Respuesta :
Given:
[tex]T_{1}[/tex] = 50°C = 273+50 =323 K
[tex]T_{2}[/tex] = 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:
[tex]\Delta S = cln\frac{T_{f}}{T_{i}}[/tex]
After the bars brought in contact with each other,
final temperature, [tex]T_{f}[/tex] = [tex]\frac{T_{1}+T_{2}}{2}[/tex]
final temperature, [tex]T_{f}[/tex] = [tex]\frac{323+573}{2}[/tex] = 448K
Now, entropy for first gold bar, using eqn-1
[tex]\Delta S = cln\frac{T_{f}}{T_{1}}[/tex]
[tex]\Delta S_{1} = 0.129ln\frac{448}{323}[/tex] =0.042 J/K
[tex]\Delta S_{2}[/tex] = 0.129ln [tex]\frac{448}{573}[/tex] = - 0.032 J/K
Total entropy generation,
[tex]\Delta S_{1}[/tex] = [tex]\Delta S_{1}[/tex] + [tex]\Delta S_{2}[/tex]
[tex]\Delta S[/tex] = 0.042 + (- 0.032) = 0.010 J/K
