Question

1 kg of liquid UO_2 at 3100 K is mixed with 5 kg of UO2 at 3600 K. a. Find the temperature after mixing. b. Find the entropy changes of the system and surroundings. c. Is the process spontaneous? Assume C_p = 100 J/(mole-K) and is constant

Answer 1

Step-by-step explanation:

(a). Let us assume that
`T_(f)` be the final temperature after mixing.

Hence,
`m_(1)C_(p)(T_(1) - T_(f)) = m_(2)C_(p)(T_(f) - T_(2))`

`T_(f) = ([T_(1) + (m_(2))/(m_(1)) * T_(2)])/(1 + (m_(2))/(m_(1)))`

The given data is as follows.

`m_(1)` = 1 kg,
`m_(2)` = 5 kg

`T_(1)` = 3100 K,
`T_(2)` = 3600 K

`C_(p)` = 100 J/mol K

Hence, putting the given values into the above formula as follows.

`T_(f) = ([T_(1) + (m_(2))/(m_(1)) * T_(2)])/(1 + (m_(2))/(m_(1)))`

`T_(f) = ([3100 + ((5)/(1)) * 3600])/([1 + ((5)/(1)))]`

`T_(f)` = 3516.7 K

(b). As, entropy change of
`UO_(2)` with
`m_(1)` = 1 kg at 3100 K to attain 3516.7 K. Therefore, change in entropy will be calculated as follows.

`\Delta S_(1) = m_(1) C_(p) ln (T_(f))/(T_(1))`

=
`1 * 100 * ln ((3516.7)/(3100))`

= 12.22 J/molK

Now, entropy change of
`UO_(2)` with
`m_(2)` = 5kg at 3600 K to attain 3516.7 K. Hence,

`\Delta S_(2) = m_(2)C_(p) ln (T_(f))/(T_(2))`

=
`5 * 100 * ln((3516.7)/(3600))`

= -11.63 J/mol k

So, entropy of the total mass of
`UO_(2)` will be as follows.

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

= 12.22 + (-11.63) J/mol K

= 0.59 J/mol K

(c). As, there is occurring an increase in the entropy. Therefore, the process is spontaneous.