Enthalpy change for supercooled water freezing at 258 K

[Joel Martis]

1.At constant temperature of 258K and constant pressure of 1.000 atm , super cooled liquid water (2.00mol) freezes irreversibly. Find the change in enthalpy(ΔH) by assuming that C_p,m of liquid water and ice are 75.48 KJ/mol and 37.15KJ/mol and are constant. (this problem was from my assignment sheet so there might be errors in the units of C_p,m. By the way , C_p,m is the specific heat capacity at constant pressure.)



2.the only equation i could think of is ΔH=ΔU+ΔPV

 

[Chestermiller]

Hi Joel Martis. Welcome to Physics Forums.

In terms of the enthalpy and temperature, what is the defining equation for the heat capacity at constant pressure?

Chet

 

[Joel Martis]

I think it is C_p,m= (∂H/∂T)_p

 

[Chestermiller]

I think it is C_p,m= (∂H/∂T)_p

Good. So now you have to define a path form the starting point to the final point along which you know all the properties needed to calculate the ΔH. The starting point is liquid water at -15C, and the end point is ice at -15 C. But, you only know the heat of melting/fusion at 0C. So, how do you get form the starting point to the final point if you only know the heat of fusion/melting at 0 C, and the heat capacities for the liquid water and the water ice?

Chet

 

[Joel Martis]

Well, how about heating the supercooled water from 258K to 273K , freezing it, and then cooling the ice back to 258K?

 

[Chestermiller]

Well, how about heating the supercooled water from 258K to 273K , freezing it, and then cooling the ice back to 258K?

Sounds great.

Chet