Respuesta :
Answer:
[tex] T_C = 27+273.15 = 300.15 K[/tex]
[tex] T_H = 477+273.15 = 750.15 K[/tex]
And replacing in the Carnot efficiency we got:
[tex] e= 1- \frac{300.15}{750.15}= 0.59988 = 59.98 \%[/tex]
[tex] W_{max}= e* Q_H = 0.59988 * 65000 \frac{KJ}{min}= 38992.2 \frac{KJ}{min}[/tex]
Explanation:
For this case we can use the fact that the maximum thermal efficiency for a heat engine between two temperatures are given by the Carnot efficiency:
[tex] e = 1 -frac{T_C}{T_H}[/tex]
We have on this case after convert the temperatures in kelvin this:
[tex] T_C = 27+273.15 = 300.15 K[/tex]
[tex] T_H = 477+273.15 = 750.15 K[/tex]
And replacing in the Carnot efficiency we got:
[tex] e= 1- \frac{300.15}{750.15}= 0.59988 = 59.98 \%[/tex]
And the maximum power output on this case would be defined as:
[tex] W_{max}= e* Q_H = 0.59988 * 65000 \frac{KJ}{min}= 38992.2 \frac{KJ}{min}[/tex]
Where [tex] Q_H[/tex] represent the heat associated to the deposit with higher temperature.
