Answer:
The power output of this engine is [tex]P = 17.5 W[/tex]
The the maximum (Carnot) efficiency is [tex]\eta_c = 0.7424[/tex]
The actual efficiency of this engine is [tex]\eta _a = 0.46[/tex]
Explanation:
From the question we are told that
The temperature of the hot reservoir is [tex]T_h = 1250 \ K[/tex]
The temperature of the cold reservoir is [tex]T_c = 322 \ K[/tex]
The energy absorbed from the hot reservoir is [tex]E_h = 1.37 *10^{5} \ J[/tex]
The energy exhausts into cold reservoir is [tex]E_c = 7.4 *10^{4} J[/tex]
The power output is mathematically represented as
[tex]P = \frac{W}{t}[/tex]
Where t is the time taken which we will assume to be 1 hour = 3600 s
W is the workdone which is mathematically represented as
[tex]W = E_h -E_c[/tex]
substituting values
[tex]W = 63000 J[/tex]
So
[tex]P = \frac{63000}{3600}[/tex]
[tex]P = 17.5 W[/tex]
The Carnot efficiency is mathematically represented as
[tex]\eta_c = 1 - \frac{T_c}{T_h}[/tex]
[tex]\eta_c = 1 - \frac{322}{1250}[/tex]
[tex]\eta_c = 0.7424[/tex]
The actual efficiency is mathematically represented as
[tex]\eta _a = \frac{W}{E_h}[/tex]
substituting values
[tex]\eta _a = \frac{63000}{1.37*10^{5}}[/tex]
[tex]\eta _a = 0.46[/tex]
