How many BTUs are removed from 2000 lbs. of ice at 32 F to give ice at 28 F?

To determine how many BTUs are removed from 2000 lbs. of ice when cooling it from 32°F to 28°F, it is essential to consider the specific heat of ice. The specific heat of ice is 0.5 BTU/lb°F.

First, calculate the temperature change, or delta T, which is the difference between the initial and final temperatures. In this case, the temperature change is 32°F - 28°F, resulting in a delta T of 4°F.

Next, to find the total BTUs removed, use the formula:

BTUs = weight (lbs) × specific heat (BTU/lb°F) × temperature change (°F).

Substituting the known values:

BTUs = 2000 lbs × 0.5 BTU/lb°F × 4°F.

BTUs = 2000 × 0.5 × 4 = 4000 BTUs.

This calculation shows that 4000 BTUs are removed from the ice to cool it from 32°F to 28°F, making this the correct answer.