Water at 100C^{\circ} \mathrm{C}, quality 50% in a rigid box is heated to 110C^{\circ} \mathrm{C}. How do the properties (P, v, x, u, and s) change (increase, stay about the same, or decrease)?


Step 1
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We are given following data for Water in rigid tank:

T1=100 CT_1=100\text{ C}

T2=110 CT_2=110\text{ C}


From Saturated Water table B.1.1. corresponding to T1=100CT_1=100 C we can find stage 1 pressure saturation and evaporation specific volume, internal energy and entropy:

P1=101.3 kPaP_1=101.3\text{ kPa}

uf=418.91 kJ kgu_f=418.91\frac{\text{ kJ}}{\text{ kg}}

ufg=2087.58 kJ kgu_{fg}=2087.58\frac{\text{ kJ}}{\text{ kg}}

νf=0.001044 m3 kg\nu_f=0.001044\frac{\text{ m}^3}{\text{ kg}}

νfg=1.67185 m3 kg\nu_{fg}=1.67185\frac{\text{ m}^3}{\text{ kg}}

sf=1.3068 kJ kg Ks_f=1.3068\frac{\text{ kJ}}{\text{ kg K}}

sfg=6.0480 kJ kg Ks_{fg}=6.0480\frac{\text{ kJ}}{\text{ kg K}}

Calculating initial specific internal energy:

u1=uf+x1ufg=418.91+0.52087.58=1462.70 kJ kgu_1=u_f+x_1\cdot u_{fg}=418.91+0.5\cdot 2087.58=1462.70\frac{\text{ kJ}}{\text{ kg}}

Calculating initial specific volume:

ν1=νf+x1νfg=0.001044+0.51.67185=0.837 m3 kg\nu_1=\nu_f+x_1\cdot \nu_{fg}=0.001044+0.5\cdot 1.67185=0.837\frac{\text{ m}^3}{\text{ kg}}

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