15.4 Carnot’s perfect heat engine: the second law of thermodynamics  (Page 4/7)

$\text{403}\text{º}\text{C}$

(a) $2\text{44}\text{º}\text{C}$

(b) $\text{477}\text{º}\text{C}$

(c)Yes, sinceautomobiles enginescannot gettoo hotwithout overheating,their efficiencyis limited.

(a) ${\mathit{\text{Eff}}}_{\text{1}}=1-\frac{T_{\text{c,1}}}{T_{\text{h,1}}}=1-\frac{\text{543 K}}{\text{723 K}}=0\text{.}\text{249}\text{or}\text{24}\text{.}\mathrm{9\%}$

(b) ${\mathit{\text{Eff}}}_2=1-\frac{\text{423 K}}{\text{543 K}}=0\text{.}\text{221}\text{or}\text{22}\text{.}\mathrm{1\%}$

(c) $\begin{array}{l}{\mathit{\text{Eff}}}_1=1-\frac{T_{\text{c,1}}}{T_{\text{h,1}}}\Rightarrow T_{\text{c,1}}=T_{\text{h,1}}\left(1,-,{\mathit{\text{eff}}}_1\right)\end{array}$ $\begin{array}{l}\text{similarly,}{T}_{\text{c,2}}=T_{\text{h,2}}\left(1-{\mathit{\text{Eff}}}_2\right)\end{array}$ $\begin{array}{l}\text{using}{T}_{\text{h,2}}=T_{\text{c,1}}\text{in}\text{above}\text{equation}\text{gives}\end{array}$ $\begin{array}{l}{T}_{\text{c,2}}=T_{\text{h,1}}\left(1-{\mathit{\text{Eff}}}_1\right)\left(1-{\mathit{\text{Eff}}}_2\right)\equiv T_{\text{h,1}}\left(1-{\mathrm{Eff}}_{\text{overall}}\right)\\ \therefore \left(1-{\mathrm{Eff}}_{\text{overall}}\right)=\left(1-{\mathit{\text{Eff}}}_1\right)\left(1-{\mathit{\text{Eff}}}_2\right)\\ {\mathrm{Eff}}_{\text{overall}}=1-\left(1-0\text{.}\text{249}\right)\left(1-0\text{.}\text{221}\right)=\text{41}\text{.}\mathrm{5\%}\end{array}$

(d) ${\text{Eff}}_{\text{overall}}=1-\frac{\text{423 K}}{\text{723 K}}=0\text{.}\text{415}\text{or}\text{41}\text{.}\mathrm{5\%}$

The heat transfer to the cold reservoir is $Q_{\text{c}}=Q_{\text{h}}-W=\text{25}\text{kJ}-\text{12}\text{kJ}=\text{13}\text{kJ}$ , so the efficiency is $\mathit{Eff}=1-\frac{Q_{\text{c}}}{Q_{\text{h}}}=1-\frac{\text{13}\text{kJ}}{\text{25}\text{kJ}}=0\text{.}\text{48}$ . The Carnot efficiency is ${\mathit{\text{Eff}}}_{\text{C}}=1-\frac{T_{\text{c}}}{T_{\text{h}}}=1-\frac{\text{300}\text{K}}{\text{600}\text{K}}=0\text{.}\text{50}$ . The actual efficiency is 96% of the Carnot efficiency, which is much higher than the best-ever achieved of about 70%, so her scheme is likely to be fraudulent.

(a) $\text{–56.3ºC}$

(b) The temperature is too cold for the output of a steam engine (the local environment). It is below the freezing point of water.

(c) The assumed efficiency is too high.