Respuesta :
Answer:
a) [tex]\eta=0.9998\ or\ 99.98\%[/tex]
b) [tex]\eta_c=0.7490\ or\ 74.9\% [/tex]
[tex]\rm \eta>\eta_c\ is\ not\ possible[/tex]
Explanation:
Given:
- temperature of source reservoir, [tex]T_H=660+273=933\ K[/tex]
- temperature of sink reservoir, [tex]T_L=-38.9+273=234.1\ K[/tex]
- quantity of aluminium frozen by the engine during 1 cycle, [tex]m_a=0.003\ kg[/tex]
- quantity of mercury melted by the engine during 1 cycle, [tex]m_m=0.045\ kg[/tex]
- latent heat of fusion of aluminium, [tex]L_a=3.97\times 10^5\ J.kg^{-1}[/tex]
- latent heat of fusion of mercury, [tex]L_m=1.18\times 10^4\ J.kg^{-1}[/tex]
a)
Heat absorbed by the engine:
[tex]Q_H=m_a.L_a[/tex]
[tex]Q_H=0.045\times (3.97\times 10^{5})[/tex]
[tex]Q_H=178650\ J[/tex]
Heat rejected by the engine:
[tex]Q_L=m_m.L_m[/tex]
[tex]Q_L=0.003\times (1.18\times 10^{4})[/tex]
[tex]Q_L=35.4\ J[/tex]
Now the efficiency of the engine:
[tex]\eta=\frac{Q_H-Q_L}{Q_H}[/tex]
[tex]\eta=\frac{178650-35.4}{178650}[/tex]
[tex]\eta=0.9998\ or\ 99.98\%[/tex]
b)
Now the Carnot efficiency of the engine:
[tex]\eta_c=1-\frac{T_L}{T_H}[/tex]
[tex]\eta_c=1-\frac{234.1}{933}[/tex]
[tex]\eta_c=0.7490\ or\ 74.9\% [/tex]
[tex]\rm \eta>\eta_c\ is\ not\ possible[/tex]
