Answer in Molecular Physics | Thermodynamics for Rhyna Mae Taghap #144882
February 21st, 2023
“textsf{Thermal Efficiency }(eta) = dfrac{W}{Q_H}”
(i) W = 5000J – 4500J = 500J
“Q_H” = 5000J
“eta = dfrac{500J}{5000J} = 0.1”
(ii) W = 2000J
“Q_H” = 25000J
“eta = dfrac{2000J}{25000J} = 0.08”
(iii) W = 400J
“Q_H” = 400J + 2800J = 3200J
“eta = dfrac{400J}{3200J} = 0.125”
“therefore” Engine number (iii) with 0.125 thermal efficiency has the highest efficiency, and Engine number (ii) with 0.08 thermal efficiency has the lowest efficiency.