The correct answer is -24.3°C.
What is Thermal expansion?
Thermal expansion refers to the process whereby an object increases in size as a result of temperature changes. Since most materials expand when heated and contract when cooled, this phenomenon is commonly observed. When materials are heated, they expand; as they are cooled, they contract. Because of the way molecules interact with one another, the expansion and contraction of materials are linked to changes in their internal energy. Consideration of thermal expansion is essential when designing everything from buildings to bridges and from satellites to coffee cups.
Explanation:We'll use the formula for linear thermal expansion to calculate the temperature change.ΔL = αLΔT,
whereΔL = change in length
L = original length
ΔT = change in temperature
α = coefficient of linear expansion
The formula can be rearranged to solve for ΔT.ΔT = ΔL / αL
From Table 5.2 in the eBook, the coefficient of linear expansion for aluminum is 24.0 × 10^-6 (°C)^-1.ΔL = 1 mm = 0.001 mL = 26 mΔT = ΔL / αLΔT = (0.001 m) / (24.0 × 10^-6 (°C)^-1 × 26 m)ΔT = 16.03°C
Now, we must find the temperature at which the aluminum wing is 1 mm shorter.
ΔT = T2 - T1ΔT = (T2 - 24.3°C) = 16.03°CT2 = 40.33°C -24.3°C = T2 - 24.3°CT2 = -24.3°C + 40.33°CT2 = 16.03°C
Therefore, at -24.3°C, the aluminum wing on a passenger jet is 1 mm shorter.
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Measurements of the radioactivity of a certain isotope tell you that the decay rate decreases from 8335 decays per minute to 3105 decays per minute over a period of 5.00 days. What is the half-life T_1/2 of this isotope? Express your answer to ,in days,three significance
"
The half-life T1/2 of the isotope is approximately 1.87 days.
The half-life T1/2 of the isotope in question, with a decay rate that decreases from 8335 decays per minute to 3105 decays per minute over 5.00 days, can be determined using the formula
T1/2 = tln(2) / ln(N₀/N), where
t represents the elapsed time,
N₀ is the initial number of undecayed nuclei, and
N is the final number of undecayed nuclei.
Given:
t = 5.00 days
N₀ = 8335 decays per minute
N = 3105 decays per minute
Substituting the given values into the formula, we obtain:
T1/2 = 5.00ln(2) / ln(8335/3105)
Evaluating the equation, we find:
T1/2 ≈ 1.87 days (rounded to three significant figures)
Hence, the half-life T1/2 of the isotope is approximately 1.87 days.
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A monoethylene glycol/water mixture is used as a cooling liquid which flows through a metallic tube. The liquid has a density of 1036 kg/m", a dynamic viscosity of 311x10- Ns/m², a thermal conductivity of 0.5 W/mK, and Cp = 3.87 kJ/kg K. Calculate the Prandtl number. What information can you get about the thickness of the hydrodynamic boundary layer compared to the thickness of the thermal boundary layer?
The Prandtl number is 2.416 and the thickness of the thermal boundary layer will be greater than the thickness of the hydrodynamic boundary layer.
Given: Density (ρ) = 1036 kg/m³Dynamic Viscosity (μ) = 311 x 10^-6 Ns/m²Thermal conductivity (K) = 0.5 W/mKSpecific Heat capacity (Cp) = 3.87 kJ/kg KTo find: Prandtl Number (Pr)Prandtl Number:Prandtl number (Pr) is a dimensionless number used to determine the relative importance of momentum diffusivity (kinematic viscosity) and thermal diffusivity. It is defined as the ratio of momentum diffusivity to thermal diffusivity.
Mathematically,Pr = μCp/KWhere,μ = Dynamic ViscosityCp = Specific Heat CapacityK = Thermal Conductivity Substitute the given values,Pr = 311 x 10^-6 x 3.87 x 10³ / 0.5Pr = 2.416 > 1This indicates that the heat transfer takes place by both convection and conduction. Therefore, the thickness of the thermal boundary layer will be greater than the thickness of the hydrodynamic boundary layer.The Prandtl number is 2.416 and the thickness of the thermal boundary layer will be greater than the thickness of the hydrodynamic boundary layer.
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Calculate the
average atomic
mass for this
element.
Isotope -
Percent
Mass Abundance
X-84
0.56
X-86
9.86
X-87
7.00
X-88
82.58
[?] amu
Enter your rounded answer to the hundredths place.
Enter
Answer:
88 AMU
Explanation:
Average AMU = (84 x 0.0056) + (86 x 0.0986) + (87 x 0.0700) + (88 x 0.8258)
= 0.4704 + 8.4796 + 6.09 + 72. 64
= 87.68 AMU
= 88 AMU (sig figs)
Answer:
87.71
Explanation: acellus
20. An electron in a hydrogen atom drops from a potential energy of −1.5eV to −3.4 eV. What is the frequency of the emitted photon? a. 8.58×10 ^14Hz b. 1.55×10 ^12Hz c. 1.62×10^19Hz d 4.58×10^14Hz
The frequency of the emitted photon is d. (4.58×10^14Hz).
When an electron in a hydrogen atom drops from a potential energy of -1.5 eV to -3.4 eV, the frequency of the emitted photon can be calculated using the formula:
ΔE = hv where ΔE is the change in energy, h is Planck's constant, and v is the frequency of the emitted photon. Since the electron drops from a higher energy level to a lower one, ΔE is negative.
Thus, ΔE = -3.4 eV - (-1.5 eV) = -1.9 eV = -1.9 × 1.602 × [tex]10^{-19}[/tex] J (since 1 eV = 1.602 × [tex]10^{-19}[/tex] J)
Now, we can use the above equation to calculate the frequency:
v = ΔE/h = (-1.9 × 1.602 × [tex]10^{-19}[/tex] J)/6.626 × [tex]10^{-34}[/tex] J s= 4.57 × [tex]10^{14}[/tex] Hz
Therefore, the frequency of the emitted photon is 4.57 × 10^14 Hz. Answer: d (4.58×[tex]10^{14}[/tex] Hz)
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In fluidized-bed combustion of coal, what is mixed with coal to remove which major air pollutant?
a) sand; nitrogen oxides
b) limestone; carbon dioxide
c) sand; sulfur oxides
d) limestone; sulfur oxides
e) water; sulfur oxides
The major air pollutant sulfur oxide is removed from the flue gas by limestone, which is mixed with coal during the fluidized bed combustion process.
In fluidized-bed combustion of coal, limestone is mixed with coal to remove the major air pollutant sulfur oxides.
