Question

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

Answer 1

$Q = m(u_2 -u_1)$

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

$200 = 4(u_2 -83.9) \\ u_2 = 134 \;kJ/kg$

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