How much heat must be removed to cool 1000 lbs of water from 72 degrees F to -18 degrees F?

To determine how much heat must be removed to cool 1000 lbs of water from 72 degrees F to -18 degrees F, we need to account for both the cooling of the water from 72 degrees F to 32 degrees F (the freezing point) and then the additional heat removal needed to freeze the water and cool it down to -18 degrees F.

First, calculate the heat removed to cool the water from 72 degrees F to 32 degrees F using the specific heat of water, which is approximately 1 BTU/lb°F:

1. Cooling from 72°F to 32°F:

Temperature change = 72 - 32 = 40°F

Heat removed for cooling = mass × specific heat × temperature change

= 1000 lbs × 1 BTU/lb°F × 40°F

= 40,000 BTU

Next, we need to calculate the heat removed to freeze the water at 32 degrees F. The latent heat of fusion for water is about 144 BTU/lb:

2. Freezing the water:

Heat removed to freeze 1000 lbs of water = mass × latent heat of fusion

= 1000