Answer
a) $T_2=39.4°C$
b) $m_i=2.40\ kg$
c) $Q_i=899\ kJ$
Work Step by Step
From tables A-11 to A-13:
Initial ($x_1=0.55, T_1=14°C$): $v_1=0.02427\ m³/kg,\ u_1=162.57\ kJ/kg$
Final ($P_2=1\ MPa, x_2=1$): $v_2=0.02033\ m³/kg,\ u_2=250.71\ kJ/kg$
Entering ($P_i=1.4\ MPa, T_i=100°C$): $h_i=330.322\ kJ/kg$
a) $T_2=T_{sat}=39.4°C$
b) Given the tank volume of 0.3 m³, and $m=V/v$:
$m_1=12.36\ kg,\ m_2=14.76\ kg$
From the material balance:
$m_i=m_2-m_1$
$m_i=2.40\ kg$
c) From the energy balance:
$Q_i+m_ih_i=m_2u_2-m_1u_1$
$Q_i=899\ kJ$