What is fluidized bed combustion?
Fluidized bed combustion (FBC) is a process that burns solid fuel in the presence of a fluidized air stream. This combustion method is similar to circulating fluidized bed combustion (CFBC).
A bed of solid particles is maintained in a state of suspension and turbulence by an upward velocity of the fluid, typically air or an air and fuel mixture. Coal, biomass, and waste products are the most common fuels used in fluidized bed combustion.
What is the purpose of limestone in fluidized bed combustion?
The flue gas from fluidized bed combustion contains major air pollutants such as sulfur oxides (SOx), nitrogen oxides (NOx), and carbon dioxide (CO2). The use of limestone, a calcium-rich mineral, in fluidized bed combustion technology aids in the removal of sulfur oxides (SOx).
Limestone is used as a reagent, which reacts with the sulfur in the coal to form calcium sulfate. It is then captured and eliminated as a solid by the combustion process.
Therefore, the major air pollutant sulfur oxide is removed from the flue gas by limestone, which is mixed with coal during the fluidized bed combustion process.
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The quality control officer at a chemical plant wants to know what proportion of the chemicals produced contain some kind of impurity. Company guidelines require a ' 99 ′
% confidence level and a margin of error of ' 2%. Past audits have found impurities in ' ′
% of the chemicals. A public health official responding to an outbreak of measles needs to estimate the vaccination rate in the community. The official will use a confidence interval of ' 95 ′
% and a margin of error of ' 2%, but they do not have an estimate for the population proportion.
To determine the proportion of chemicals with impurities, the quality control officer at the chemical plant can use a confidence interval. The company guidelines require a 99% confidence level with a margin of error of 2%. Based on past audits, impurities were found in '′% of the chemicals.
The quality control officer wants to know the proportion of chemicals produced that contain impurities. This can be
Since the officer doesn't provide the sample size or an estimate for the population proportion, it is not possible to provide a specific confidence interval in this case. However, the officer can use these steps to calculate the confidence interval once the necessary information is available.
Keep in mind that confidence intervals are used to estimate population parameters based on sample data. They provide a range of values within which the true population parameter is likely to fall. The confidence level represents the level of certainty or confidence associated with the interval. A larger confidence level requires a wider interval, resulting in a larger margin of error. Similarly, a smaller margin of error requires a larger sample size or a higher level of confidence.
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3 moles of an ideal gas with CV=3R/2 undergo a quasistatic adiabatic expansion from Plabs=8 atm, V1=6L to V2=101. a) Find the initial temperature, T1. b) Find the final pressure and temperature, P2 and T2. c) Sketch the process on a P−V diagram. d) Find Q,ΔU, and Wk for this process (give your answers in atmL).
a) Initial temperature, T1 = 480 Kb)
Final pressure, P2 = 1 atm
Final temperature, T2 = 157.5 KC)
Sketch the process on a
P-V diagram
D) Q = 0ΔU = nCv(T2 - T1)Wk = nCv(T2 - T1) - nRT
1) Initial temperature, T1 = 480 K
Number of moles, n = 3CV = 3R/2R = 8.314 J/K mo
lHence, CV = (3/2) x 8.314 = 12.471 J/K mol
PV = nRT1 x 8 = n x 8.314 x 480
on substituting the value of n = 3,
we get
P1V1 = 11717.632 Jor P1V1 = 11.717632 L.atm
Q = 0 (since the process is adiabatic)
ΔU = nCV(T2 - T
1)on substituting the value of n = 3, CV = 12.471 J/K mol, T2 = 157.5 K, and T1 = 480 K, we getΔU = - 13138.601 J or - 12.66 L.atm
Wk = ΔU - nRT1
on substituting the value of ΔU = - 13138.601 J, n = 3, R = 0.082 L.atm/mol
K, T1 = 480 K, we get
Wk = - 4100.41 J or - 3.95 L.atm
b) Final pressure, P2 = 1 atm
Final volume, V2 = 101 LPVγ = constant
(where γ is the ratio of specific heats
)For 1 mole of ideal gas, γ = CP/CV
where CP = CV + Rwhere CV = 3R/2, and R = 0.082 L.atm/mol
KCP = 5/2 Rγ = CP/CV = (5/2)/ (3/2) = 5/3
Now, P1V1γ = P2V2γSo, P2 = P1(V1/V2)γ
on substituting the values of P1 = 8 atm, V1 = 6 L, V2 = 101 L, and γ = 5/3, we getP2 = 0.0568 atm
Final temperature, T2 = (P2V2/nR)
on substituting the values of P2 = 0.0568 atm, V2 = 101 L, n = 3, and R = 0.082 L.atm/mol K, we getT2 = 157.5 Kc)
The graph of the process is shown below:
d) Q = 0ΔU = nCv(T2 - T1)Wk = ΔU - nRT1
On substituting the values of ΔU = - 13138.601 J, n = 3, Cv = 12.471 J/K mol, T2 = 157.5 K, T1 = 480 K, and R = 0.082 L.atm/mol K, we getΔU = - 13138.601 J or - 12.66 L.atm
Wk = - 4100.41 J or - 3.95 L.atm
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While solving a problem, we use a system in which mass (kg), force (N), and length (m) are the base units. What would you recommend for this system from the following: A new system of units will have to be formulated b. The above situation is not feasible Only the unit of time have to be changed from second to something else a No changes are required A Moving to another question will save this response.
The correct answer for this question is "A new system of units will have to be formulated."
When the mass (kg), force (N), and length (m) are the base units, it forms the SI system.
Therefore, the SI system is not appropriate in this scenario since it has meter, kilogram, and second as its base units. The only option remaining is to form a new system of units that would support these base units in order to resolve the problem.The Metric system is a popular system that is used all around the world. It is quite simple and has three main base units which include mass (gram), length (meter), and time (second). However, it is not acceptable for the aforementioned scenario. Instead, a new system of units will have to be formulated that would support the base units of mass, force, and length.
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H94. In the Bohr model of the hydrogen atom, an electron orbits a proton at \( 2.18 \times 10^{6} \mathrm{~m} / \mathrm{s} \) on a circle with radius \( 5.29 \times 10^{-11} \mathrm{~m} \). Calculate
The centripetal force acting on the electron in the Bohr model of the hydrogen atom is approximately [tex]\( 7.8476 \times 10^{-9} \, \text{N} \).[/tex].
