Answer :
Answer:
[tex]\eta=0.0512\ or\ 5.12\%[/tex]
Explanation:
Heat engine is device operating in a continuous cycle between two reservoirs such that it transfers heat from a high temperature reservoir to a low temperature reservoir giving some work output in synchronization to Kelvin-Plank statement of the second law of thermodynamics.
Given that:
- temperature of higher temperature reservoir, [tex]T_H=20+273=293\ K[/tex]
- temperature of lower temperature reservoir, [tex]T_L=5+273=278\ K[/tex]
For maximum efficiency:
[tex]\eta=1-\frac{T_L}{T_H}[/tex]
were:
[tex]T_H\ \&\ T_L[/tex] are the temperatures of higher temperature reservoir and lower temperature reservoir respectively.
[tex]\eta=1-\frac{278}{293}[/tex]
[tex]\eta=0.0512\ or\ 5.12\%[/tex]
