Question

Three bodies at initial temperatures of 200 K, 250 K and 540 K. What is the maximum amount of work that can be extracted in a process in which these bodies are brought to a final common temperature? Each of these bodies satisfy equation U= CT. where C is heat capacity, V is the internal energy. (Take C = 8.4 kJ/K)

• A. 960 kJ
• B. 1090 kJ
• C. 657 kJ
• D. 756 kJ

Answer 1

The correct option is D 756 kJ

Let the final common temperature of 3 reservoirs be $T_f$

given: $U=CT$

$\Rightarrow$ $dU=CdT$

Now, $dS=\frac{CdT}{T}$

$\Rightarrow$ $\Delta S=Cln\frac{T_f}{T_i}$

$T_i=$ Initial temperature

For maximum work, $\Sigma \Delta S=0$

$\Rightarrow$ $\Delta S_1+\Delta S_2+\Delta S_3=0$

$\Rightarrow$ $Cln\frac{T_f}{200}+Cln\frac{T_f}{250}+Cln\frac{T_f}{540}=0$

$\Rightarrow$ $T_f^3=200\times 250\times 540$

$\Rightarrow$ $T_f=300K$

$W_{max}=-[\Delta U_1+\Delta U_2+\Delta U_3]$

$=-\left[C\right[T_f-T_1]+C[T_f-T_2]+C[T_f-T_3\left]\right]$

$=-\left[8.4\right(300-200)+8.4(300-250)+8.4(300-540\left)\right]$

$\Rightarrow$ $W_{max}=756kJ$