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:
The temperature of the strip as it exits the furnace is 819.15 °C
Explanation:
The characteristic length of the strip is given by;
The Biot number is given as;
< 0.1, thus apply lumped system approximation to determine the constant time for the process;
The time for the heating process is given as;
Apply the lumped system approximation relation to determine the temperature of the strip as it exits the furnace;
Therefore, the temperature of the strip as it exits the furnace is 819.15 °C
The rotor is situated inside the stator and is mounted on the AC motor's shaft. It is the rotating part of the AC motor. And while we know this, the major function of the rotor and the stator is helping the motor shaft rotate.
Answer:
P=- 88.41 KW
Negative sign indicates that power is given to the system.
Explanation:
Given that
P₂=500 KPa
T₂=130°C
V₂=100 m/s
mass flow rate ,m= 0.8 kg/s
Lets take inlet condition for air
T₁=25°C
P₁=100 KPa
V₁=0 m/s
We know that
Heat capacity for air Cp=1.005 KJ/kg.k
We know that for air change in enthalpy only depends only on temperature
Now from first law for open system
W=-110.52 KJ/kg
Shaft power P = m .W
P = -110.52 x 0.8
P=- 88.41 KW
Negative sign indicates that power is given to the system.