LAB 10
2 years ago 10
Report102.docx
Exp10PHY2048LExp10DataandInstructions28129.pdf
PHY2048LExp10Theory2.pdf
Report102.docx
Experiment 10:
Student name:
Pre-lab section:
1) Introduction: Explain the theory behind this experiment in a paragraph between 150 and 250 words. (2 Points)
Suppose you are using external resources; include the reference. It would be best if you had any relevant formulas and explanations of each term. You may use the rich formula tools embedded here.
2) Hypothesis: In an If /Then statement, highlight the purpose of the experiment. (1 point)
Post-lab section:
3) Attach an image of your signed data sheet here. (1.5 Points)
4) Attach your analysis here, including any table, chart, or plot image. (7.5 points)
This should include:
Table 1: 1.5 points
Table 2: 1.5 points
Table 3: 1.5 points
Table 4; 1 point
Table 4: 1 point
Table 6: 1 point
5) Attach the image of samples of your calculation here. (1 point)
6) In a paragraph between 100 and 150 words, explain what you Learn. What conclusion can you draw from the results of this lab assignment? (2 points)
7) In one sentence, compare the results of the experiment with your Hypothesis. Why? (1 point)
8) Attach your response to the questions in the lab manual here. (4 points)
Question 1: 1 point for each part, 0.5 points
Question 2: 1 point
Question 3: 0.5 points
Question 4: 1.5 points
Exp10PHY2048LExp10DataandInstructions28129.pdf
Provided data for Exp 10 and instructions for data analysis and lab report
1. Provided data for Exp 10
Table 1 Measured data for specific heat of aluminum
aluminum m (g) 1T (C)
Wm (g) WT (C)
2T (C) c (cal/gC) (measured)
Trial 1
21.17
100
100
19.4
22.7
Trial 2 22.4
Trial 3 22.1
Table 2 Measured data for specific heat of steel
steel m (g) 1T (C)
Wm (g) WT (C)
2T (C) c (cal/gC) (measured)
Trial 1
66.67
100
100
19.4
23.9
Trial 2 24.0
Trial 3 24.3
Table 3 Measured data for specific heat of brass
brass m (g) 1T (C)
Wm (g) WT (C)
2T (C) c (cal/gC) (measured)
Trial 1
114.17
100
100
19.4
26.3
Trial 2 25.5
Trial 3 26.2
2. Instruction for data analysis in Exp 10
(a) Use the measured data in Tables 1 to 3 to calculate the measured specific heat (including the average
value and deviation) of the three metals. Record the results in Tables 4 to 6 (in the lab manual).
(b) Calculate the % errors between the measured and the literature values of specific heats of the
three metals. Record the results in Tables 4 to 6 (in the lab manual).
3. Instructions for Exp 10 lab report
(a) Tables 1 to 6 should be included in your lab report.
(b) It is required that the answers or solutions to the 4 questions (at the end of the lab manual)
should be included in your lab report.
(c) The required other contents and format for your lab report can be found in the syllabus
PHY2048LExp10Theory2.pdf
Concept behind the experiment
Conservation of Energy: For an isolated system, total energy of system remains constant. Only transfer of energy occurs among the components of the system.
Applying to heat energy: If a hot (at higher temperature) and cold (at lower temperature) body insulated away, they exchange heat to reach a final equilibrium temperature, that is same for both bodies. When equilibrium is reached, we have: Total heat energy lost by hot body while reaching equilibrium = Total heat energy gained by cold body while reaching equilibrium
Key Idea for experiment: Take a body at 100 ̊C , immerse in water at room temperature and insulate this system and let it reach equilibrium through exchange of heat energy.
Specific heat of a material
Definition: The amount of heat needed to change the temperature of an object of 1 unit mass, made of some material, through 1 unit temperature is said to be the specific heat c of the material. It is constant for a material. The amount of heatΔQ needed to change temperature by ΔT
of massm of a substance is then: ΔQ = cmΔT (1) In c.g.s system, unit of heat is specific heat is cal /g ̊C . In SI
system, unit of specific heat is J /kg K . We will use the c.g.s system. It takes ΔQ = 1cal amount of heat to raise temperature of 1g
of water from 14.5 ̊C to 15.5 ̊C . Hence, specific heat of water
is: cw = ΔQ mΔT
= 1cal
(1 g)(1 ̊C ) = 1cal /g ̊C .
Experimental setup and quantities
We have the following different quantity measurements needed for the experiment:
Mass of water in cup: mw = 100 g Initial temperature of water in cup:Tw Specific heat of water: cw = 1cal /g ̊C Mass of object: m Initial temperature of object: T
1 = 100 ̊C
Specific heat of object: c Final equilibrium temperature of water and object in cup after heat exchange: T
2
Deriving equation
Heat lost by object to reach equilibrium: ΔQobject = cm(T
1 − T
2 ) . (2)
Heat gained by water in cup to reach equilibrium: ΔQH
2 O = cwmw(T 2 − Tw) (3)
According to conservation of energy, we then have: ΔQobject = ΔQH 2O
(4)
Then, substituting Eqn. (2) & (3) in (4), we get: cm(T
1 − T
2 ) = cwmw(T 2 − Tw)
→ c = mw(T 2− Tw)
m(T 1 − T
2 ) cw (5)
End of Theory
Report102.docx
Experiment 10:
Student name:
Pre-lab section:
1) Introduction: Explain the theory behind this experiment in a paragraph between 150 and 250 words. (2 Points)
Suppose you are using external resources; include the reference. It would be best if you had any relevant formulas and explanations of each term. You may use the rich formula tools embedded here.
