Answer:
C. [tex]W = 115.12\,Btu[/tex]
Explanation:
Thermodynamically speaking, a Carnot engine represents an entirely reversible thermal process and its energy efficiency represents the maximum theoretical efficiency that thermal machines can reach. The efficiency of the ideal thermal process ([tex]\eta[/tex]), no unit, is:
[tex]\eta = \left(1-\frac{T_{L}}{T_{H}} \right)[/tex] (1)
Where:
[tex]T_{L}[/tex] - Temperature of the cold reservoir, measured in Rankine.
[tex]T_{H}[/tex] - Temperature of the hot reservoir, measured in Rankine.
If we know that [tex]T_{H} = 1809.67\,R[/tex] and [tex]T_{L} = 584.67\,R[/tex], then the energy efficiency of the ideal thermal process is:
[tex]\eta = 0.678[/tex]
By First Law of Thermodynamics, we calculate the work output:
[tex]W = Q_{H}-Q_{L}[/tex]
[tex]W = \frac{W}{\eta} -Q_{L}[/tex] (By definition of efficiency)
[tex]Q_{L} = \frac{W}{\eta}-W[/tex]
[tex]Q_{L} = \left(\frac{1}{\eta}-1 \right)\cdot W[/tex](2)
Where:
[tex]Q_{H}[/tex] - Heat received by the engine, measured in Btu.
[tex]Q_{L}[/tex] - Heat rejected by the engine, measured in Btu.
[tex]W[/tex] - Work output, measured in Btu.
If we know that [tex]\eta = 0.678[/tex] and [tex]Q_{L} = 55\,Btu[/tex], then the work output of the Carnot engine is:
[tex]W = \frac{Q_{L}}{\frac{1}{\eta}-1 }[/tex]
[tex]W = 115.807\,Btu[/tex]
The work output of the Carnot engine is 115.807 Btu. (Answer: C)
