Answer to Question #152385 in Molecular Physics | Thermodynamics for Ashish puraviya

Question

(a) Consider an isolated system that contains two pieces of copper sepa-
rated by an (internal) insulating wall. Initially, the first piece is at 500K
and the second is at 300K. Calculate the entropy change in the sys-
tem when the insulating wall is removed; assume that each piece has
half a mole of copper in it. Given: Specific heat capacity of Cu is
CP = 22.6JK−1mol−1
.

Answer 1

let two pieces are A and B.

$T_A=500K, n_A=0.5 mol ,\space T_B=300K \space\&\space n_B=0.5mol\\ C_p=22.6JK^{-1}mol^{-1}$

Let final common temperature is T_c.

Since system is isolated, due conservation of heat, we have...

$$n_AC_p( T_A-T_c)=n_BC_p(T_c- T_B)$$

Putting all values in above equation we have,

$$T_c=400K$$

Now change in entropy of pieces A and B are...

$$\Delta S_A=\int_{500}^{400}\cfrac{n_AC_p dT}{T}\\ \Delta S_A=\cfrac{C_p}{2}\ln{\cfrac{4}{5}}\\ \Delta S_B=\cfrac{C_p}{2}\int_{300}^{400}\cfrac{dT}{T}\\ \Delta S_B=\cfrac{C_p}{2}\ln{\cfrac{4}{3}}\\$$

Net Change in Entropy of the system is..

$$\Delta S=\Delta S_A+\Delta S_B\\ \Delta S= \cfrac{22.6}{2}\ln{\cfrac{16}{15}}\\ \Delta S=0.729 J K^{-1}...Ans$$