The efficiency is 0.22 (22%)
Explanation:
The efficency of an engine can be written as:
[tex]\eta = 1-\frac{Q_{out}}{Q_{in}}[/tex]
where
[tex]\eta[/tex] is the efficiency
[tex]Q_{out}[/tex] is the heat rejected
[tex]Q_{in}[/tex] is the heat taken in
For the engine in this problem, we have:
[tex]Q_{in} = 2400 J[/tex] is the heat in input
[tex]Q_{out}=1880 J[/tex] is the heat rejected
Solving the equation, we find the efficiency:
[tex]\eta = 1-\frac{1880}{2400}=0.22[/tex]
So, the efficiency is 22%.
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