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in Thermodynamics by (15 points)

Consider four heat reservoirs A, B, C, D an engine working between A and C has an efficiency which is mean of efficiencies of same engine working between AB and AD. Then. Absolute temperature of C will be

(1) Geometric mean of temperature of A & D (2) Arithmetic mean of temperature of B & D (3) Harmonic mean of temperature of A & D (4) Sum of temperatures of A and D

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1 Answer

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by (44.0k points)

Correct option is (2) Arithmetic mean of temperature of B & D

We know that, the efficiency of engine

\(\eta=1-\frac{T_2}{T_1}\)

where T2 = sink temperature

T1 = source temperature

\(\eta_{AB} = 1-\frac{T_B}{T_A}\) 

\(\eta_{AD} = 1-\frac{T_D}{T_A}\)

\(\eta_{AC} = 1-\frac{T_C}{T_A}\) 

Given

\(\eta_{AC}=\frac{\eta_{AB}+\eta_{AD}}{2}\) 

1 - \(\frac{T_C}{T_A}\) = \(\cfrac{1-\frac{T_B}{T_A}+1-\frac{T_D}{T_A}}2\) 

1 - \(\frac{T_C}{T_A}\) = \(\frac12-\frac{T_B}{2A}+\frac12-\frac{T_D}{2T_A}\) 

1 - \(\frac{T_C}{T_A}\) = 1 - \(\frac{T_B}{2T_A}-\frac{T_D}{2T_A}\)

\(-\frac{T_C}{T_A}=-\frac1{T_A}(\frac{T_B+T_D}2)\) 

TC = \(\frac{T_B+T_D}2\)

Arithmetic mean of temperature of B and D.

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