Thermodynamics of a perfect liquid, help please

[kawf]

Homework Statement

Water, assumed to be a perfect liquid, has a density of 1000 kg/m3 and a specific heat of 4184 J/kgK. The water undergoes an adiabatic, frictionless process in which its pressure is raised from 200 kPa to 1500 kPa. Find:
1. change in temperature
2. change in specific internal energy
3. change in specific enthalpy
4. temperature change required for isobaric process to change the enthalpy that same amount as in part 3

Homework Equations

ds = c ln(P2/P1)
Tds = de
Tds = dh

The Attempt at a Solution

I have done the following, is this correct logic?

1. ds = c ln(P2/P1) = 4184 ln(1500/200) = 8430
Tds=cdT => dT = Tds/c = (298*8430)/4184 = 600K
so change in temp dT = 600K

2. Tds = de
de = 298 * 8430 = 2512 kJ/kg

3. Tds = dh
dh = 298 * 8430 = 2512 kJ/kg

4. I am not quite sure how to do this part.

Any help or suggestions are appreciated.

 

[KataKoniK]

Thank you for your time.

Hello! Your approach for parts 1-3 looks good. For part 4, you can use the formula for enthalpy change in an isobaric process, which is dh = c(T2-T1). You already have the change in enthalpy from part 3 (2512 kJ/kg), so you can plug that in and solve for the change in temperature. This will give you the temperature change required for an isobaric process to produce the same change in enthalpy as in part 3. Hope this helps!

 

[Mene]

Thank you.



Your approach to the first three parts is correct. For the fourth part, you can use the equation for enthalpy change in an isobaric process: dH = Cp*dT, where Cp is the specific heat at constant pressure. Since the process is adiabatic and frictionless, the enthalpy change will be the same as the internal energy change. So, you can set the two equations equal to each other and solve for the temperature change (dT) required to produce the same change in enthalpy as in part 3. This will give you the temperature change required for an isobaric process to produce the same enthalpy change as in the adiabatic process.