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## Homework Statement

I have a problem that I need assistance with:

There is a well insulated rigid tank with 3 kg of saturated liquid-vapor mixture of water at 200 kPa. Initially, half of the mass is in the liquid phase. An electric resistor placed in the tank is connected to a 120-V source, and a current of 6 A flows throught the resistor when the switch is turned on. How long will it take to vaporize all the liquid in the tank?

## Homework Equations

## The Attempt at a Solution

State 1:

P = 200 kPa

x = .5

Therefore T = 120.21 C

v = v

_{f}+xv

_{fg}

v

_{f}= 0.001061 m

^{3}/kg

v

_{fg}= 0.88578 - 0.001061 m

^{3}/kg

Multiply v by m to obtain V = 1.33206 m

^{3}

Therefore V

_{2}= 1.33206 and since the mass does not change, v

_{2}= .4423595 m

^{3}/kg = v

_{g}

u

_{1}= u

_{fg}@ 200 kPA = 2024.6 kJ/kg

W

_{e}= Delta U => VI(Delta T) = m(u

_{2}- u

_{1})

Looking for the v

_{2}, it is assumed to be between 400 kPa and 450 kPa. Therefore, use the v values to interpolate the u

_{g}which is 2554.6669 kJ/kg.

Therefore

((1 kJ/s)/(1000 VA))720 VA (Delta T) = 3 kg(2554.6669- 2024.6 kJ/kg)

Delta T = 2208.6 s or .6135 hours.

I just want to know if I made a mistake anywhere in this problem. I'm not the best with thermodynamics and usually wind up missing some minute detail that is vital to the rest of the problem.