necessaryAnswer:
a)[tex]n=0,5175[/tex]
b)[tex]Q_{c}=379.37J[/tex]
c)[tex]dS_{h}=0.3234J/K[/tex]
d)[tex]dS_{c}=1.017J/K[/tex]
e) 3787.6 cycles
Explanation:
As this engine is ideal, the efficiency depends only on the temperatura of the reservoirs(in kelvin scale):
[tex]n=1-\frac{T_{c} }{T_{h}}=1-\frac{373}{773}=0.5175[/tex]
the efficiency is also related to the heat exchanged between the reservoirs:
[tex]n=1-\frac{Q_{c} }{Q_{h}}[/tex]
Solving for [tex]Q_{c}[/tex] and having in mind that [tex]Q_{h}=250J[/tex]:
[tex]Q_{c}=(n+1)*Q_{h}=1.5175*250J=379.37J[/tex]
The change of entropy for the hot reservoir:
[tex]dS_{h}=\frac{250J}{773K} =0.3234J/K[/tex]
Similar for the cold reservoir:
[tex]dS_{c}=\frac{379.37J}{373K} =1.017J/K[/tex]
From the efficiency we can calculate the work that the engine can make for a cycle:
[tex]n=\frac{W}{Q_{h}}[/tex]
solving for W:
[tex]W=n*Q_{h}=0.5175*250J=129.37J[/tex]
[tex]E=m*g*h=500kg*9.8m/s^{2} *100m=490000J[/tex]
It will take the engine:
[tex]\frac{490000J}{129.37J/cycle}=3787.6cycles[/tex]