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

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Q. The figure shows two identical copper blocks of mass $ 0.5 \,kg $ . 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 $ = 386\, J/kg-K $ )Physics Question Image

AMUAMU 2013

Solution:

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 $(T_2)_R]$
$= 193(2m \,313 -$ In $333 - $ In $293]$
$= 193(2 \times 5.746 - 5.8 - 5.68]$
$= 193(11.492 - 5.8 - 5.68]$
$= 193(11.492-11.48)$
$= 2.316 \,J/K - 2.4 \,J/K$