Answer:
D) ΔU = -153.43 kJ and ΔH = -146.00 kJ
Explanation:
Given;
heat energy released by the exothermic reaction, ΔH = -146 kJ
number of gas mol, n = 3 mol
temperature of the gas, T = 298 K
Apply first law of thermodynamic
Change in the internal energy of the system, ΔU;
ΔU = ΔH- nRT
where;
R is gas constant = 8.314 J/mol.K
ΔU = -146kJ - (3 x 8.314 x 298)
ΔU = -146kJ - 7433 J
ΔU = -146kJ - 7.433 kJ
ΔU = -153.43 kJ
Therefore, the enthalpy change of the reaction ΔH is -146 kJ and change in the internal energy of the system is -153.43 kJ
D) ΔU = -153.43 kJ and ΔH = -146.00 kJ
Answer:
1. To bring down the pressure of the refrigerant
2. To meet up with the load to be refrigerated (the amount of heat to be evacuated)
Explanation:
1. To bring down the pressure of the refrigerant
The high pressure of the refrigerant coming from the condenser require reduction to enable vaporization in the evaporator at the proper temperature
The throttling valve as a small aperture through which the refrigerant flows that lowers the pressure of the refrigerant to a point at which the refrigerant vaporize of which the refrigerant then passes into the evaporator in a partly as liquid and vapor at a low temperature and pressure
2. To meet up with the load to be refrigerated (the amount of heat to be evacuated)
The throttling valve allows more refrigerant to flow through it when there is an increased load at a higher temperature to be refrigerated
Similarly, in a condition of reduced refrigeration load, hence, a lesser amount of heat to be evacuated, the throttling valve restricts the amount of flow of the refrigerant through it.
Answer:
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Answer:
a) 53 MPa, 14.87 degree
b) 60.5 MPa
Average shear = -7.5 MPa
Explanation:
Given
A = 45
B = -60
C = 30
a) stress P1 = (A+B)/2 + Sqrt ({(A-B)/2}^2 + C)
Substituting the given values, we get -
P1 = (45-60)/2 + Sqrt ({(45-(-60))/2}^2 + 30)
P1 = 53 MPa
Likewise P2 = (A+B)/2 - Sqrt ({(A-B)/2}^2 + C)
Substituting the given values, we get -
P1 = (45-60)/2 - Sqrt ({(45-(-60))/2}^2 + 30)
P1 = -68 MPa
Tan 2a = C/{(A-B)/2}
Tan 2a = 30/(45+60)/2
a = 14.87 degree
Principal stress
p1 = (45+60)/2 + (45-60)/2 cos 2a + 30 sin2a = 53 MPa
b) Shear stress in plane
Sqrt ({(45-(-60))/2}^2 + 30) = 60.5 MPa
Average = (45-(-60))/2 = -7.5 MPa
Based on the percent moisture content of the dried product, the mass of dried casein produced os 852.3 kg.
<h3>What is the mass of casein in wet casein?</h3>
The mass of casein in 1000 Kg of wet casein is 75% 1000 kg = 750 Kg
Mass of water 250 kg
The mass of casein is constant while the moisture content can be changed.
At 12% moisture content;
750 kg = 88%%
100 % = 100 ×750/88 = 852.27 kg
Therefore, the mass of dried casein produced os 852.3 kg.
Learn more about mass at: brainly.com/question/24658038
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