Respuesta :
Explanation:
Relation between entropy change and specific heat is as follows.
[tex]\Delta S = C_{p} log (\frac{T_{2}}{T_{1}})[/tex]
The given data is as follows.
mass = 500 g, [tex]C_{p}[/tex] = 24.4 J/mol K
[tex]T_{h}[/tex] = 500 K, [tex]T_{c}[/tex] = 250 K
Mass number of copper = 63.54 g /mol
Number of moles = [tex]\frac{mass}{/text{\molar mass}}[/tex]
= [tex]\frac{500}{63.54}[/tex]
= 7.86 moles
Now, equating the entropy change for both the substances as follows.
[tex]7.86 \times 24.4 \times [T_{f} - 250][/tex] = [tex]7.86 \times 24.4 \times [500 -T_{f}][/tex]
[tex]T_{f} - 250 = 500 - T_{f}[/tex]
[tex]2T_{f}[/tex] = 750
So, [tex]T_{f}[/tex] = [tex]375^{o}C[/tex]
- For the metal block A, change in entropy is as follows.
[tex]\Delta S = C_{p} log (\frac{T_{2}}{T_{1}})[/tex]
= [tex]24.4 log [\frac{375}{500}][/tex]
= -3.04 J/ K mol
- For the block B, change in entropy is as follows.
[tex]\Delta S = C_{p} log (\frac{T_{2}}{T_{1}})[/tex]
= [tex]24.4 log [\frac{375}{250}][/tex]
= 4.296 J/Kmol
And, total entropy change will be as follows.
= 4.296 + (-3.04)
= 1.256 J/Kmol
Thus, we can conclude that change in entropy of block A is -3.04 J/ K mol and change in entropy of block B is 4.296 J/Kmol.