The given problem involves the Bohr model of the hydrogen atom, where an electron orbits a proton in a circular path. We are provided with the velocity of the electron ([tex]\(2.18 \times 10^6 \, \text{m/s}\[/tex])) and the radius of the circular path[tex](\(5.29 \times 10^{-11} \, \text{m}\)).[/tex]
To solve the problem, we need to calculate the centripetal force acting on the electron.
The centripetal force is given by the formula:
[tex]\[ F = \frac{mv^2}{r} \][/tex]
where
[tex]\( F \)[/tex]is the centripetal force,
[tex]\( m \)[/tex]is the mass of the electron,
[tex]\( v \)[/tex]is the velocity of the electron, and
[tex]\( r \)[/tex]is the radius of the circular path.
Since we are dealing with the hydrogen atom, we know that the mass of the electron[tex](\( m \))[/tex] is approximately [tex]\( 9.11 \times 10^{-31} \, \text{kg} \).[/tex]
Let's substitute the given values into the formula:
[tex]\[ F = \frac{(9.11 \times 10^{-31} \, \text{kg})(2.18 \times 10^6 \, \text{m/s})^2}{5.29 \times 10^{-11} \, \text{m}} \][/tex]
Simplifying the expression:
[tex]\[ F = \frac{9.11 \times 10^{-31} \, \text{kg} \times (2.18 \times 10^6 \, \text{m/s})^2}{5.29 \times 10^{-11} \, \text{m}} \][/tex]
Calculating the numerator:
[tex]\[ 9.11 \times 10^{-31} \, \text{kg} \times (2.18 \times 10^6 \, \text{m/s})^2 = 4.157218 \times 10^{-19} \, \text{kg} \cdot \text{m}^2/\text{s}^2 \][/tex]
Substituting the value of the numerator into the formula:
[tex]\[ F = \frac{4.157218 \times 10^{-19} \, \text{kg} \cdot \text{m}^2/\text{s}^2}{5.29 \times 10^{-11} \, \text{m}} \][/tex]
Simplifying the expression:
[tex]\[ F = \frac{4.157218 \times 10^{-19} \, \text{kg} \cdot \text{m}^2/\text{s}^2}{5.29 \times 10^{-11} \, \text{m}} \][/tex]
Calculating the division:
[tex]\[ F \approx 7.8476 \times 10^{-9} \, \text{N} \][/tex]
Therefore, the centripetal force acting on the electron in the Bohr model of the hydrogen atom is approximately [tex]\( 7.8476 \times 10^{-9} \, \text{N} \).[/tex]
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M photons at a wavelength of λ
1
are incident on a solution in a cuvette. The concentration of a specific chemical in the solution is N and the molar extinction of the solution (per unit length) is α. Derive an equation for the number of the emitted photons at wavelength λ
2
, if the quantum efficiency is η. Show details of your derivation and assumption.
The number of emitted photons at wavelength λ2 is directly proportional to the intensity of light at that wavelength.
The equation for the number of the emitted photons at wavelength λ2 is derived by using the Beer-Lambert law, which gives the relationship between the intensity of light and the concentration of the absorbing species in the solution.
The number of emitted photons at wavelength λ2 can be derived using the Beer-Lambert law, which gives the relationship between the intensity of light and the concentration of the absorbing species in the solution. The equation is given by
I = I0e-αl,
where I is the intensity of light after passing through a solution of thickness l, I0 is the initial intensity of light, α is the molar extinction coefficient (per unit length), and l is the path length of the solution. The molar extinction coefficient α is related to the concentration of the absorbing species
N by α = εN,
where ε is the molar extinction coefficient (per unit concentration).Thus, the equation for the number of emitted photons at wavelength λ2 is given by
N2 = η(I2/I1)(λ1/λ2),
where N2 is the number of emitted photons at wavelength λ2, η is the quantum efficiency, I1 is the initial intensity of light at wavelength λ1, and I2 is the intensity of light at wavelength λ2. The assumption made is that the quantum efficiency η is independent of the wavelength. Therefore, the number of emitted photons at wavelength λ2 is directly proportional to the intensity of light at that wavelength.
The equation for the number of emitted photons at wavelength λ2 is derived using the Beer-Lambert law, which gives the relationship between the intensity of light and the concentration of the absorbing species in the solution. The equation is given by
N2 = η(I2/I1)(λ1/λ2),
where N2 is the number of emitted photons at wavelength λ2, η is the quantum efficiency, I1 is the initial intensity of light at wavelength λ1, and I2 is the intensity of light at wavelength λ2. The assumption made is that the quantum efficiency η is independent of the wavelength. Therefore, the number of emitted photons at wavelength λ2 is directly proportional to the intensity of light at that wavelength.
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Which of the following is false about DSC: A. A metal pan is placed on each disc. One of the pans contains a sample and the other is containing copper as reference. B. Both the sample and the reference are heated at a constant temperature rate (°C/min C. The transition temperatures for a sample are obtained using a standard reference sample 12 A fluctuating stress caused a material to fail at a stress lower than the yield strength due: A creep B, tension C. fatigue D. hardness 13 DBTT value should be below the working environment temp of the designed object: A. False B. True A test that is conducted on a sample such that different parts of a specimen experience different angular displacements is called? D. hardness BTT value should be below the working environment temp of the designed object: A. False B. True test that is conducted on a sample such that different parts of a specimen experie fferent angular displacements is called? A. Compressive test B. Torsion test C. Flexural test D. Fatigue test
Fatigue is a form of failure that occurs when a material is subjected to a fluctuating stress load. A torsion test is a mechanical test that is used to determine a material's mechanical properties under torsional loads. The test is conducted on a sample such that different parts of a specimen experience different angular displacements.
Differential scanning calorimetry (DSC) is a thermal analysis method that examines the differences in the amount of heat required to raise the temperature of a sample and reference as a function of temperature or time. Here's the answer to your questions.1. Which of the following is false about DSC: A metal pan is placed on each disc. One of the pans contains a sample, and the other contains copper as a reference.
A fluctuating stress caused a material to fail at a stress lower than the yield strength due to: Answer: fatigueFatigue is the answer to this question. Fatigue is a form of failure that occurs when a material is subjected to a fluctuating stress load. The stress load is below the yield strength, but it causes the material to fail. Fatigue is a common cause of failure in engineering materials, and it can lead to unexpected and catastrophic failures if not correctly accounted for during design.3. DBTT value should be below the working environment temp of the designed object. Answer: TrueDBTT value should be below the working environment temp of the designed object.
