The amount of heat removed to lower the temperature of 15 lbs of water from 48°F to 39°F is:

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Multiple Choice

The amount of heat removed to lower the temperature of 15 lbs of water from 48°F to 39°F is:

Explanation:
When you want to know how much heat is removed to change the temperature of a liquid, multiply how much liquid you have by how much energy is needed to change the temperature of one unit of that liquid, and by the temperature change itself. For water, the specific heat is about 1 BTU per pound per degree Fahrenheit. Here, you have 15 pounds of water and you’re cooling it from 48°F to 39°F, a drop of 9°F. So the heat removed is: Q = m × Cp × ΔT = 15 lb × 1 BTU/(lb·°F) × 9°F = 135 BTU. So the total heat removed is 135 BTU. The other options either express energy per pound or use incorrect values, which would not match the total energy for the 15-pound sample.

When you want to know how much heat is removed to change the temperature of a liquid, multiply how much liquid you have by how much energy is needed to change the temperature of one unit of that liquid, and by the temperature change itself. For water, the specific heat is about 1 BTU per pound per degree Fahrenheit.

Here, you have 15 pounds of water and you’re cooling it from 48°F to 39°F, a drop of 9°F. So the heat removed is:

Q = m × Cp × ΔT = 15 lb × 1 BTU/(lb·°F) × 9°F = 135 BTU.

So the total heat removed is 135 BTU. The other options either express energy per pound or use incorrect values, which would not match the total energy for the 15-pound sample.

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