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Two bolcks to the same metal having same...

Two bolcks to the same metal having same mass and at temperature `T_1` and `T_2` respectively ,are brought in contact with each other and allowed to attain thermal equilibrium at constant pressure .The change in entropy ,`Delta S` for this process is :

A

`2C_(p) ln[(T_(1)+T_(2))/(2T_(1)T_(2))]`

B

`2C_(p) ln[(T_(1)+T_(2))/(4T_(1)T_(2))]`

C

`2C_(p) ln[((T_(1)+T_(2))^(1//2)/(T_(1)T_(2)))]`

D

`C_(p)ln[((T_(1)+T_(2))^(2)/(4T_(1)T_(2)))]`

Text Solution

Verified by Experts

The correct Answer is:
D
Heat released by one block at `T_(2)`= Heat absorbed by block at `T_(1)`
`-ms(T-T_(2)) = ms(T-T_(1))` (where `T_(2) gt T_(1)`). T is equilibrium temperature
`(T_(2)-T_(1))=(T-T_(1)),T= (T_(1) +T_(2))2`
`DeltaS = C_(p) ln T/T_(1) + C_(p)ln T/T_(2) = C_(p) ln (T/T_(1) xx T/T_(2)) rArr DeltaS = C_(p) ln T^(2)/(T_(1)T_(2))`
`DeltaS = C_(p) ln (T_(1)-T_(2))^(2)/(4T_(1)T_(2))`
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