Respuesta :
Answer:
a. [tex]P=18.61\ W[/tex]
b. [tex]\eta_c=0.6976=69.76\%[/tex]
c. [tex]\eta=0.4718=47.18\%[/tex]
Explanation:
Given:
- temperature of the hotter reservoir, [tex]T_H=1250\ K[/tex]
- temperature of the colder reservoir, [tex]T_C=378\ K[/tex]
- heat absorbed by the engine, [tex]Q_H=1.42\times 10^{5}\ J[/tex]
- heat rejected to the cold reservoir, [tex]Q_L=7.5\times 10^4\ J[/tex]
- time duration of the energy transfer, [tex]t=1\ hr=3600\ s[/tex]
Now the work done by the engine:
Using energy conservation,
[tex]W=Q_H-Q_L[/tex]
[tex]W=14.2\times 10^4-7.5\times 10^4[/tex]
[tex]W=6.7\times 10^4\ J[/tex]
a.
Hence the power output:
[tex]P=\frac{W}{t}[/tex]
[tex]P=\frac{6.7\times 10^4}{3600}[/tex]
[tex]P=18.61\ W[/tex]
b.
[tex]\eta_c=1-\frac{T_L}{T_H}[/tex]
[tex]\eta_c=1-\frac{378}{1250}[/tex]
[tex]\eta_c=0.6976=69.76\%[/tex]
c.
now actual efficiency:
[tex]\eta=\frac{W}{Q_H}[/tex]
[tex]\eta=\frac{6.7\times 10^4}{1.42\times 10^{5}}[/tex]
[tex]\eta=0.4718=47.18\%[/tex]