The statement is true. The DBTT value should be lower than the temperature of the working environment in which the object is being used. This is because the DBTT value is the temperature at which the material's ductility becomes brittle. A material with a high DBTT value is more prone to brittle fractures, which can be catastrophic in a working environment.4. A test that is conducted on a sample such that different parts of a specimen experience different angular displacements is called?The test is conducted on a sample such that different parts of a specimen experience different angular displacements. The results of the test are used to determine the material's shear modulus, which is a measure of the material's resistance to shear deformation.
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atexteotesutury) (6 mank) 6. Is a misture of ibuprofen enantiomers, (5) - tuprofen showed an ontical rotation of +34.1−54.5 How mach of eachenantiomer is present in the mistote? Pure (S) enantiorner fist a wecific rotation of +54.5(destrototatoryh i 6 marks)
19.29% of (S)-enantiomer and 80.71% of (R)-enantiomer are present in the mixture.
Optical rotation of a mixture of ibuprofen enantiomers, (5)-tuprofen = +34.1.
Specific rotation of pure (S)-enantiomer of ibuprofen, [α]D° = +54.5.
The molecular formula of ibuprofen is C13H18O2. Ibuprofen belongs to the class of organic compounds known as phenylpropionic acids. It is a member of the phenylpropionic acid class of molecules which are essentially analogues of ibuprofen. Thus, the molecular structure of ibuprofen is: IUPAC Name: 2-(4-isobutylphenyl)propanoic acid.
The observed specific rotation of the mixture of enantiomers is +34.1° and that of the pure (S)-enantiomer is +54.5°.
As we know, Specific rotation (α) = observed rotation (α°) / (l x c), where,
l = length of the cell, c = concentration of the sample (g/mL)
From the given data, we can write: α(observed) = α(mixture)α(mixture) = +34.1°α(S-enantiomer) = +54.5°
Now, we can use the following relation to calculate the amount of each enantiomer in the mixture:
Enantiomeric excess (e.e.) = {[α](observed) - [α](pure)] / [α](pure)} × 100%where [α] = specific rotation.
So, the enantiomeric excess (e.e.) of the mixture can be given as: e.e. = {[α](observed) - [α](pure)] / [α](pure)} × 100%
= {[+34.1° - (+54.5°)] / (+54.5°)} × 100%
= -37.43%
Now, the formula to calculate the amount of each enantiomer is:
Amount of (S)-enantiomer = (e.e. + 100%) / 2 × Total amount of ibuprofen
Amount of (R)-enantiomer = (100% - e.e.) / 2 × Total amount of ibuprofen
Putting the values, Amount of (S)-enantiomer = (-37.43% + 100%) / 2 × 6 = 19.29%
Amount of (R)-enantiomer = (100% - (-37.43%)) / 2 × 6 = 80.71%
Therefore, 19.29% of (S)-enantiomer and 80.71% of (R)-enantiomer are present in the mixture.
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what is the longest wavelength light capable of ionizing a hydrogen atom in the n=7 state?
The longest wavelength light capable of ionizing a hydrogen atom in the n=7 state is approximately 913.4 nanometers.
To determine the longest wavelength of light capable of ionizing a hydrogen atom in the n=7 state, we need to consider the energy levels of hydrogen atoms and the ionization process.
In hydrogen atoms, electrons occupy different energy levels or shells, labeled by the principal quantum number (n). The energy of an electron in a specific energy level is inversely proportional to the square of the principal quantum number. This means that higher energy levels have lower binding energies.
The ionization of a hydrogen atom occurs when an electron is completely removed from the atom, breaking the electrostatic attraction between the electron and the proton in the nucleus. Ionization requires supplying enough energy to overcome the binding energy of the electron.
The energy required to ionize a hydrogen atom in the n=7 state is equal to the energy difference between the n=7 energy level and the ionization energy level, which corresponds to the electron being completely removed from the atom.
The ionization energy of hydrogen is approximately 13.6 eV (electron volts). Using the energy equation E = hc/λ, where E is the energy of a photon, h is Planck's constant, c is the speed of light, and λ is the wavelength of light, we can calculate the wavelength of the longest wavelength light capable of ionizing the hydrogen atom.
First, we convert the ionization energy from electron volts to joules:
1 eV =[tex]1.602 \times 10^{-19[/tex] J
Ionization energy = 13.6 eV × 1.602 ×[tex]10^{-19[/tex] J/eV = 2.179 × 10^-18 J
Next, we rearrange the energy equation to solve for wavelength:
λ = hc/E
Plugging in the known values:
λ = (6.626 × 10^-34 J·s × 3.00 × [tex]10^8[/tex] m/s) / (2.179 × [tex]10^{-18[/tex] J)
Calculating this equation yields:
λ ≈ 913.4 nm
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Explain emissivity and state its range. (2) (b) Surface A is coated with white paint and is maintained at a temperature of 200°C. It is located directly opposite surface B which is considered a black body and is maintained at a temperature of 800°C. Calculate the amount of heat that needs to be removed from surface A per unit area to maintain its constant temperature. Assume the following: emissivity = 0.97, view factor = 1; Stefan Boltzmann constant = 56.7 x 10-9 W/m².K4
Therefore, the amount of heat that needs to be removed from surface A per unit area to maintain its constant temperature is 296.5 W/m².
Emissivity refers to the ability of an object to radiate energy compared to a black body at the same temperature.
The range of emissivity is between 0 and 1. A perfectly reflecting surface has an emissivity of 0, whereas a black body has an emissivity of 1.
(b) Surface A, which is coated with white paint, has an emissivity of 0.97 and is maintained at a temperature of 200°C.
It is located directly opposite surface B, which is considered a black body and is maintained at a temperature of 800°C.
Using the Stefan-Boltzmann law, we can calculate the amount of heat that needs to be removed from surface A per unit area to maintain its constant temperature.
The Stefan-Boltzmann law states that the rate of heat transfer by radiation is proportional to the fourth power of the absolute temperature and is given by:
q/A = εσ(Ta⁴ - Tb⁴)F12
where q/A is the rate of heat transfer per unit area, Ta and Tb are the temperatures of surfaces A and B, ε is the emissivity of surface A, σ is the Stefan-Boltzmann constant, and F12 is the view factor, which is equal to 1 in this case
since the surfaces are directly opposite each other.
