Question

Find an expression for the change in entropy when two blocks of the same substance and of equal mass, one at the temperature Th and the other at Tc, are brought into thermal contact and allowed to reach equilibrium. Evaluate the change in entropy for two blocks of copper, each of mass 500 g, with Cp,m = 24.4 J K−1 mol−1, taking Th = 500 K and Tc = 250 K.

Answer 1

Step-by-step explanation:

The relation between change in entropy and
`C_(p)` is as follows.

`\Delta S = C_(p) log ((T_(2))/(T_(1)))`

The given data is as follows.

mass = 500 g

`C_(p)` = 24.4 J/mol K

`T_(h)` = 500 K

`T_(c)` = 250K

As, atomic mass of copper = 63.54 g /mol. Therefore, number of moles of copper will be calculated as follows.

Number of moles =
`\frac{mass}{\text{molar mass}}`

=
`(500 g)/(63.54 g/mol)`

= 7.86 moles

Hence,
`7.86 * 24.4 * [T_(f) - 250] = 7.86 * 24.4 * [500 -T_(f)]`

`T_(f) - 250 = 500 - T_(f)`

`2T_(f)` = 750

`T_(f) = 375^(o)C`

For the metal block A,

`\Delta S = C_(p) log ((T_(2))/(T_(1)))`

=
`24.4 * log [(375)/(500)]`

= -3.04 J/ K mol

For the block B,

`\Delta S = 24.4 * log[(375)/(250)]`

= 4.296

Therefore, calculate the change in entropy as follows.

Total entropy change = 4.296 + (-3.04)

= 1.256 J/Kmol

Thus, we can conclude that the change in entropy for given two blocks of copper is 1.256 J/Kmol .