How many BTUs must be removed to lower the temperature of 2000 lbs of ice from 32 degrees F to 29 degrees F?

To determine how many BTUs must be removed to lower the temperature of 2000 lbs of ice from 32 degrees F to 29 degrees F, it's important to consider the specific heat of ice and the amount of mass involved.

The specific heat capacity of ice is approximately 0.5 BTU/lb°F. This means that it takes 0.5 BTUs of energy to lower the temperature of one pound of ice by one degree Fahrenheit.

In this scenario, we are lowering the temperature of 2000 lbs of ice by 3 degrees (from 32°F to 29°F). Therefore, the amount of heat energy removed can be calculated using the formula:

Heat Removed (in BTUs) = Mass (in lbs) × Specific Heat (in BTU/lb°F) × Temperature Change (in °F).

Plugging in the numbers:

Heat Removed = 2000 lbs × 0.5 BTU/lb°F × 3°F

Heat Removed = 2000 × 0.5 × 3 = 3000 BTUs.

Thus, the correct answer is 3000 BTUs. This calculation correctly accounts for both the mass of the ice and the amount of temperature decrease, leading to