Plugging in the given values, we get:
q/A = 0.97 × 56.7 × 10⁻⁹ × (473.15⁴ - 1073.15⁴)
q/A = 296.5 W/m²
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Find a to f using formula
b) Nominal \( \mathrm{GOP}_{2023}= \) ? c) Real \( \mathrm{GbP}_{2022}= \) ? d) Real GOP \( 2023= \) ? e) Economic growith \( (2022-23)= \) ? f) In how many yeass will this arf \( \mathrm{} \)
a) In order to provide answers to b), c), d), e), and f), we need the missing information or formulas.
b) Nominal GOP (Gross Domestic Product) for 2023 cannot be determined without the necessary information or formula.
c) Real GbP (Gross British Pound) for 2022 cannot be determined without the necessary information or formula.
d) Real GOP (Gross Domestic Product) for 2023 cannot be determined without the necessary information or formula.
e) The economic growth rate between 2022 and 2023 cannot be determined without the necessary information or formula.
f) The time required for "arf" cannot be determined without further clarification on what "arf" refers to. Please provide more context or specify the term you are referring to.
a) In order to provide answers to b), c), d), e), and f), we need the missing information or formulas. Please provide the necessary details or formulas so that I can assist you accurately.
b) Nominal GOP (Gross Domestic Product) for 2023 cannot be determined without the necessary information or formula.
c) Real GbP (Gross British Pound) for 2022 cannot be determined without the necessary information or formula.
d) Real GOP (Gross Domestic Product) for 2023 cannot be determined without the necessary information or formula.
e) The economic growth rate between 2022 and 2023 cannot be determined without the necessary information or formula.
f) The time required for "arf" cannot be determined without further clarification on what "arf" refers to. Please provide more context or specify the term you are referring to.
Without the relevant information or formulas, it is not possible to provide specific answers to b), c), d), e), and f). Please provide the required details or formulas so that I can assist you accurately.
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By how many carbon atoms does each member of a homologous series differ from the previous member? (1) 1, (2) 2, (3) 3, (4) 4.
The correct answer is that each member of a homologous series typically differs from the previous member by one carbon atom.
n a homologous series, each member differs from the previous member by the same functional group and a constant increment of carbon atoms. This constant increment is known as the "carbon atom difference" or "carbon atom increment."
In the given options, the correct answer is (1) 1. Each member of a homologous series typically differs from the previous member by adding or subtracting one carbon atom.
For example, consider the homologous series of alkanes:
Methane (CH4)
Ethane (C2H6)
Propane (C3H8)
Butane (C4H10)
In this series, each member differs from the previous member by adding one carbon atom.
Methane has one carbon atom, ethane has two carbon atoms (one more than methane), propane has three carbon atoms (one more than ethane), and so on.
Therefore, the correct answer is that each member of a homologous series typically differs from the previous member by one carbon atom.
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what are buffers and why are they important to life
Buffers are solutions that can resist changes in pH upon the addition of acidic or basic components.
In a physiological context, buffers play a crucial role in maintaining stable pH levels in various bodily fluids. They are vital to life because changes in pH can significantly impact biological processes, and excessive changes can lead to cell damage and death. For example, enzymes are highly sensitive to pH levels and require a narrow range of acidity to function properly. Therefore, the ability of buffers to resist changes in pH is critical to maintaining the integrity and function of enzymes within the body.In addition to maintaining pH levels, buffers are also important in other biological processes. For instance, they can help regulate metabolic reactions by balancing the concentrations of various ions in the body. They can also help stabilize DNA, RNA, and proteins by preventing changes in their charge, structure, and function. Overall, buffers are essential to life because they play an integral role in maintaining pH levels, regulating metabolic reactions, and stabilizing biological molecules.
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IP A gas has a temperature of 290 K and a pressure of 105kPa. Aswuring the gas molecules can be approwimated as wmal spheres of darmeter 30×10
−19
en determine the fraction of the volume found h part A that is acoupled by the maleciles Express your antwer using two shanificant figures:
Given,Temperature of the gas, T = 290 KPressure of the gas, P = 105 kPaDiameter of the gas molecules, d = 30 × 10^-19 mFirst we have to find the volume of one molecule of the gas. It can be given as:V = (4/3) × π(d/2)^3V = (4/3) × π(15 × 10^-19 m)^3V = 1.41 × 10^-57 m^3Now, we have to find the fraction of the volume occupied by the gas molecules. It can be given as:Fraction of the volume occupied by the gas molecules = (Volume of the gas molecules × Number of gas molecules) / Volume of the containerNumber of gas molecules can be calculated using the ideal gas equation as:n = PV / RTn = (105 × 10^3 Pa × 1 m^3) / (8.31 J/K mol × 290 K)≈ 0.00412 molNumber of gas molecules = 0.00412 × 6.022 × 10^23 = 2.48 × 10^21Fraction of the volume occupied by the gas molecules = (V × n) / VcontainerFraction of the volume occupied by the gas molecules = n = 2.48 × 10^21 ≈ 2.5 × 10^21 (approx)Therefore, the fraction of the volume found in part A that is occupied by the molecules is 2.5 × 10^21.
(a) Explain the connection between the ‘Greenhouse effect’ and blackbody radiation
(b) List the basic assumptions used in the Bohr model of the atom.
(c) Sketch and label the experimental arrangement that can be used to measure the photoelectric effect.
(d) Calculate the longest wavelength that could be used to excite an electron from the valence band of Si to its conduction band, given that the band gap of Si is 1.1eV.
I want all questions answered but could you explain the concept of question (d) especially.
Thank you.
a) The greenhouse effect and blackbody radiation are interrelated because the latter plays a vital role in the former.
Blackbody radiation is the term used to define the emission of electromagnetic radiation from an object when it is heated.
These radiations have different wavelengths, and their intensities are determined by Planck's Law.
The greenhouse gases present in the earth's atmosphere have the property of absorbing and emitting radiation.
When solar radiation hits the earth's surface, it is absorbed and re-emitted as infrared radiation.
The greenhouse gases present in the atmosphere absorb this infrared radiation, thereby raising the temperature of the earth's atmosphere.
Hence, the greenhouse effect is the phenomenon where the presence of greenhouse gases causes the earth's temperature to rise.
b) The Bohr model of the atom has the following assumptions:
An electron in an atom moves in a circular orbit around the nucleus.
The energy of the electron in an atom is quantized, i.e., it can have certain discrete values only.
The angular momentum of the electron is quantized, i.e., it can have certain discrete values only.
An electron can move from one energy level to another by either absorbing or emitting radiation.
