Question

A reversible heat engine operates between temperatures t1 and t2, the energy rejected by this engine is received by a second reversible engine at temperature t2 and rejected to a to a reservoir at temperature t3. if the efficiencies engine are same then the relationship between t1, t2 and t3 is given by

Answer 1

The relation between the temperature is T_2^{2}=T_1*T_3

Efficiency of the reversible heat engine is given as

η=1-(t2/T1)

Since the efficinecy of both the heat engine is same therefore

1-(t2/T1)=1-(t3/T2)

$T_2^{2}=T_1*T_3$

Hence, the relation between the temperatures is T_2^{2}=T_1*T_3

Answer 2

question is incomplete , A complete question is -----> : A reversible heat engine operates between temperature T1 and T (T1>T). The energy rejected from this engine is received by s second reversible heat engine at the same temperature T. the second engine rejects energy at a temperature T2 (T2<T1). Show that temperature T is the arithmetic mean of temperature T1 and T2 if the engine produce the same amount of work output.


solution :- a/c to question,
same amount of work output,
e.g., $W_1=W_2$

we know, $W=Q_1-Q_2$
where $Q_1$ is heat from source of Temperature $T_1$ and $Q_2$ is heat to sink of temperature $T_2$.

so, $W_1=Q_1-Q$........(1)
$W_2=Q-Q_2$..............(2)

$W_1=W_2$

$Q_1-Q=Q-Q_2$

$Q_1+Q_2=2Q$

we also know,From carnot heat engine.
$\frac{Q_1}{T_1}=\frac{Q}{T}\\\\Q_1=Q\left(\frac{T_1}{T}\right)$

similarly,
$\frac{Q}{T}=\frac{Q_2}{T_2}\\\\Q_2=Q\left(\frac{T_2}{T}\right)$


now, $Q_1+Q_2=2Q$

$Q\left(\frac{T_1}{T}\right)+Q\left(\frac{T_2}{T}\right)=2Q$

$T=\left(\frac{T_1+T_2}{2}\right)$
hence, proved