HW 4.1

[Baoying Li 1B]

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]

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

[Rodrigo2J]

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]

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.