Answer:
The temperature of the steam during the heat rejection process is 42.5°C
Explanation:
Given the data in the question;
the maximum temperature T
in the cycle is twice the minimum absolute temperature T
in the cycle
T = 0.5T
now, we find the efficiency of the Carnot cycle engine
η
= 1 - T/T
η = 1 - T/0.5T
η = 0.5
the efficiency of the Carnot heat engine can be expressed as;
η = 1 - W
/Q
where W is net work done, Q is is the heat supplied
we substitute
0.5 = 60 / Q
Q = 60 / 0.5
Q = 120 kJ
Now, we apply the first law of thermodynamics to the system
W = Q - Q
60 = 120 - Q
Q = 60 kJ
now, the amount of heat rejection per kg of steam is;
q = Q/m
we substitute
q = 60/0.025
q = 2400 kJ/kg
which means for 1 kilogram of conversion of saturated vapor to saturated liquid , it takes 2400 kJ/kg of heat ( enthalpy of vaporization)
q = h
= 2400 kJ/kg
now, at h = 2400 kJ/kg from saturated water tables;
T = 40 + ( 45 - 40 ) ( 
)
T = 40 + (5) × (0.5)
T = 40 + 2.5
T = 42.5°C
Therefore, The temperature of the steam during the heat rejection process is 42.5°C
Answer:
false
Explanation:
I think I am right with this
I think the answer is false
The shock wave from the meteoroid in the lower atmosphere has a Mach angle of 0.948°.
(a) The meteoroid's speed 
Air sound wave speed 
Speed of the shock wave in Mach 
Hence, 0.948° is the Mach angle of the shock wave from the meteoroid in the lower atmosphere.
<h3>What is the speed of the meteoroid?</h3>
A meteoroid's speed can be loosely broken down into three categories: slow, medium, and fast.
- Slow meteors move around the sun at a leisurely pace of about 32 kilometers per second (20 miles per second).
- Medium-speed meteors travel around the sun at approximately 50 kilometers per second (30 miles per second),
- while fast meteors zoom past at over 120 kilometers per second (75 miles per second)!
To learn more about meteoroid, visit:
brainly.com/question/1939309
#SPJ4
It is a scaler because it’s only fully describes by a magnitude and a numerical alone