2) Hypothesis: In an If /Then statement, highlight the purpose of the experiment. (1 point)
Post-lab section:
3) Attach an image of your signed data sheet here. (1.5 Points)
4) Attach your analysis here, including any table, chart, or plot image. (7.5 points)
This should include:
Table 1: 1.5 points
Table 2: 1.5 points
Table 3: 1.5 points
Table 4; 1 point
Table 4: 1 point
Table 6: 1 point
5) Attach the image of samples of your calculation here. (1 point)
6) In a paragraph between 100 and 150 words, explain what you Learn. What conclusion can you draw from the results of this lab assignment? (2 points)
7) In one sentence, compare the results of the experiment with your Hypothesis. Why? (1 point)
8) Attach your response to the questions in the lab manual here. (4 points)
Question 1: 1 point for each part, 0.5 points
Question 2: 1 point
Question 3: 0.5 points
Question 4: 1.5 points
Exp10PHY2048LExp10DataandInstructions28129.pdf
Provided data for Exp 10 and instructions for data analysis and lab report
1. Provided data for Exp 10
Table 1 Measured data for specific heat of aluminum
aluminum m (g) 1T (C)
Wm (g) WT (C)
2T (C) c (cal/gC) (measured)
Trial 1
21.17
100
100
19.4
22.7
Trial 2 22.4
Trial 3 22.1
Table 2 Measured data for specific heat of steel
steel m (g) 1T (C)
Wm (g) WT (C)
2T (C) c (cal/gC) (measured)
Trial 1
66.67
100
100
19.4
23.9
Trial 2 24.0
Trial 3 24.3
Table 3 Measured data for specific heat of brass
brass m (g) 1T (C)
Wm (g) WT (C)
2T (C) c (cal/gC) (measured)
Trial 1
114.17
100
100
19.4
26.3
Trial 2 25.5
Trial 3 26.2
2. Instruction for data analysis in Exp 10
(a) Use the measured data in Tables 1 to 3 to calculate the measured specific heat (including the average
value and deviation) of the three metals. Record the results in Tables 4 to 6 (in the lab manual).
(b) Calculate the % errors between the measured and the literature values of specific heats of the
three metals. Record the results in Tables 4 to 6 (in the lab manual).
3. Instructions for Exp 10 lab report
(a) Tables 1 to 6 should be included in your lab report.
(b) It is required that the answers or solutions to the 4 questions (at the end of the lab manual)
should be included in your lab report.
(c) The required other contents and format for your lab report can be found in the syllabus
PHY2048LExp10Theory2.pdf
Concept behind the experiment
Conservation of Energy: For an isolated system, total energy of system remains constant. Only transfer of energy occurs among the components of the system.
Applying to heat energy: If a hot (at higher temperature) and cold (at lower temperature) body insulated away, they exchange heat to reach a final equilibrium temperature, that is same for both bodies. When equilibrium is reached, we have: Total heat energy lost by hot body while reaching equilibrium = Total heat energy gained by cold body while reaching equilibrium
Key Idea for experiment: Take a body at 100 ̊C , immerse in water at room temperature and insulate this system and let it reach equilibrium through exchange of heat energy.
Specific heat of a material
Definition: The amount of heat needed to change the temperature of an object of 1 unit mass, made of some material, through 1 unit temperature is said to be the specific heat c of the material. It is constant for a material. The amount of heatΔQ needed to change temperature by ΔT
of massm of a substance is then: ΔQ = cmΔT (1) In c.g.s system, unit of heat is specific heat is cal /g ̊C . In SI
system, unit of specific heat is J /kg K . We will use the c.g.s system. It takes ΔQ = 1cal amount of heat to raise temperature of 1g
of water from 14.5 ̊C to 15.5 ̊C . Hence, specific heat of water
is: cw = ΔQ mΔT
= 1cal
(1 g)(1 ̊C ) = 1cal /g ̊C .
Experimental setup and quantities
We have the following different quantity measurements needed for the experiment:
Mass of water in cup: mw = 100 g Initial temperature of water in cup:Tw Specific heat of water: cw = 1cal /g ̊C Mass of object: m Initial temperature of object: T
1 = 100 ̊C
Specific heat of object: c Final equilibrium temperature of water and object in cup after heat exchange: T
2
Deriving equation
Heat lost by object to reach equilibrium: ΔQobject = cm(T
1 − T
2 ) . (2)
Heat gained by water in cup to reach equilibrium: ΔQH
2 O = cwmw(T 2 − Tw) (3)
According to conservation of energy, we then have: ΔQobject = ΔQH 2O
(4)
Then, substituting Eqn. (2) & (3) in (4), we get: cm(T
1 − T
2 ) = cwmw(T 2 − Tw)
→ c = mw(T 2− Tw)
m(T 1 − T
2 ) cw (5)
End of Theory