The efficiency of the device is 30 %
Explanation:
The efficiency of a heat engine is given by:
[tex]\eta = \frac{W}{Q_{in}}[/tex]
where
W is the work done by the engine
[tex]Q_{in}[/tex] is the heat in input to the engine
For the device in this problem, we have:
W = 120 J is the work done
[tex]Q_{in} = 400 J[/tex] is the heat in input
Substituting, we find the efficiency:
[tex]\eta=\frac{120}{400}=0.30[/tex]
which corresponds to an efficiency of 30%.
Learn more about work:
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