(A) Air temperature is at 350 K, Heat transfer= dQ= 1050 J
Entropy generation,
"S_{gen}= \\Delta S_{sys}+ \\frac{dQ}{T}"
Here, in case of ambient condititon change in entropy will be zero
"S_{gen}=\\frac{1050}{350}=3 \\frac{J}{K}"
(B) As per the second case we know that for carnot heat engine
"\\frac{Q_1}{Q_2}= \\frac{T_1}{T2}"
"\\frac{1050}{Q_2}= \\frac{350.001}{349.999}"
Q2=1049.99 J
and entropy generation
"S_{gen}=\\frac{1049.999}{349.999}= 2.999 J\/K"
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