Question

Two one-kilogram bars of gold are initially at 50°C and 300°C. The bars are brought in contact with each other. What is their final temperature, the entropy change of each block, and the total entropy generation in the process.

Answer 1

Given:

`T_(1)` = 50°C = 273+50 =323 K

`T_(2)` = 300°C = 273+300 =573 K

Solution:

We know that:

specific heat for gold, c = 0.129 J/g°C

Also, change in entropy, ΔS is given by:

`\Delta S = cln(T_(f))/(T_(i))`

After the bars brought in contact with each other,

final temperature,
`T_(f)` =
`(T_(1)+T_(2))/(2)`

final temperature,
`T_(f)` =
`(323+573)/(2)` = 448K

Now, entropy for first gold bar, using eqn-1

`\Delta S = cln(T_(f))/(T_(1))`

`\Delta S_(1) = 0.129ln(448)/(323)` =0.042 J/K

`\Delta S_(2)` = 0.129ln
`(448)/(573)` = - 0.032 J/K

Total entropy generation,

`\Delta S_(1)` =
`\Delta S_(1)` +
`\Delta S_(2)`

`\Delta S` = 0.042 + (- 0.032) = 0.010 J/K