Two Carnot engines A and B are operated in series. The first one, A receives heat at T1(.=600K.) and rejects to a reservoir at temperature T2. The second engine B receives heat rejected by the first engine and, in turn, rejects to a heat reservoir at T3(.=400K.). Calculate the temperature T2 if the work outputs of the two engines are equal

Question

Two Carnot engines A and B are operated in series. The first one, A receives heat at T1(.=600K.) and rejects to a reservoir at temperature T2. The second engine B receives heat rejected by the first engine and, in turn, rejects to a heat reservoir at T3(.=400K.). Calculate the temperature T2 if the work outputs of the two engines are equal (A) 500K (B) 300K (C) 600K (D) 400K

Tardigrade Admin · Accepted Answer

First case, W=Q1-Q2 Second case, W=Q2-Q3 Given Q1-Q2=Q2-Q3 Q1+Q3=2Q2 (Q1/Q2)+(Q3/Q2)=2 (T1/T2)+(T3/T2)=2⇒ T2=(T1 + T3/2)=500K

$500 K$

$300 K$

$600 K$

$400 K$

First case, $W = Q_{1} - Q_{2}$
Second case, $W = Q_{2} - Q_{3}$
Given $Q_{1} - Q_{2} = Q_{2} - Q_{3}$
$Q_{1} + Q_{3} = 2 Q_{2}$
$\frac{Q _{1}}{Q _{2}} + \frac{Q _{3}}{Q _{2}} = 2$
$\frac{T _{1}}{T _{2}} + \frac{T _{3}}{T _{2}} = 2 \Rightarrow T_{2} = \frac{T _{1} + T _{3}}{2} = 500 K$

500K

300K

600K

400K

First case, W=Q1​−Q2​ Second case, W=Q2​−Q3​ Given Q1​−Q2​=Q2​−Q3​ Q1​+Q3​=2Q2​ Q2​Q1​​+Q2​Q3​​=2 T2​T1​​+T2​T3​​=2⇒T2​=2T1​+T3​​=500K

$500 K$

$300 K$

$600 K$

$400 K$

First case, $W = Q_{1} - Q_{2}$
Second case, $W = Q_{2} - Q_{3}$
Given $Q_{1} - Q_{2} = Q_{2} - Q_{3}$
$Q_{1} + Q_{3} = 2 Q_{2}$
$\frac{Q _{1}}{Q _{2}} + \frac{Q _{3}}{Q _{2}} = 2$
$\frac{T _{1}}{T _{2}} + \frac{T _{3}}{T _{2}} = 2 \Rightarrow T_{2} = \frac{T _{1} + T _{3}}{2} = 500 K$