Aclosed, rigid tank contains 1.50 kg of water, initially a two phase liquid-vapor mixture at t1 = 80°c. heat transfer occurs until the tank contains only saturated vapor at t2 = 120°c. determine the initial pressure, p1, in kpa, and the heat transfer for the process, in kj.

Explanation:

Given data:

m = 1.5 kg

degree celcius

degree celcius

from saturated water table

at T = 120 degree C

Enthalapy hg = 2528.9 kj/kg

Initial state is 2 phase liquid

at 80 degree c

At 80 degree C

vf = 0.001029 m^3/kg

vg = 3.4053 m^/kg

0.89133 = 0.001029 + x1 = ( 3.4053-0.001029)

at 80 degree C

= 334.97 + 0.261 (2146.6)

           = 0 + ( 2528.9 - 896.35)

answer:

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explanation: