A 0.5-m3 rigid tank contains refrigerant-134a initially at 200 kPa and 40 percent quality. Heat is transferred now to the refrigerant from a source at 358C until the pressure rises to 400 kPa. Determine (a) the entropy change of the refrigerant, (b) the entropy change of the heat source, and (c) the total entropy change for this process.
Answers
Answer:
(a)
(b)
(c)
Explanation:
Hello,
(a) In this case, such refrigerant, we can notice that at the given conditions, the initial entropy from property tables (Cengel 7th ed) is:
Now, for the final condition, we first need to compute the initial specific volume as it remains the same (rigid tank) after the thermodynamic process:
Then, at 400 kPa we evaluate the given volume that is also between the liquid and vapor specific volume, thus, we calculate the quality at the end of the process:
With it, we compute the final entropy:
Finally, entropy change for the refrigerant turns out:
(b) In this case, by using the first law of thermodynamics we compute the acquired heat by the refrigerant from the heat source by computing the initial and final internal energy respectively (no work is done):
Hence:
Finally, the entropy change of the heat source (which release the heat, therefore it is negative):
(c) Then, the total entropy change or the entropy generation for the process is:
Which has thermodynamic agreement as it is positive
Regards.
Final answer:
The entropy changes in this process can be partially calculated using principles from thermodynamics. However, without the exact heat transfer, not all values can be determined.
Explanation:
The calculation of the entropy change in this thermodynamic process involves principles from thermodynamics and requires steps to determine the initial and final states of the refrigerant. First, we would need to find the entropy at the initial and final states using the refrigerant properties table for refrigerant-134a and the provided information (200 kPa and 40% quality initially, 400 kPa finally). The entropy change of the refrigerant is the difference between the final and initial entropy.
Next, the entropy change of the heat source is calculated as the heat transfer divided by the absolute temperature of the source. However, the problem does not provide the amount of heat transferred from the source, making it impossible to determine this value directly.
Finally, in an isolated system, the total entropy change of the process is the sum of the entropy changes of the refrigerant and the heat source. Here, the precise values cannot be calculated due to a lack of specific data including exact heat transfer.
Learn more about Entropy Change here:
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Answers
Explanation:
1 Pascal = 1 N/m²
To convert Pa to lbf/ft²
So, the conversion of N to pound force (lbf) is shown below as:
1 N = 0.224809 pound force (lbf)
The conversion of m² to ft² is shown below:
1 m² = 10.7639 ft²
So,
[tex]1\ Pa=\frac {1\ N}{1\ m^2}=\frac {0.224809\ lbf}{10.7639\ ft^2}
1 Pa = 0.02089 lbf / ft²
Hence proved.
The chemist uses 11 g of CaF2 and an excess of H2SO4, and the reaction produces 2.2 g of HF.
(a) Calculate the theoretical yield of HF.
(b) Calculate the percent yield of HF.
Answers
Answer:
39.3%
Explanation:
CaF2 + H2SO4 --> CaSO4 + 2HF
We must first determine the limiting reactant, the limiting reactant is the reactant that yields the least number of moles of products. The question explicitly says that H2SO4 is in excess so CaF2 is the limiting reactant hence:
For CaF2;
Number of moles reacted= mass/molar mass
Molar mass of CaF2= 78.07 g/mol
Number of moles reacted= 11g/78.07 g/mol = 0.14 moles of Calcium flouride
Since 1 mole of calcium fluoride yields two moles of 2 moles hydrogen fluoride
0.14 moles of calcium fluoride will yield 0.14×2= 0.28 moles of hydrogen fluoride
Mass of hydrogen fluoride formed (theoretical yield) = number of moles× molar mass
Molar mass of hydrogen fluoride= 20.01 g/mol
Mass of HF= 0.28 moles × 20.01 g/mol= 5.6 g ( theoretical yield of HF)
Actual yield of HF was given in the question as 2.2g
% yield of HF= actual yield/ theoretical yield ×100
%yield of HF= 2.2/5.6 ×100
% yield of HF= 39.3%
Answers
Answer: The empirical formula of compound is .
Explanation:
Mass of Sample= 0.702 g
Mass of = 1.4 g
Mass of = 0.86 g
Mass of = 0.478 g
First we have to calculate moles of,
and
formed.
1. Moles of
Now , Moles of carbon == Moles of = 0.032
2. Moles of =0.048mol
Now , Moles of hydrogen = Moles of
=
3. Moles of mol
Now , Moles of silicon = Moles of = 0.008 moles
Therefore, the ratio of number of moles of C : H : Si is = 0.032 : 0.096 : 0.008
For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C=
For H =
For Si=
Thus, C: H: Si = 4 : 12 : 1
The simplest ratio represent empirical formula.
Hence, the empirical formula of compound is .
Answers
Answer:
The correct answer is D.
Explanation:
Water can evaporate, and if it does, the density decreases
Answers
Answer:
Go ahead and plug in the percentages and time to find the answer.
Explanation:
The amount of a substance with half-life h, that remains after time t is 0.5t/h
Since 26% has decomposed, 74% remains.
So .74 = 0.580/h
ln .74 = (80/h) ln 0.5
h/80 = ln 0.5 / ln .74
h = 80 ln 0.5 / ln .74
h = 184.16 minutes
completely neutralize 150.0 milliliters of 0.100 M
NaOH(aq)?
A. 62.5 mL
B. 125 ml
C.
180. mL
D. 360. mL
Answers
Answer:
B) 125 mL
Explanation:
M1V1=M2V2
(0.120M)(x)=(150.0 mL)(0.100M)
x= 125 mL
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