Correct Answer - Option 3 : 281.8 kJ
Concept:
\({\rm{Available\;energy\;}}\left( {{\rm{AE}}} \right) = m{C_p}[\left( {T - {T_0}} \right) - {T_0}\;ln\frac{{{T_0}}}{{{T_0}}}\)
Calculation:
Given:
T0 = 15°C = 288 K, m1 = 25 kg m2 = 35 kg
After mixing if T is the final temperature of mixture then
m1 Cp (95 – T1) = m2 Cp (t - 35)
25 (95 –T1) = 35 (t - 35)
∴ T = 60°C
\({\rm{Available\;energy\;}}\left( {{\rm{AE}}} \right) = m{C_p}[\left( {T - {T_0}} \right) - {T_0}\;ln\frac{{{T_0}}}{{{T_0}}}\)
For 25 kg water at 95°C
\(A{E_1} = 25 \times 4.2\;\left[ {\left( {368 - 288} \right) - 288\;ln\;\frac{{368}}{{288}}} \right]\)
∴ AE1 = 987- 49 kJ
For 35 kg water at 35°C
\(A{E_2} = 35 \times \left[ {\left( {308 - 288} \right) - 288\;ln\;\frac{{308}}{{288}}} \right]\)
∴ AE2 = 97.9 kJ
For 60 kg water in 60°C
\(A{E_3} = 60 \times 4.2\;\left[ {\left( {333 - 288} \right) - 288\;ln\frac{{333}}{{288}}} \right]\)
∴ AE3 = 803.27 kJ
So decrease in available energy = AE1 + AE2 – AE3
= 987.49 + 97.59 – 803.27 = 281.80 kJ
∴
Decrease in available energy is 281.80 kJ