How many BTUs must be removed to cool 100 animals from 85 °F to 45 °F if the average weight per animal is 750 pounds?

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Multiple Choice

How many BTUs must be removed to cool 100 animals from 85 °F to 45 °F if the average weight per animal is 750 pounds?

To determine how much heat to remove, multiply the total mass by the temperature change and by the material’s specific heat. For animal tissue, a simple approximation uses a specific heat of about 1 BTU per pound per degree Fahrenheit. Total mass = 100 animals × 750 lb each = 75,000 lb. Temperature drop = 85°F − 45°F = 40°F. Heat to remove = 75,000 lb × 1 BTU/lb·°F × 40°F = 3,000,000 BTU. So, 3,000,000 BTU must be removed.

To determine how much heat to remove, multiply the total mass by the temperature change and by the material’s specific heat. For animal tissue, a simple approximation uses a specific heat of about 1 BTU per pound per degree Fahrenheit.

Total mass = 100 animals × 750 lb each = 75,000 lb.

Temperature drop = 85°F − 45°F = 40°F.

Heat to remove = 75,000 lb × 1 BTU/lb·°F × 40°F = 3,000,000 BTU.

So, 3,000,000 BTU must be removed.

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