Respuesta :
Explanation:
(a). Let us assume that [tex]T_{f}[/tex] be the final temperature after mixing.
Hence, [tex]m_{1}C_{p}(T_{1} - T_{f}) = m_{2}C_{p}(T_{f} - T_{2})[/tex]
[tex]T_{f} = \frac{[T_{1} + \frac{m_{2}}{m_{1}} \times T_{2}]}{1 + \frac{m_{2}}{m_{1}}}[/tex]
The given data is as follows.
[tex]m_{1}[/tex] = 1 kg, [tex]m_{2}[/tex] = 5 kg
[tex]T_{1}[/tex] = 3100 K, [tex]T_{2}[/tex] = 3600 K
[tex]C_{p}[/tex] = 100 J/mol K
Hence, putting the given values into the above formula as follows.
[tex]T_{f} = \frac{[T_{1} + \frac{m_{2}}{m_{1}} \times T_{2}]}{1 + \frac{m_{2}}{m_{1}}}[/tex]
[tex]T_{f} = \frac{[3100 + (\frac{5}{1}) \times 3600]}{[1 + (\frac{5}{1})}][/tex]
[tex]T_{f}[/tex] = 3516.7 K
(b). As, entropy change of [tex]UO_{2}[/tex] with [tex]m_{1}[/tex] = 1 kg at 3100 K to attain 3516.7 K. Therefore, change in entropy will be calculated as follows.
[tex]\Delta S_{1} = m_{1} C_{p} ln \frac{T_{f}}{T_{1}}[/tex]
= [tex]1 \times 100 \times ln (\frac{3516.7}{3100})[/tex]
= 12.22 J/molK
Now, entropy change of [tex]UO_{2}[/tex] with [tex]m_{2}[/tex] = 5kg at 3600 K to attain 3516.7 K. Hence,
[tex]\Delta S_{2} = m_{2}C_{p} ln \frac{T_{f}}{T_{2}}[/tex]
= [tex]5 \times 100 \times ln(\frac{3516.7}{3600})[/tex]
= -11.63 J/mol k
So, entropy of the total mass of [tex]UO_{2}[/tex] will be as follows.
[tex]\Delta S = \Delta S_{1} + \Delta S_{2}[/tex]
= 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.
