How do i solve this using Thermochemistry?

Question

Determine the amount of Heat (in kJ) needed to change 72.1 g of ice (solid water) from -23.3 C to 87.9 C.  (round to hundredths place)

Given:

Heat of Vaporization of water = 2260 J/g

Heat of Fusion of water = 334 J/g

Specific Heat of H2O(s) = 2.030 J/g C

Specific Heat of H2O(l) = 4.184 J/g C

Specific Heat of H2O(g) = 2.000 J/g C

1000 J = 1 kJ

Answer 1

Hello, Emma,

We have 72.1 grams of ice at -23.3C that requires energy to reach 87.9C. The specific heat of ice is different than that of water, and we are going through a phase change from ice to water, This phase change is a third energy requirement - that of breaking apart the ice structure into a liquid. This "Heat of Fusion" occurs with no temperature change. 72.1 grams of ice at 0 C becomes 72.1 grams of water at 0 C.

So we have three heat calculations:

1. Raise the temperature of ice [H₂O_(s)] from -23.3 C to 0 C. Specific Heat of H₂O_(s) = 2.030 J/g C
2. (72.1 grams)*(2.030 J/g*C)*(0 C -(-23.3 C)) = 3410 J
3. Break the crystalline structure. Heat of Fusion of water = 334 J/g
4. (72.1 grams)*(334 J/g) = 24081 J
5. Raise the temperature of water [H₂O_(l)] from0 to 87.9 C. Specific heat of H₂O_(l) = 4.184 J/gC
6. (72.1 grams)*(4.184 J/gC)*(87.9C - 0C) = 26517 J

Now add all three heat calculations to find the energy needed to change ice into water at the temperatures indicated. That should be around 54008 J. Express it in kJ: 54.008 kJ Make it 54.o kJ for 3 sig figs.

We didn't exceed the boiling point of water, so had no need of the heat of vaporization and specific heat of steam (H₂O_(g)) for this problem. However, it is interesting to see the vast differences in the heat requirements at each step.

Bob