Question #265082

a rigid tank holds 4 kg of water at 20 degree Celcius. If 200 kJ of heat is added to this system, what will be the final temperature? how much work is done in this process


1
Expert's answer
2021-11-15T13:04:55-0500

Q=m(u2u1)Q = m(u_2 -u_1)

Q = heat

m = mass of water

u1 = the initial specific internal energy

u2 = the final specific internal energy for liquid

From the Table C-1 of “Properties of Saturated H2O-Temperature table” for initial temperature

T1 = 20 °C , note the initial specific internal energy u1 = 83.9 kJ/kg

m= 4 kg

Q = 200 kJ

200=4(u283.9)u2=134  kJ/kg200 = 4(u_2 -83.9) \\ u_2 = 134 \;kJ/kg

From the Table C-1 of “Properties of Saturated H2O-Temperature table” for final specific internal energy u2 = 134 kJ/kg note the final temperature T2 = 32 °C

The final temperature is 32 °C

Since the container is rigid one and there is no possible for work done.

W=0


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