The frequency of the absorbed/emitted radiation is directly proportional to the energy difference between the two levels.
c) An experimental arrangement to measure the photoelectric effect can be sketched as follows:
Light Source
|
V
Collimator
|
V
Monochromator
|
V
Photoelectric
Material
|
V
Anode Plate
|
V
Ammeter
Light Source:
Provides a source of light, usually a monochromatic (single-frequency) light such as a laser or a mercury lamp.
The frequency of the light can be controlled.
Collimator:
A device that ensures the light beams emitted from the light source are parallel and concentrated into a narrow beam.
Monochromator:
A device that selects a specific wavelength or frequency of light from the collimated beam, allowing for precise control of the incident light's frequency.
Photoelectric Material:
A metallic surface or a semiconductor material (photocathode) that exhibits the photoelectric effect.
It is placed in the path of the selected monochromatic light.
Anode Plate:
A positively charged electrode placed near the photoelectric material to collect the emitted electrons (photoelectrons).
The anode is connected to an ammeter to measure the photoelectric current.
Ammeter:
A device used to measure the magnitude of the photoelectric current. It indicates the flow of electrons from the photoelectric material to the anode plate.
The experimental setup allows the experimenter to vary the frequency (or wavelength) of the incident light and measure the corresponding photoelectric current.
By systematically changing the frequency and observing the current, one can investigate the threshold frequency, determine the relationship between frequency and kinetic energy of emitted electrons, and study other properties of the photoelectric effect.
d) The bandgap energy is given by:
Eg = hc/λ
where
Eg is the bandgap energy,
h is Planck's constant,
c is the speed of light, and
λ is the wavelength of light.
The bandgap energy of Si is given as 1.1 eV.
Therefore, we can find the maximum wavelength that could excite an electron from the valence band to the conduction band of Si using the following equation:
Eg = hc/λ1.1 eV
= 4.14 × 10-15 eV s × 3 × 108 m/s / λλ
= (4.14 × 10-15 eV s × 3 × 108 m/s) / 1.1 eVλ
= 1.21 × 10-6 m or 1210 nm
Therefore, the longest wavelength that could be used to excite an electron from the valence band of Si to its conduction band is 1210 nm.
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Convert a density of 55.3 lbs/ft3 into g/mL.
The density of 55.3 lbs/ft³ is 0.884 g/mL.
To convert density from pounds per cubic foot (lbs/ft³) to grams per milliliter (g/mL), we can use the following conversion factors:
1 pound = 453.59237 grams
1 foot = 30.48 centimeters (cm)
1 inch = 2.54 centimeters (cm)
1 milliliter (mL) = 1 cubic centimeter (cm³)
First, we convert pounds to grams:
55.3 lbs = 55.3 lbs * 453.59237 g/lb = 25050.364 grams
Next, we convert cubic feet to milliliters:
1 ft³ = (30.48 cm)³ = 28316.8466 cm³
1 cm³ = 1 mL
Finally, we calculate the density in g/mL:
Density = (25050.364 g) / (28316.8466 mL) ≈ 0.884 g/mL
Therefore, the density of 55.3 lbs/ft³ is approximately 0.884 g/mL.
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What is the specific volume of a sample of dry air from Earth’s atmosphere if the pressure is 50 kPa and
the temperature is 263 K?
To determine the specific volume of a sample of dry air, we can use the ideal gas law. The ideal gas law equation is:
PV = nRT
Where:
P = Pressure (in Pascal)
V = Volume (in cubic meters)
n = Number of moles of the gas
R = Ideal gas constant (8.314 J/(mol·K))
T = Temperature (in Kelvin)
To convert the given pressure and temperature to the SI units (Pascal and Kelvin), we have:
Pressure = 50 kPa = 50,000 Pa
Temperature = 263 K
Now, we can rearrange the ideal gas law equation to solve for volume (V):
V = (nRT) / P
Since we're dealing with a specific volume, we need to determine the volume per unit mass. Therefore, we'll consider one kilogram (kg) of dry air.
To calculate the number of moles (n) of air in one kilogram, we need to know the molar mass of dry air. The molar mass of dry air can be approximated as 28.97 g/mol.
1 kg of air = 1000 g
Number of moles (n) = (mass of air) / (molar mass of air)
n = (1000 g) / (28.97 g/mol)
Now we can substitute the values into the equation:
V = [(1000 g) / (28.97 g/mol)] * (8.314 J/(mol·K)) * (263 K) / (50,000 Pa)
V/n ≈ 0.0434 m³/mol is the specific volume of the dry air sample.
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why is the soap derived from coconut oil so soluble
Soap derived from coconut oil is highly soluble in water due to its unique molecular structure and composition.
Coconut oil soap is one of the most commonly used soaps in the world. Coconut oil is transformed into soap using a chemical process known as saponification. Coconut oil is a natural triglyceride, which means it is a compound made up of glycerol and three fatty acid chains. During saponification, the triglyceride is converted into soap, which is a combination of fatty acid salts and glycerol. The fatty acid salts produced by saponification are the key to the high solubility of coconut oil soap.
The polar (hydrophilic) head of the soap molecule interacts with water, while the nonpolar (hydrophobic) tail interacts with grease and oils. Because of the way the soap molecules are organized, they are readily dissolved in water, allowing them to be easily washed away along with dirt, oil, and other impurities. This is what makes coconut oil soap an excellent cleanser and degreaser.Coconut oil soap also creates a lather that is rich and creamy, and it can be used to wash dishes, clothes, and even your skin. Because it is so gentle, coconut oil soap is often used as a natural alternative to commercial soaps that contain harsh chemicals.
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Methane gas (CHA) is burned in an adiabatic combustor with a given percentage of excess air. The pressure and temperature of both the air and fuel is 101 kPa and 298 K respectively. Assume that the mole fractions are 79% nitrogen and 21 % oxygen for air (use M=28.97 kg/kmol and R=0.287 kJ/kg-K) and that water is a vapor in the exhaust. The adiabatic flame temperature is the exhaust temperature that would satisfy the relationship Hproducts = Hreactants.
Given the values below, determine the following:
--Given Values--
m_fuel (kg) = 103
Excess Air = 44%
Determine the air fuel ratio. (kmol_air/kmol_fuel)
The air-fuel ratio is 4.5 kg/kg fuel.
Given Values: m_fuel = 103 kg, Excess Air = 44%.
To determine the air-fuel ratio, we need to determine the mass of air (kg_air) required for the combustion of 1 kg of fuel by stoichiometric combustion and then calculate the total mass of air by adding the excess air to it.
