How much heat must be removed to convert 10 lb. of steam at 218°F to ice at 12°F?

To determine how much heat must be removed to convert steam at 218°F to ice at 12°F, a series of calculations needs to be made. This involves considering different stages in the phase change and cooling process.

First, you need to condense the steam into water at 218°F. This requires removing the latent heat of vaporization, which for water is approximately 1,000 BTUs per pound. For 10 pounds of steam, this amounts to 10,000 BTUs.

Next, the water must be cooled from 218°F to 32°F (the freezing point of water). The specific heat of water is about 1 BTU per pound per degree Fahrenheit. To calculate the heat removal needed for this cooling, you would use the temperature difference (218°F - 32°F), which is 186°F. Thus, for 10 pounds of water, the heat removal is 10 pounds × 186°F × 1 BTU/lb°F = 1,860 BTUs.

Then, once the temperature reaches 32°F, the water needs to freeze into ice. This involves removing another latent heat of fusion, which is approximately 144 BTUs per pound. For 10 pounds, this