Question

Please help me. I will medal and fan.

Calculate the amount of heat released when 27.0 g H2O is cooled from a liquid at 314 K to a solid at 263 K. The melting point of H2O is 273 K. Other useful information about water is listed below.

Cp, liquid = 4.184 J/(g•K)
Cp, solid = 2.093 J/(g•K)
Hfusion = 40.7 kJ/mol

Answer 1

I know the steps of the process:
1. cool the liquid from 314K to 273K
2. the liquid freezes
3. the solid cools to 263K
...problem is I don't know how to find the amount of heat released.

If it helps, I am currently learning about Enthalpy and Phase Changes.

Answer 2

Have you diagrammed the problem to account for all the changes in state?
ex: H2O (l) @314K ----> H20(l) at 273K etc...

Answer 3

@surry99
I am very sorry. I completely forgot I even had this question open, and I never got a notification that anyone had replied to it which is odd.

Answer 4

Would this be the completed diagram of the problem?
H2O (l) at 314K ----> H20(l) at 273K ----> H20(s) ---->H20(s) at 263K

Answer 5

H2O (l) at 314K ----> H20(l) at 273K ----> H20(s) at 273K ---->H20(s) at 263K
1 2 3

ok so now do you know what equations to calculate q for each step, 1,2 and 3?

Answer 6

It looks like this right?\[q=mC _{p}\Delta T\]

Answer 7

okay that will work for steps 1 and 3 but what about step 2?