Open System, Heat Transfer & Thermo I

[beetlejuice]

1. Homework Statement
We want to heat up 200 liters of water in a bathtub from 15 ◦C to 30 ◦C. The temperature is increased by adding marble stones to the water. The initial temperature of the marble stones is 773, 15 K.
Assume that the marble is rigid, with the heat capacity cm = 0.88 J ·K−1 · g −1 and the water is incompressible, with cp = 4.2 J · K−1 · g −1 and the density ρ = 1000 kg · m−1 .
Neglect all heat losses to the environment or any change in kinetic or potential energy. After the process, the system is in balance.
Calculate the minimal mass of marble needed for the process. Approach the question making the following assumptions:
consider an instationary open system. Balance the system of marble stones and water as a whole. Hint: Begin with the first law in its general form, integrate over the time and consider the definition of enthalpy h and its connection to the inner energy u.
2. Relevant equations
H= E+PV, deltaH=q, E=3/2*RT, E=q+w, w=-pdeltaV3. The attempt at a solution
Open systems are new to me. Qualitatively, I assume the heat flux and volume changes work the same way that they would for a closed system? Heat flux moves from the warm marbles to the colder water. Volume decreases because temperature increases, and they are inversely related. I assume since the system is open we are working with something where w=0. Therefore heat given off/absorbed comprises the energy of the system. So we then focus on H= E+PV. Am I on the right track here?

 

[SteamKing]

1. Homework Statement
We want to heat up 200 liters of water in a bathtub from 15 ◦C to 30 ◦C. The temperature is increased by adding marble stones to the water. The initial temperature of the marble stones is 773, 15 K.
Assume that the marble is rigid, with the heat capacity cm = 0.88 J ·K−1 · g −1 and the water is incompressible, with cp = 4.2 J · K−1 · g −1 and the density ρ = 1000 kg · m−1 .
Neglect all heat losses to the environment or any change in kinetic or potential energy. After the process, the system is in balance.
Calculate the minimal mass of marble needed for the process. Approach the question making the following assumptions:
consider an instationary open system. Balance the system of marble stones and water as a whole. Hint: Begin with the first law in its general form, integrate over the time and consider the definition of enthalpy h and its connection to the inner energy u.
2. Relevant equations
H= E+PV, deltaH=q, E=3/2*RT, E=q+w, w=-pdeltaV3. The attempt at a solution
Open systems are new to me. Qualitatively, I assume the heat flux and volume changes work the same way that they would for a closed system? Heat flux moves from the warm marbles to the colder water. Volume decreases because temperature increases, and they are inversely related. I assume since the system is open we are working with something where w=0. Therefore heat given off/absorbed comprises the energy of the system. So we then focus on H= E+PV. Am I on the right track here?

If the liquid water is going to be assumed to be incompressible, then dV ≈ 0 over this small temperature change. Since it's an open system, P = constant = atmospheric.

Since all heat losses to the environment are to be neglected, then the bath tub is functioning essentially like a large calorimeter.

 

[beetlejuice]

Okay, I see how those things add up. Thank you!

What would I do to calculate the number of marbles needed to calculate that change? For the closed system, I used the first law. Does the same method still apply?

 

[SteamKing]

Okay, I see how those things add up. Thank you!

What would I do to calculate the number of marbles needed to calculate that change? For the closed system, I used the first law. Does the same method still apply?

Well, this problem has been so restricted that I think applying the First Law would be correct.