Answer:
Heat
Explanation:
Carnot cycle:
Carnot cycle is the ideal cycle for all working engine .Carnot cycle all processes are reversible.It have fore process Out of two are constant temperature process and other two are isentropic process(reversible adiabatic).
We know that area under T-s diagram represents the heat.
So 
From cycle we can say that
Answer: a 8143.71 kJ/kg
b 393.15 K
Explanation:
This system is an isobaric process in which there is no change in pressure a quasistatic process where a pressure distribution exists
a since no change in pressure =0 the system does work thus
FOR HELIUM properties in standard thermodynamic chart
cv = 3.1 kJ/kgK
M = Molar mass = 4 kg/kmol
R = Universal gas constant = 8.314 kJ/kg K
cp ≈ cv +R /M = 3.1 + 8.314 /4 = 5.1785 kJ/kgK
Cp = cp * M = 5.1785 kJ/kgK * 4 kg/kmol = 20.714 kJ/kgkmol
T = 120 °C to Kelvin = 120 + 273.15k = 393.15 K
W =n Cp ΔT = 1 kmol * 20.714 kJ/kg kmol* 393.15 K = 8143.71 kJ/kg
b convert T °C = T K thus 120 + 273.15 K = 393.15 K
P₁/T₁ = P₂/T₂
200 kPa/ 393.15 K = 200 kPa/T₂
T₂ = 200 kPa * 393.15 K/ 200 kPa = 393.15 K or 120 k
Answer: At time 18.33 seconds it will have moved 500 meters.
Explanation:
Since the acceleration of the car is a linear function of time it can be written as a function of time as
Integrating both sides we get
Now since car starts from rest thus at time t = 0 ; v=0 thus c=0
again integrating with respect to time we get
Now let us assume that car starts from origin thus D=0
thus in the first 15 seconds it covers a distance of
Thus the remaining 125 meters will be covered with a constant speed of
in time equalling 
Thus the total time it requires equals 15+3.33 seconds
t=18.33 seconds
Answer:
The engine's thermal efficiency is 0.32
Explanation:
Thermal efficiency = work done ÷ quantity of heat supplied
Work done = 210 J
Quantity of heat supplied = work done + waste heat = 210 + 440 = 650 J
Thermal efficiency = 210 ÷ 650 = 0.32
