Answer:
70 %
9.33331 kW
Explanation:
Q = Heat supplied = 800 kJ/min
[tex]T_h[/tex] = Hot reservoir = 1000 K
[tex]T_c[/tex] = Cold reservoir = 300 K
Converting kJ/min to kW
[tex]800\times \frac{1}{60}=13.3333\ kW[/tex]
Thermal efficiency
[tex]\eta=1-\frac{T_c}{T_h}\\\Rightarrow \eta=1-\frac{300}{1000}\\\Rightarrow \eta=0.7[/tex]
The thermal efficiency is 70%
Efficiency
[tex]\eta=\frac{W}{Q}\\\Rightarrow W=\eta Q\\\Rightarrow W=0.7\times 13.3333\\\Rightarrow W=9.33331\ kW[/tex]
The power output of the engine is 9.33331 kW
