How Is Equilibrium Temperature Calculated in Isolated Thermodynamic Systems?

[berthanas]

hi I'm a mechanical engineering student. I've 2 questions as homework for tomorrow but since I've missed couple of classes i couldn't solve them. i'd appreciate if you may help me with these:

1.) An isolated system consists of a 10 kg copper block, initially at 30 0C , and 0.2 kg of saturated water vapor, initially at 130 0C.Assuming no volume change, determine the final equilibrium temperature of the isolated system.

2.) A cylinder fitted with a piston restrained by a linear spring has a cross-sectional area of 0.05 m2.The cylinder contains ammonia at 1 MPa ,60 0C, and a volume of 20L.The spring constant is 150 kN/m.Heat is rejected from the system, and the piston moves until 6,25 kJ of work has been done on the ammonia.

a.) Find the final temperature of the ammonia.

b.) Calculate the heat transfer for the process.


thanx already...

 

[Pyrrhus]

Well the policy here is to show your work, why don't you read the chapter of your book on this, and then post your attempt?

 

[piercas]

Hi there! I am happy to help you with your thermodynamics questions. Let's take a look at each question and work through them together.

1.) For this problem, we have an isolated system consisting of a 10 kg copper block and 0.2 kg of saturated water vapor. Since it is isolated, we know that there will be no heat or mass transfer with the surroundings. This means that the total energy of the system will remain constant. We also know that there is no change in volume, so the work done by the system will be zero.

To solve for the final equilibrium temperature, we can use the First Law of Thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. Mathematically, this can be written as:

ΔU = Q - W

Since there is no work done, we can simplify this to:

ΔU = Q

We can also assume that the copper block and water vapor will reach thermal equilibrium, meaning that they will have the same final temperature. We can write the energy equation for each substance as:

ΔU1 = m1c1ΔT1

ΔU2 = m2c2ΔT2

Where m is the mass, c is the specific heat, and ΔT is the change in temperature. Since the final temperature will be the same for both substances, we can set ΔU1 equal to ΔU2 and solve for the final temperature, T:

m1c1ΔT1 = m2c2ΔT2

m1c1(Tf - 30) = m2c2(Tf - 130)

Solving for Tf, we get:

Tf = (m1c1(30) + m2c2(130))/(m1c1 + m2c2)

Substituting in the values given, we get:

Tf = (10 kg * 386 J/kg*K * 30 K + 0.2 kg * 2000 J/kg*K * 130 K)/(10 kg * 386 J/kg*K + 0.2 kg * 2000 J/kg*K)

Tf = 34.8 °C

Therefore, the final equilibrium temperature of the isolated system will be 34.8 °C.

2.) For this problem, we have