Answer:
[tex]\eta=0.303[/tex]
Explanation:
Given that
Temperature of the hot reservoir ,T₁ = 970 K
Temperature of the cold reservoir ,T₂ = 480 K
We know that only Carnot engine is the ideal engine which gives us the maximum power.The efficiency of Carnot engine is given as
[tex]\eta_{max}=1-\dfrac{T_2}{T_1}[/tex]
[tex]\eta_{max}=1-\dfrac{480}{970}[/tex]
[tex]\eta_{max}=0.505[/tex]
Therefore the efficiency of the given engine will be
[tex]\eta=\dfrac{3}{5}\eta_{max}[/tex]
[tex]\eta=\dfrac{3}{5}\times 0.505[/tex]
[tex]\eta=0.303[/tex]
