A rigid tank contains liquid water and water vapor with an initial total interna

Question

A rigid tank contains liquid water and water vapor with an initial total internal energy of 500 kJ. During a process 400 kJ of energy transferred into the system from a burner and 2,000 coulombs of electrical charge move through the resistance at a potential difference of 25 V. As a result, 500 kJ are transferred from water to the surrounding air. What kind of work interaction is present in this process Is this process adiabatic Will the volume of the tank change after this process Please calculate the final internal energy of the water contained in the tank.

Explanation / Answer

8a)What kind of work interaction is present in this process?

we assume a perfect seal (no mass flow in or out of the system), no loss due to friction, and quasi-equilibrium processes are maintained. By convention positive work is that done by the system on the surroundings, and negative work is that done by the surroundings on the system, Thus since negative work results in an increase in internal energy of the system, this explains the negative sign in the above energy equation.

one coulomb is the equivalent of one ampere-second.

2000 Coloumbs = 2000 Ampere

Potentail Difference : 25 V

Power = V*I = 2000*25 = 50000 J/s = 50kW

U2 = 500+400+50 = 950kW

Q = (U2 - U1)+W

Q = Qs-Qr = 400-500 = -100kW

-100 = (950-400)+W

W = -100-550 = -650KJ

8b) Is this process Adiabatic?

In Adiabatic process, Heat energy transferred will be equal to Zero. Here in this Process, Heat is being added of 400kJ and rejected from water to surroundings is 500kJ.Hence this process is no more Adiabatic.

8c) Will the volume of Tank Change as after this process?

Since it is considered as a rigid tank for this process, the volume of the tank will not change during this process.

8d) Please calculate the final internal energy of the water contained in the tank

Potentail Difference : 25 V

P = V*I = 2000*25 = 50000 J/s = 50kJ

U2 = 500+400+50 = 950kJ

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