Answer:
See explaination
Explanation:
LM358 is the useful IC which works as buffer. It enables circuit to remove overloading effect on each other. Image is in attachment.
We can define a light-emitting diode (LED) as a semiconductor light source that emits light when current flows through it. Electrons in the semiconductor recombine with electron holes, releasing energy in the form of photons
See attached file for detailed solution of the given problem.
Answer:
The amount of heat transferred to the air is 340.24 kJ
Explanation:
From P-V diagram,
Initial temperature T1 = 27°C
Initial pressure P1 = 100 kPa
final pressure P3 = P2 = 300 kPa
volume at point 2, V2 = V1 = 0.4 m³
final temperature T2 = T3 = 1200 K
To determine the final pressure V3, use ideal gas equation
PV = mRT
Where R is the specific gas constant = 0.2870 KPa m³ kg K
But,
from initial condition, mass m = PV/RT
m = (P1*V1)/R*T1
T1 = 27+273 = 300K
m = (100*0.4)/(0.2870*300) = 0.4646 kg
Then;
Final volume V3 = mRT3/P3
V3 = (0.4646*0.2870*1200)/300
V3 = 0.5334 m³
Total work done W is determined where there is volume change which from point 2 to 3.
W = P3*(V3-V2)
W = 300*(0.5334-0.4) = 40.02 kJ
To get the internal energy, the heat capacity at room temperature Cv is 0.718 kJ/kg K
∆U = m*Cv*(T2-T1)
∆U = 0.4646*0.718(1200-300)
∆U = 300.22 kJ
The heat transfer Q = W + ∆U
Q = 40.02 + 300.22 = 340.24 kJ
Determine the amount of heat transferred to the air, in kJ, while increasing the temperature to 1200 K is 340.24 kJ
The attached file shows the Pressure - Volume relationship (P -V graph)
