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Physics Question on carnot cycle

Three Carnot engines operate in series between a heat source at a temperature T1T_1 and a heat sink at temperature T4T_4 (see figure). There are two other reservoirs at temperature T2T_2, and T3T_3, as shown, with T2>T2>T3>T4T_2 > T_2 > T_3 > T_4 . The three engines are equally efficient if:

A

T2=(T12T4)1/3;T3=(T1T42)1/3T_{2} =\left(T_{1}^{2} T_{4}\right)^{1/3} ; T_{3} =\left(T_{1} T_{4}^{2}\right)^{1/3}

B

T2=(T1T42)1/3;T3=(T12T4)1/3T_{2} =\left(T_{1} T_{4}^{2} \right)^{1/3} ; T_{3} =\left(T_{1}^{2} T_{4}\right)^{1/3}

C

T2=(T13T4)1/4;T3=(T1T43)1/4T_{2} =\left(T_{1}^{3} T_{4}\right)^{1/4} ; T_{3} =\left(T_{1} T_{4}^{3}\right)^{1/4}

D

T2=(T1T4)1/2;T3=(T12T4)1/3T_{2} =\left(T_{1} T_{4}\right)^{1/2} ; T_{3} =\left(T_{1}^{2} T_{4}\right)^{1/3}

Answer

T2=(T12T4)1/3;T3=(T1T42)1/3T_{2} =\left(T_{1}^{2} T_{4}\right)^{1/3} ; T_{3} =\left(T_{1} T_{4}^{2}\right)^{1/3}

Explanation

Solution

t1=1T2T1=1T2T2=1T4T3t_{1} = 1- \frac{T_{2}}{T_{1}} = 1- \frac{T_{2}}{T_{2}} = 1 - \frac{T_{4}}{T_{3}} T2T1=T3T4=T4T3\Rightarrow \frac{T_{2} }{T_{1}} = \frac{T_{3}}{T_{4}}= \frac{T_{4}}{T_{3}} T2=T1T2=T1T2T4 \Rightarrow T_{2} = \sqrt{T_{1}T_{2}} = \sqrt{T_{1} \sqrt{T_{2}T_{4}}} T3=T2T4T_{3} = \sqrt{T_{2}T_{4}} T23/4=T11/2T41/4T_{2}^{3/4} = \sqrt{T_{1}^{1/2}} T^{1/4}_{4} T2=T12/3T41/3T_{2} =T_{1}^{2/3} T_{4}^{1/3}