement. So, we can get the value of calorimeter capacity:
Relative error value:
The data of C cal calculation:
= 16,713
= -14,87
= 21,61
2nd Experiment
Determine the Specific Heat of Solid Substance (cb)
| No | Mad (gr) | Mb (gr) | T1 (0C) | T2 (0C) | T3 (0C) | Cb (cal g-1‑ 0C |
| 1. 2. 3. | 50,08 50,04 50,63 | 100,41 100,41 100,41 | 16 18 15 | 80 78 75 | 34 35 29 | 0,37 0,37 0,29 |
The formula to determine Cb
Error theory which used to calculate the specific heat of solid matter (copper) error is:
ü Equipment Error
∆ mad : 0,01 g mb : 100,41 g
∆ t1 : 0,5 x 1 = 0,50C s mb : 0,025 g
∆ t2 : 0,50C C cal : 43,87
∆ t3 : 0,50C ∆ C cal : 6,67
The differential of ∆ Cb formula which get and the example from data number 1.
= -0,37
= 0,02
= 0,0265
ü Rambat Error
=1,24.10-1 Cal g-1 0C-1
=0,124 Cal g-1 0C-1
∆Cb2=1,599.10-1 Cal g-1 0C-1
∆Cb3= 9,693.10-2 Cal g-1 0C-1
Mean C is got by the weight mean formula:
= 3,936.10-2
The mean variation (∆Ckb)
Formula to determine ∂cb ia explained before, so that value is:
∆Cb1=0,125 Cal g-1 0C-1
∆Cb2=1,608 Cal g-1 0C-1
∆Cb3=0,098 Cal g-1 0C-1
Rambat error of Cb, determine by weight mean formula is:
= 3,976.10-2 Cal g-1 0C-1
Based on the data above, we can determine the relative error value is:
VII. DISCUSSION
1) - Heat is flow of energy that can move from a matter with high temperature to a matter with lower temperature when 2 of matter have a contact. Heat movement will occurred until 2 matter have equilibrium temperature (thermal temperature).
- Heat is flow of energy that can move from a matter with high temperature to a matter with lower temperature when 2 of matter have a contact. Heat movement will occurred until 2 matter have equilibrium temperature (thermal temperature).
- Temperature is hot or cold condition of a matter.
- Specific heat of a matter is the amount of heat needed to increase the temperature in every mass unit of matter in every unit of temperature change.
- Heat capacity of matter is the amount of heat needed to increase temperature of matter in every unit of temperature change.
2) The measure of heat capacity in calorimeter is based on “Black Principle” so it doesn’t use the measure of calorimeter specific heat.
3) To determine the specific heat of a matter in mixing method, the specific heat of a matter which heated until certain temperature (t2) must known first. Then put it in calorimeter which has fill by cold water with t1 temperature. This mixing will produce the equilibrium temperature (t3). The factor that must be pay attention in using this method is must be in adiabatic condition. It means that there is no heat flow which go in and out of the system. So, me must avoid all phenomena which bigger the possibility of heat exchange between calorimeter system and the environment.
4) Based on the data which get, the solid matter is copper which has specific heat is 390 J/kg or 0,09 cal/g 0C. But, if we compare this value with the measure of matter specific heat in this experiment, we get the value of matter specific heat (3,976+4,836) cal/kg K. The significant difference value of copper specific heat is caused by mistake in measurement. So, it needs high carefulness in this experiment.
5) Based on 1st experiment in determine the heat capacity of calorimeter we get the relative value 15,2%. In this experiment, we always repeat the observation in several times. We also do that in 2nd experiment to determine the specific heat of solid matter (copper) and get relative value 12,2%.
6) It is allowed to assumed that specific heat of water is constant during the experiment as long there no influence from outside environment. And the temperature of water is must be lower than room temperature. It’s formulated as:
Specific heat of a matter beside based on the number of increasing temperature, it also can be known by the number of absorbed heat or absorb heat and the mass of matter.
Information :
Q : Absorbed heat or absorb heat (joule)
m : Mass of a matter (kg)
c : Specific heat of matter (J/kg 0C)
Dt : Change of temperature of matter (K)
Heat capacity of a matter is equal with the amount of heat which needed and the change of temperature. Systematically, it is written as:
Q = C.Dt
Because of Q = m . c . Dt, so:
c = heat capacity (JK-1)
C = specific heat of matter
In movement of heat obtain Black Principle, “ The amount of heat release of a matter (Q1) equal with the amount of heat absorbed by the other matter (Q2)”
Q1 = Q2
m. c1.Dt1 = m . c2 . Dt2
Assignment
1. If the volume of water in calorimeter compare than the mass of solid matter which put in it is too much, so the temperature of t1 will too large and the t2 is constant, so the value of t3 will nearly same with the value of t1. The differences t3 and t2 will be so small. So, the movement of heat is difficult to observed. The 2nd possibility, the specific heat of solid matter > specific heat liquid matter.
2. Known:
Mb = 100 gram t1= 00C Cb= 0,04 cal/g0C
Mad= 400 gram t3= 10,10C Cad=1
Ask: t2=?
Answer:
Q release = Q absorbed
Mb . cb . Dtb = Mad . cad . Dtad
100 . 0,04 . (t2-10,1) = 400 . 1 . (10,1-0)
4t2 – 40,4 = 40404
t2 = 1020
VIII. CONCLUSION AND SUGGESTION
Conclusion:
a. Temperature is a degree of heat or coldness of a matter.
b. Heat capacity of a matter is the amount of heat which needed to increase the temperature of a matter every temperature change unit.
c. Calorimeter experiment is based on “Black Principle” which state that the amount of heat released is equal with the amount of heat absorbed.
d. Heat is energy which move from higher temperature system to lower temperature system.
e. Specific heat of matter is the amount of heat needed to increase the temperature every unit of substances mass and temperature change.
f. In this experiment we get calorimeter capacity (Ccal) as (Ck+ DCk) = (43,87 + 6,67) cal/g0C and specific heat of solid matter/copper (Cb) as (Cb + DCb) = (3,976 + 4,838.10-3) cal/g0C.
Suggestion:
a. The equilibrium temperature afford to reach thermal condition.
b. When put the heat water or copper into calorimeter, it should be as fast as possible. It is in order to decrease the heat exchange from calorimeter system and environment.
IX. REFERENCES
Alonso, M Finn, EJ. 1992. Dasar-dasarFisika Universitas Jilid I Mekanika dan Termodinamika. Jakarta: Penerbit Erlangga.
Halliday. 1992. Fisika Jilid I. Jakarta: Penerbit Erlangga.
Kanginan, Marthen. 2004. Fisika untuk SMA X. Jakarta: Penerbit Erlangga.
White, Meaning. 1954. Experimental College Physics. M Book Company.
Yoedono. 1991. Pedoman Praktikum Fisika Universitas. Bandung: Armico.
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