Question

The figure shows two identical copper blocks of mass $0.5kg$ . When they were not in contact, block $L$ was at temperature $60^{\circ }C$ and block $R$ was at temperature $20^{\circ }C$ . But, when the blocks bring in contact, they come to the equilibrium temperature $40^{\circ }C$ . What is the net entropy change of the two block system during the irreversible process ? (Specific heat of copper $=386J/kg-K$ )

• A. 2.4 J/K
• B. 3.6 J/K
• C. 4.2 J/K
• D. 5.2 J/K

Answer 1

Answer: (A)

Change in entropy of the system
$=\Delta S=\Delta S_L+\Delta S_R$
$=m_L$ ln $\frac{T_{eq}}{(T_1{)}_L}+m_{R}S$ ln $\frac{T}{(T_1{)}_R}$
$=0.5\times 386$ [ln $T_{eq}-ln(T_1)L+$ ln $T_{eq}-$ ln $(T_1)R]$
$=193[2$ In $T_{eq}$- ln $(T_1{)}_L$ -ln $(T_2{)}_R]$
$=193(2m313-$ In $333-$ In $293]$
$=193(2\times 5.746-5.8-5.68]$
$=193(11.492-5.8-5.68]$
$=193(11.492-11.48)$
$=2.316J/K-2.4J/K$