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[tex]_H[/tex] in the cycle is twice the minimum absolute temperature T[tex]_L[/tex] in the cycle
T[tex]_H[/tex] = 0.5T[tex]_L[/tex]
now, we find the efficiency of the Carnot cycle engine
η[tex]_{th[/tex] = 1 - T[tex]_L[/tex]/T[tex]_H[/tex]
η[tex]_{th[/tex] = 1 - T[tex]_L[/tex]/0.5T[tex]_L[/tex]
η[tex]_{th[/tex] = 0.5
the efficiency of the Carnot heat engine can be expressed as;
η[tex]_{th[/tex] = 1 - W[tex]_{net[/tex]/Q[tex]_H[/tex]
where W[tex]_{net[/tex] is net work done, Q[tex]_H[/tex] is is the heat supplied
we substitute
0.5 = 60 / Q[tex]_H[/tex]
Q[tex]_H[/tex] = 60 / 0.5
Q[tex]_H[/tex] = 120 kJ
Now, we apply the first law of thermodynamics to the system
W[tex]_{net[/tex] = Q[tex]_H[/tex] - Q[tex]_L[/tex]
60 = 120 - Q[tex]_L[/tex]
Q[tex]_L[/tex] = 60 kJ
now, the amount of heat rejection per kg of steam is;
q[tex]_L[/tex] = Q[tex]_L[/tex]/m
we substitute
q[tex]_L[/tex] = 60/0.025
q[tex]_L[/tex] = 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[tex]_L[/tex] = h[tex]_{fg[/tex] = 2400 kJ/kg
now, at h[tex]_{fg[/tex] = 2400 kJ/kg from saturated water tables;
T[tex]_L[/tex] = 40 + ( 45 - 40 ) ( [tex]\frac{2400-2406.0}{2394.0-2406.0}\\}[/tex] )
T[tex]_L[/tex] = 40 + (5) × (0.5)
T[tex]_L[/tex] = 40 + 2.5
T[tex]_L[/tex] = 42.5°C
Therefore, The temperature of the steam during the heat rejection process is 42.5°C
