A) \[2{{C}_{p}}\ln \left[ \frac{{{T}_{1}}+{{T}_{2}}}{2{{T}_{1}}{{T}_{2}}} \right]\]
B) \[{{C}_{p}}\ln \left[ \frac{{{({{T}_{1}}+{{T}_{2}})}^{2}}}{4{{T}_{1}}{{T}_{2}}} \right]\]
C) \[2{{C}_{p}}\ln \left[ \frac{({{T}_{1}}+{{T}_{2}})}{4{{T}_{1}}{{T}_{2}}} \right]\]
D) \[2{{C}_{p}}\ln \left[ \frac{{{({{T}_{1}}+{{T}_{2}})}^{\frac{1}{2}}}}{{{T}_{1}}{{T}_{2}}} \right]\]
Correct Answer: B
Solution :
\[{{T}_{final}}=\frac{{{T}_{1}}+{{T}_{2}}}{2}\] \[\Delta {{S}_{1}}={{C}_{p}}\ln \frac{{{T}_{f}}}{{{T}_{1}}},\Delta {{S}_{2}}={{C}_{p}}\ln \frac{{{T}_{f}}}{{{T}_{2}}}\] \[\Delta S=\Delta {{S}_{1}}+\Delta {{S}_{2}}={{C}_{p}}\ln \frac{T_{f}^{2}}{{{T}_{1}}{{T}_{2}}}={{C}_{p}}\ln \left[ \frac{{{({{T}_{1}}+{{T}_{2}})}^{2}}}{4{{T}_{1}}{{T}_{2}}} \right]\]You need to login to perform this action.
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