The stoichiometric combustion of Methane can be given as follows:
CH4 + 2(O2+3.76N2) --> CO2 + 2H2O + 7.52N2
The coefficient of oxygen in the above reaction is 2 kmol_air / kmol_CH4.
Now we can calculate the mass of air required for the stoichiometric combustion of 1 kg of fuel as:
mass of air per kmol of CH4 = 2*(32+3.76*28)*0.02897 = 49.92 kg/kmol_
mass of air per kg of CH4 = 49.92/16 = 3.12 kg/kg
We know that 103 kg of fuel is burned;
hence the mass of air required for stoichiometric combustion would be given by:
mass of air required for 103 kg of fuel = 103*3.12 = 321.36 kg/kg of fuel
Now, the mass of air required for the combustion of 1 kg of fuel with 44% excess air would be:m_{air} = m_{stoichometric air}*(1 + EA/100) = 321.36*(1+44/100) = 463.75 kg/kg of fuel
Thus, the air-fuel ratio would be given as:
AFR = (mass of air required for combustion of 1 kg of fuel) / 1 kg of fuel= 463.75/103 = 4.5 kg/kg of fuel
Hence, the air-fuel ratio is 4.5 kg/kg fuel.
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Calculate the pressure exerted by 4 mol of a perfect gas that occupies a volume of 9dm
3
at a temperature of 34
∘
C. Express your answer in units of bar and with no decimals. For carbon dioxide, CO
2
, the value of the second virial coefficient, B, is −142 cm
3
mol
−1
at 273 K. Use the truncated form of the virial equation to calculate the pressure exerted by carbon dioxide gas at this temperature if the molar volume is 299 cm
3
mol
−1
. Report your answer in units of MPa and to two decimal places.
1. Pressure exerted by 4 mol of a perfect gas
Calculate the pressure exerted by 4 mol of a perfect gas that occupies a volume of 9 dm³ at a temperature of 34 °C. Ideal gas law:
PV = nRT
where
P = pressureV = volume (in liters)
n = number of moles
R = gas constant (0.0821 atm L/mol K)
T = temperature (in Kelvin)
Convert 9 dm³ to liters= 9 L, n = 4 mol
Rearrange the ideal gas law to solve for pressure (P):
P = nRT/V
Substitute the given values and convert temperature to Kelvin:
P = (4 mol)(0.0821 atm L/mol K)(307 K)/(9 L)P = 11.2 atm
Therefore, the pressure exerted by 4 mol of a perfect gas that occupies a volume of 9 dm³ at a temperature of 34 °C is 11.2 atm.2. Pressure exerted by carbon dioxide gasFor carbon dioxide, CO2, the value of the second virial coefficient, B, is −142 cm³/mol⁻¹ at 273 K.
Use the truncated form of the virial equation to calculate the pressure exerted by carbon dioxide gas at this temperature if the molar volume is 299 cm³/mol⁻¹.
Truncated virial equation:
P = RT/Vm + BP/(RT)²whereP = pressureR = gas constant (8.314 J/mol K)T = temperature (in Kelvin)Vm = molar volumeB = second virial coefficientP/(RT/Vm) = 1 + BP/(RT)²
Rearrange the equation to solve for pressure (P):
P = RT/(Vm - B)
Substitute the given values and convert molar volume to liters:P = (8.314 J/mol K)(273 K)/(0.299 L/mol - (-0.142 cm³/mol⁻¹))(1 atm/101.3 kPa)(10⁶ Pa/1 atm)P = 7.94 MPaTo two decimal places,
the pressure exerted by carbon dioxide gas at this temperature is 7.94 MPa.
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In this section you simply need to click on the picture for each of the red food coloring solutions whose concetrations are known (2.00 ppm, 4.00 ppm, 6.00 ppm, 8.00 ppm, and 10.00 ppm)) and record the "Absorbance at max" for that solution exactly as given in the picture. You also need to do the same for the unknown solution
In this section, you will be working with different red food coloring solutions with known concentrations (2.00 ppm, 4.00 ppm, 6.00 ppm, 8.00 ppm, and 10.00 ppm).
Your task is to click on each picture representing these solutions and record the "Absorbance at max" value exactly as given in the picture. Additionally, you need to do the same for an unknown solution.
To begin, locate the red food coloring solutions with known concentrations (2.00 ppm, 4.00 ppm, 6.00 ppm, 8.00 ppm, and 10.00 ppm). Click on each picture and note down the "Absorbance at max" value provided in the picture. The "Absorbance at max" is a measure of how much light is absorbed by the solution at its maximum absorption wavelength.
Repeat the same process for the unknown solution. Click on the picture representing the unknown solution and record its "Absorbance at max" value. This value will help determine the concentration of the unknown solution.
By comparing the absorbance values of the known solutions with their corresponding concentrations, you can create a calibration curve. This curve will allow you to determine the concentration of the unknown solution based on its absorbance value. The relationship between concentration and absorbance is typically linear, so you can use this calibration curve to find the concentration of the unknown solution.
Remember to record the "Absorbance at max" values accurately, as they will be crucial for determining the concentration of the unknown solution.
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explain why the measured pressure of a gaseous system under conditions that are very close to those that ewould result in condensation will be lowered than what the ideal gas law would redict
The deviation from the ideal gas law prediction is approximately 150 times when a gas is near its boiling point.
The ideal gas law is dependent on the notion that there is no intermolecular attraction between gas particles. As a result, ideal gas law predictions differ from the actual measured pressure of a gaseous system under conditions that are very close to those that would result in condensation by about 150 times.
According to the ideal gas law, PV=nRT. It implies that the product of the pressure and volume of a gas is proportional to the number of molecules, absolute temperature, and gas constant. But when gas is compressed, its pressure grows and the density of molecules becomes high.
As the volume decreases, the gas particles come closer, and they start to attract each other with intermolecular forces.This intermolecular attraction between gas particles lowers the measured pressure of a gaseous system under conditions that are very close to those that would result in condensation.
Therefore, the ideal gas law cannot explain why there is a reduction in pressure under such circumstances.
The deviation from the ideal gas law prediction is approximately 150 times when a gas is near its boiling point.
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given nickel's position in the periodic table, nickel is more likely to _____ electrons in a chemical reaction and become isoelectronic with _____ because _____.
The nickel is more likely to lose electrons in a chemical reaction and become isoelectronic with argon because it is energetically favorable.
Nickel is more likely to lose electrons in a chemical reaction and become isoelectronic with argon because of the following reason.
