Hello, Bri,
The ice cube goes through three steps as the temperature goes from -11.6 C to 32.0 C. The ice warms from -11.6 C to 0 C. The ice then melts at 0 C. The temperature does not change until all the ice is melted. The is a phase change from solid to liquid (the ice melts). Then the liquid water goes from 0 C to 32.0 C. Heat is being absorbed in all three steps, but the three calculations require different specific heats.
We have 12.4 g of ice.
Make a table to keep track of the steps. Note that the specific heats are all different, depending on which phase to water is going through. The unit of specific heat tells us that a gram of solid water (ice) will absorb 2.087 Joules for every 1°C temperature rise. Water's specific heat of fusion does not have a unit for temperature. That is telling us that water will absorb 333.6 Joules per gram just to melt, with no temperature change. The metric term for this is "Zowie!" When the phase change to liquid water is complete, we use the specific heat of liquid water, which is 4.184 J/g °C.
To answer the question. Multiply the grams of water (12.4 g) times the specific heat for each step. We must also use the temperature change (Tf -Ti) for any calculation in which the specific heat has a °C temperature unit. All but the phase change of fusion require that we use the temperature change.
For example, we'll calculate the heat absorbed by the ice before it starts melting:
A. Ice: Raising the temperature of Ice from - 11.6 °C to 0.0 °C
(Specific Heat of Ice)*(mass)*(Tf - Ti) = Heat absorbed
(2.087 J/g)*(12.4 g)*( 0°C - (-11.6°C )) = 300 Joules
B. Heat of Fusion (melting, with no temperature change)
(333.6 J/g)*(12.4 g) = 4137 Joules [the lattice structure of ice absorbs a lot of energy, compared to the other steps]
C. Heating the Water from 0 °C to 32.0 °C:
(4.184 J/g°C)*(12.4g)*(32.0 °C - 0 °C) = 1660.2 Joules.
Now add the three numbers to find the total heat absorbed by 12.4 grams of ice from -11.6 °C to 32.0 °C.
I get 6097 Joules. 6100 Joules with 3 sig figs.
Bob
