What heat is needed to change 10 pounds of ice at 32°F into water at 80°F?

To determine the heat needed to change 10 pounds of ice at 32°F into water at 80°F, we need to calculate the heat required for two separate processes: the phase change from ice to water, and the heating of that water from 32°F to 80°F.

First, we consider the latent heat of fusion, which is the energy required to convert ice at its melting point (32°F) to water at the same temperature. The latent heat of fusion for water is approximately 144 Btu per pound. For 10 pounds of ice:

Heat required for phase change = mass × latent heat of fusion

= 10 lbs × 144 Btu/lb = 1440 Btu.

Next, we calculate the sensible heat required to raise the temperature of the resulting water from 32°F to 80°F. The specific heat of water is about 1 Btu/lb°F. The temperature change here is from 32°F to 80°F, which is a difference of 48°F:

Heat required to raise the temperature = mass × specific heat × temperature change

= 10 lbs × 1 Btu/lb°F × 48°F = 480 Btu.

Now, we add both