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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` In`[(T_1+T_2)/(2T_1T_2)]`

B

`2C_p` In `[(T_1+T_2)/(4T_1T_2)]`

C

`2C_p` In `[((T_1+T_2)^(1/2))/(T_1T_2)]`

D

`C_p` In `[((T_1+T_2)^2)/(4T_1T_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 > T_1)` , T is equilibrium temperature
`(T_2-T)=(T-T_1), T=(T_1+T_2)/2`
`DeltaS=C_p` In `T/T_1+C_p` In `T/T_2 =C_p` In `(T/T_1xxT/T_2) rArr DeltaS =C_p` In `T^2/(T_1T_2)`
`DeltaS=C_p` In `((T_1+T_2)^2)/(4T_1T_2)`
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