The correct answer is Democritus
Answer:
e_12=1-Tc/Th
This is same as the original Carnot engine.
Explanation:
For original Carnot engine, its efficiency is given by
e = 1-Tc/Th
For the composite engine, its efficiency is given by
e_12=(W_1+W_2)/Q_H1
where Q_H1 is the heat input to the first engine, W_1 s the work done by the first engine and W_2 is the work done by the second engine.
But the work done can be written as
W= Q_H + Q_C with Q_H as the heat input and Q_C as the heat emitted to the cold reservoir. So.
e_12=(Q_H1+Q_C1+Q_H2+Q_C2)/Q_H1
But Q_H2 = -Q_C1 so the second and third terms in the numerator cancel
each other.
e_12=1+Q_C2/Q_H1
but, Q_C2/Q_H2= -T_C/T'
⇒ Q_C2 = -Q_H2(T_C/T')
= Q_C1(T_C/T')
(T1 is the intermediate temperature)
But, Q_C1 = -Q_H1(T'/T_H)
so, Q_C2 = -Q_H1(T'/T_H)(T_C/T') = Q_H1(T_C/T_H) So the efficiency of the composite engine is given by
e_12=1-Tc/Th
This is same as the original Carnot engine.
Answer:
32
Explanation:
because the twp 3.5 resistors answer
