Answer
(a) $6.78\%$
Work Step by Step
We know that
$e_{max}=1-\frac{T_c}{T_h}$
We plug in the known values to obtain:
$e_{max}=1-\frac{273+2}{273+22}$
$e_{max}=1-\frac{275}{295}$
$e_{max}=0.06779$
$e_{max}=6.78\%$
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