MODERN ALCHEMIST

Ice Cubes, Water, Steam

We will discuss five processes necessary to heat 10.000 g of ice at -20.000 C to 120.00 C steam.

This problem will make use of the following:

Specific Heat of ice
Latent Heat of Fusion to convert ice to water
Specific Heat of water
Latent heat of Vaporization to convert water to steam
Specific Heat of steam

The specific heat of ice is

     cal
0.48 ---
     g K

Given the quantity of ice in grams, and the temperature change, one can calculate the energy utilized.

                      0.48 cal
10.000 g * 20.000 K * -------- = 96 cal
                         g K

Energy is required to convert the ice to water. The energy required to convert a solid to a liquid, L_F, is called the latent heat of fusion. For water at atmospheric pressure, L_F is 80 cal/g.

           80 cal 
10.000 g * ------ = 1600 cal (2 sig figs)
             g

The specific heat of water in the liquid state is

    cal
1.0 ---
    g K

Thus to heat 10.000 g of water from 0 C to 100 C we calculate:

                   1.0 cal
10.000 g * 100 K * ------- = 1000 cal (2 sig figs)
                     g K

To calculate the energy required to convert the water to steam, we use the latent heat of vaporization, L_V, 540 cal/g.

           540 cal
10.000 g * ------- = 5400 cal
              g

For the final stage, heating the steam to 120 C, we use the specific heat of steam, [I need to find the constant before I can finish the problem.]

Using the Heat and Specific Heat page at Austin Community College, we find that the specific heat of steam is 0.475.

Now, we need units for this number or we can't do the calculation. In looking at their other numbers, they list 0.48 for the specific heat of ice, and we know the specific heat of ice is 0.48 cal g^-1 K^-1.

So we assume, and we could be wrong, that the specific heat of steam is 0.47 cal g^-1 K^-1.

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