Answer:
Source 2.
Explanation:
The efficiency of the ideal reversible heat engine is given by the Carnot's power cycle:
[tex]\eta_{th} = 1 - \frac{T_{L}}{T_{H}}[/tex]
Where:
[tex]T_{L}[/tex] - Temperature of the cold reservoir, in K.
[tex]T_{H}[/tex] - Temperature of the hot reservoir, in K.
The thermal efficiencies are, respectively:
Source 1
[tex]\eta_{th} = 1 - \frac{311.15\,K}{583.15\,K}[/tex]
[tex]\eta_{th} = 0.466 \,(46.6\,\%)[/tex]
Source 2
[tex]\eta_{th} = 1 - \frac{311.15\,K}{338.15\,K}[/tex]
[tex]\eta_{th} = 0.0798 \,(7.98\,\%)[/tex]
The power produced by each device is presented below:
Source 1
[tex]\dot W = (0.466)\cdot (11000\,\frac{kJ}{min})\cdot (\frac{1\,min}{60\,s} )[/tex]
[tex]\dot W = 85.433\,kW[/tex]
Source 2
[tex]\dot W = (0.0798)\cdot (110000\,\frac{kJ}{min})\cdot (\frac{1\,min}{60\,s} )[/tex]
[tex]\dot W = 146.3\,kW[/tex]
The source 2 produces the largest amount of power.
