Answer :
The given question is incomplete. The complete question is as follows.
A 0.6-m3 rigid tank is filled with saturated liquid water at 135°C. A valve at the bottom of the tank is now opened, and one-half of the total mass is withdrawn from the tank in liquid form. Heat is transferred to water from a source of 210°C so that the temperature in the tank remains constant. Assume the surroundings to be at 25°C and 100 kPa. Determine the amount of heat transfer.
Explanation:
First, we will determine the initial mass from given volume and specific volume as follows.
[tex]m_{1} = \frac{V}{\alpha_{1}}[/tex]
= [tex]\frac{0.6}{0.001075}kg[/tex]
= 558.14 kg
Hence, the final mass and mass that has left the tank are as follows.
[tex]m_{2} = m_{out} = \frac{1}{2}m_{1}[/tex]
= [tex]\frac{1}{2} \times 558.14 kg[/tex]
= 279.07 kg
Now, the final specific volume is as follows.
[tex]\alpha_{2} = \frac{V}{m_{2}}[/tex]
= [tex]\frac{0.6}{279.07} m^{3}/kg[/tex]
= 0.00215 [tex]m^{3}/kg[/tex]
Final quality of the mixture is determined actually from the total final specific volume and the specific volumes of the constituents for the given temperature are as follows.
[tex]q_{2} = \frac{\alpha_{2} - \alpha_{liq135}}{(\alpha_{vap} - \alpha_{liq})_{135}}[/tex]
= [tex]\frac{0.00215 - 0.001075}{0.58179 - 0.001075}[/tex]
= [tex]1.85 \times 10^{-3}[/tex]
Hence, the final internal energy will be calculated as follows.
[tex]u_{2} = u_{liq135} + q_{2}u_{vap135}[/tex]
= [tex](567.41 + 1.85 \times 10^{-3} \times 1977.3) kJ/kg[/tex]
= 571.06 kJ/kg
Now, we will calculate the heat transfer as follows.
[tex]\Delta U = Q - m_{out}h_{out}[/tex]
[tex]m_{2}u_{2} - m_{1}u_{1} = Q - m_{out}h_{out}[/tex]
Q = [tex](279.07 \times 571.06 - 558.14 \times 567.41 + 279.07 \times 567.75) kJ[/tex]
= 1113.5 kJ
Thus, we can conclude that amount of heat transfer is 1113.5 kJ.