To drop the temperature of 20 pounds of water from 85 degrees F to 45 degrees F, how many BTU's need to be removed?

To determine how many BTUs need to be removed to lower the temperature of water, the formula used is:

[

\text{BTUs} = \text{Weight} \times \text{Temperature Change} \times \text{Specific Heat}

]

In this situation, the weight of the water is 20 pounds, the initial temperature is 85 degrees F, and the final temperature is 45 degrees F. The specific heat of water is about 1 BTU per pound per degree Fahrenheit.

First, calculate the temperature change:

[

\text{Temperature Change} = \text{Initial Temperature} - \text{Final Temperature} = 85 - 45 = 40 \text{ degrees F}

]

Now, plug the values into the formula:

[

\text{BTUs} = 20 \text{ lbs} \times 40 \text{ degrees F} \times 1 \text{ BTU/lb°F}

]

This simplifies to:

[

\text{BTUs} = 20 \times 40 = 800 \text{ BTUs}

]

Therefore, to lower the temperature of 20 pounds of water from 85 degrees F to