Final answer:
To freeze five pounds of water at 50°F, a total of 810 BTUs must be removed; 90 BTUs to cool the water to 32°F and 720 BTUs for the phase change from liquid to solid.
Step-by-step explanation:
To determine how many BTUs must be removed to freeze five pounds of water at 50°F, we first need to know the specific heat of water and the heat of fusion for ice. The specific heat of water is 1 BTU/(lb°F), meaning it takes 1 BTU to lower the temperature of 1 pound of water by 1°F. To bring the water from 50°F to the freezing point (32°F), we would remove 18 BTUs per pound of water (since 50°F - 32°F = 18°F), which for five pounds equals 90 BTUs (5 x 18 BTUs).
Next, the heat of fusion of ice is about 144 BTUs per pound, which is the quantity of heat to be removed to convert one pound of water at 32°F to ice at the same temperature. Therefore, for five pounds, we need to remove 720 BTUs (5 x 144 BTUs).
In total, to freeze the entire five pounds of water initially at 50°F, you would need to remove 90 + 720 = 810 BTUs.