HW 4.1

Moderators: Chem_Mod, Chem_Admin

Baoying Li 1B
Posts: 113
Joined: Sat Aug 17, 2019 12:18 am

HW 4.1

Postby Baoying Li 1B » Fri Feb 07, 2020 9:06 pm

For the Homework Problem 4.1, when we raise the temperature of the ice cube from -5.042 Celcius to 0.00 Celcius, why do we multiply the mass of the ice cube by the specific heat capacity 2.03 J per Celcius per grams instead of the specific heat of water (4.18 Celcius)?

Angus Wu_4G
Posts: 102
Joined: Fri Aug 02, 2019 12:15 am

Re: HW 4.1

Postby Angus Wu_4G » Fri Feb 07, 2020 9:21 pm

Water, depending on solid, liquid, or gas state, has different specific heats. 4.184J/gC is the specific heat capacity for liquid water. For ice, solid water, the specific heat capacity is 2.03J/gC

Posts: 102
Joined: Sat Jul 20, 2019 12:16 am

Re: HW 4.1

Postby Rodrigo2J » Fri Feb 07, 2020 9:33 pm

The problem tells you that you are raising the temperature of ice. Because Ice (H2O(s)) is the solid form of water, it will have a different heat capacity than H2O(l). You need to use the specific heat of ice which is c= 2.03J/g*c.

Osvaldo SanchezF -1H
Posts: 122
Joined: Wed Sep 18, 2019 12:21 am

Re: HW 4.1

Postby Osvaldo SanchezF -1H » Fri Feb 07, 2020 10:50 pm

Water has different specific heat depending the state that it is in. 4.18 is just for when it is in liquid form. But the problem says ice which is water in its solid form which would be 2.03. Take this into account for future problems.

Return to “Phase Changes & Related Calculations”

Who is online

Users browsing this forum: No registered users and 1 guest