122k views
5 votes
The hot reservoir for a Carnot engine has a temperature of 889 K, while the cold reservoir has a temperature of 657 K. The heat input for this engine is 4710 J. The 657-K reservoir also serves as the hot reservoir for a second Carnot engine. This second engine uses the rejected heat of the first engine as input and extracts additional work from it. The rejected heat from the second engine goes into a reservoir that has a temperature of 406 K. Find the total work delivered by the two engines.

User Jayphelps
by
5.1k points

1 Answer

2 votes

Answer:

Step-by-step explanation:

Efficiency of first engine

= T₁ - T₂ / T₁ where T₁ is temperature of hot reservoir , T₂ is temperature of cold reservoir

= (889 - 657 ) / 889

= ,261 or 26.1 %

output of work = .261 x 4710 = 1229.15 J .

Efficiency of second engine

= (657 - 406 ) / 657

= .382

Heat rejected in engine one is heat input of second engine

heat input of second engine = 4710 - 1229.15 = 3480.85

output of work of second engine

= .382 x 3480.85 = 1329.68 J

Total work delivered by two engine

= 1229.15 + 1329.68 J

= 2558.83 J

User Lkaradashkov
by
5.4k points