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When 25 kg of water at 95°C is mixed with 35 kg of water at 35°C, the pressure being taken as constant at surrounding temperature of 15°C, and Cp of water is 4.2 kJ/kg K, the decrease in available energy due to mixing will be nearly
1. 270.5 kJ
2. 277.6 kJ
3. 281.8 kJ
4. 288.7 kJ

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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

\(A{E_2} = 35 \times \left[ {\left( {308 - 288} \right) - 288\;ln\;\frac{{308}}{{288}}} \right]\)

AE= 97.9 kJ

For 60 kg water in 60°C

\(A{E_3} = 60 \times 4.2\;\left[ {\left( {333 - 288} \right) - 288\;ln\frac{{333}}{{288}}} \right]\)

AE= 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

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