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
<h2>
<u>Required</u><u> </u><u>Answer</u><u>:</u></h2>
The body will <u>stay at rest </u>(Option D). It is because a force of magnitude 50 N is pulled towards left and another force is pulling it towards right with same magnitude 50 N. So, the direction of force is opposite and magnitude is same i.e. 50 N. So, they will cancel each other and net force is 0. Hence, there would be no acceleration.
- Option A - Showing acceleration
- Option B - Showing acceleration
- Option C - Change of direction due to Net force
Hence, these options are incorrect because they are only possible when net external force is non-zero. Staying at rest i.e. Option D means there is no motion and hence no acceleration, this shows that net force is 0
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