The electronic configuration of nickel is [Ar]3d8 4s
2. In its neutral state, nickel has 28 electrons. Argon, which is located in the previous period, has an electronic configuration of [Ne]3s23p
6 and has 18 electrons in its neutral state.
In a chemical reaction, nickel is more likely to lose electrons than to gain electrons to achieve the electronic configuration of argon, which has a stable octet configuration (8 valence electrons) in its outermost energy level, due to the following reason:
Nickel has only two valence electrons in its outermost energy level, whereas argon has eight.
Therefore, it is energetically more favorable for nickel to lose its two valence electrons and become a positively charged ion (Ni2+) than to gain six electrons and become a negatively charged ion (Ni6-).
When nickel loses two electrons, it becomes isoelectronic with argon, as both nickel ion and argon have 18 electrons (Ni2+ = [Ar]3d8).
Therefore, nickel is more likely to lose electrons in a chemical reaction and become isoelectronic with argon because it is energetically favorable.
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A piston maintains nitrogen gas at a constant pressure of 2 bar and a volume of 100 L at 293 K. A 12- V source performs work on the gas with 6 amperes of current for 9 minutes such that the volume then increases to 140% its initial value. Using a constant value for cv of 0.743gKJ determine the magnitude and direction of heat transfer in J.
The magnitude of the heat transfer (Q) is approximately 1822200 J.
To determine the magnitude and direction of heat transfer, we can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat transfer into the system minus the work done by the system:
ΔU = Q - W
In this case, we are given the work done by the gas, which is performed by the 12-V source with 6 amperes of current for 9 minutes. The work done (W) can be calculated using the formula:
W = VΔP
where V is the change in volume and ΔP is the change in pressure.
Given:
Initial volume (V1) = 100 L
Final volume (V2) = 140% of V1 = 140 L
Pressure (P) = 2 bar = 200 kPa
V = V2 - V1 = 140 L - 100 L = 40 L
ΔP = P - P = 0 (since the pressure is constant)
Therefore, W = VΔP = 40 L * 0 = 0 J (no work is done)
Now, we can rearrange the first law of thermodynamics equation to solve for heat transfer (Q):
Q = ΔU + W
Since there is no work done, the equation simplifies to:
Q = ΔU
The change in internal energy (ΔU) can be calculated using the formula:
ΔU = ncvΔT
where n is the number of moles, cv is the specific heat capacity at constant volume, and ΔT is the change in temperature.
To calculate the number of moles, we can use the ideal gas law:
PV = nRT
Rearranging the equation to solve for n:
n = PV / RT
Given:
Pressure (P) = 2 bar = 200 kPa
Volume (V) = 100 L
Temperature (T) = 293 K
R (universal gas constant) = 8.314 J/(mol·K)
n = (200 kPa * 100 L) / (8.314 J/(mol·K) * 293 K) ≈ 8.524 mol
Now, we can calculate the change in internal energy:
ΔU = ncvΔT = (8.524 mol) * (0.743 gKJ/mol·K) * (293 K) = 1822.2 gJ ≈ 1822200 J
Therefore, the magnitude of the heat transfer (Q) is approximately 1822200 J.
Since the work done by the system is zero and the change in internal energy is positive, the heat transfer is in the positive direction, meaning heat is transferred into the system.
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3. The tensile strength of polymethyl methacrylate (PMMA) is about 65 MPa. This assumes the sample has perfect quality (no flaws). Calculate the tensile stress required to fail a tensile coupon that has a 0.5 mm crack on one side (see illustration below). Use the single edge notch plate model to calculate the geometric factor (Y). Assume the fracture toughness of PMMA is 1 MPa m¹/2 O 12.5 mm b 0.5 mm crack
The tensile stress required to fail a tensile coupon that has a 0.5 mm crack on one side is about 14.35 MPa.
Polymethyl methacrylate (PMMA) is a transparent thermoplastic often used as a lightweight or shatter-resistant alternative to glass. It is also used in casting, molding, and extrusion. The tensile strength of PMMA is roughly 65 MPa, but this value changes when a defect is present. The stress required to cause failure can be calculated using the single edge notch plate model to calculate the geometric factor Y.
The fracture toughness of PMMA is 1 MPa m¹/2, and the crack length is 0.5 mm. 12.5 mm is the width of the specimen.For a tensile coupon, the tensile stress required to fail it with a 0.5 mm crack on one side is calculated using the following formula:Stress = (K IC / Y √(πa)) × (b / W)where KIC is the fracture toughness, Y is the geometric factor, a is the crack length, b is the specimen width, and W is the specimen width. For a PMMA coupon with a 0.5 mm crack, a is 0.5/2 = 0.25 mm. Y = 1.12, according to the single edge notch plate model. Substituting the given values, the stress required to fail the coupon is:Stress = (1 MPa m¹/² / 1.12 √(π x 0.25 mm)) × (12.5 mm / 12.5 mm)≈ 14.35 MPa.
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The mass of CO2 is 0.061 kg in a system (with molar mass 44 kg/kmol), occupying a volume of 0.026 m 3 at 1.1 bar is compressed reversibly until the pressure is 5.78 bar. If the molar (universal) gas constant as 8.3145 kJ/kmol K, calculate the the work done on the CO2 (in joules) when the process is isothermal. To 3 d.p.
The work done on CO2 is -742 J
Given parameters:
The mass of CO2 is 0.061 kg
Molar mass of CO2 = 44 kg/kmol
Initial volume = 0.026 m³
Initial pressure, p₁ = 1.1 bar
Final pressure, p₂ = 5.78 bar
Gas constant, R = 8.3145 kJ/kmol K
The process is isothermal.
To find: Work done on the CO2 in joules.
Solution:
As per Boyle's Law,
Pressure * Volume = constant at constant temperature
p₁V₁ = p₂V₂
Therefore,
V₂ = (p₁V₁)/p₂= (1.1 * 0.026) / 5.78= 0.00496 m³
The number of moles of CO2 in the system is given byn = mass of CO2 / Molar mass= 0.061 / 44= 0.00139 kmol Gas equation,
PV = nRT
Where
T is the absolute temperature
At constant temperature,
P₁V₁ = nRT₁
P₂V₂ = nRT₂
Since the process is isothermal,
T₁ = T₂ = TnR(T₂ - T₁) * ln
(V₂/V₁)= 8.3145 * (T₂ - T₁) * ln
(V₂/V₁)Joules = 8.3145 * (273.15) * ln
(0.00496/0.026)= -741.955 J≈ -742 J
(Answer)Therefore, the work done on CO2 is -742 J when the process is isothermal.
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