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?

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?

• A. 1,500,000 BTU
• B. 2,205,000 BTU
• C. 3,000,000 BTU
• D. 750,000 BTU

Answer: (B)

To determine the amount of BTUs that must be removed to cool 100 animals from 85 °F to 45 °F, we first need to calculate the total weight of all the animals and then the heat removal required for cooling. 1. **Calculate the Total Weight of the Animals**: Each animal weighs 750 pounds, and there are 100 animals. Therefore, the total weight is: \[ \text{Total Weight} = 100 \text{ animals} \times 750 \text{ lbs/animal} = 75,000 \text{ lbs} \] 2. **Determine the Temperature Difference**: The cooling need is from 85 °F to 45 °F, which gives a temperature difference of: \[ \Delta T = 85 °F - 45 °F = 40 °F \] 3. **Calculate Total BTUs Needed**: The amount of BTUs required to lower the temperature of one pound of a substance by one degree Fahrenheit is known as a BTU. For this calculation, we consider the specific heat of body tissue, which is approximately 1 BTU/lb/°F. Therefore

To determine the amount of BTUs that must be removed to cool 100 animals from 85 °F to 45 °F, we first need to calculate the total weight of all the animals and then the heat removal required for cooling.

1. Calculate the Total Weight of the Animals:

Each animal weighs 750 pounds, and there are 100 animals. Therefore, the total weight is:

[

\text{Total Weight} = 100 \text{ animals} \times 750 \text{ lbs/animal} = 75,000 \text{ lbs}

]

2. Determine the Temperature Difference:

The cooling need is from 85 °F to 45 °F, which gives a temperature difference of:

[

\Delta T = 85 °F - 45 °F = 40 °F

]

3. Calculate Total BTUs Needed:

The amount of BTUs required to lower the temperature of one pound of a substance by one degree Fahrenheit is known as a BTU. For this calculation, we consider the specific heat of body tissue, which is approximately 1 BTU/lb/°F. Therefore