How much heat in BTUs is required to melt a 20-pound block of ice from 4°F to 45°F?

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Multiple Choice

How much heat in BTUs is required to melt a 20-pound block of ice from 4°F to 45°F?

Explanation:
To determine the amount of heat required to melt a 20-pound block of ice from 4°F to 45°F, we need to consider the heat needed for two primary processes: melting the ice and then warming the resulting water. First, we must calculate the heat required to melt the ice. The latent heat of fusion for ice is approximately 144 BTU per pound. Therefore, to melt 20 pounds of ice: \[ 20 \text{ pounds} \times 144 \text{ BTU/pound} = 2880 \text{ BTU} \] Next, we need to raise the temperature of the resulting water from 32°F (the melting point of ice) to 45°F. The specific heat capacity of water is about 1 BTU per pound per degree Fahrenheit. Thus, the temperature increase here is: \[ 45°F - 32°F = 13°F \] The heat required to raise the temperature of the water is calculated as follows: \[ 20 \text{ pounds} \times 1 \text{ BTU/pound/°F} \times 13°F = 260 \text{ BTU} \] Now, we add the heat needed

To determine the amount of heat required to melt a 20-pound block of ice from 4°F to 45°F, we need to consider the heat needed for two primary processes: melting the ice and then warming the resulting water.

First, we must calculate the heat required to melt the ice. The latent heat of fusion for ice is approximately 144 BTU per pound. Therefore, to melt 20 pounds of ice:

[

20 \text{ pounds} \times 144 \text{ BTU/pound} = 2880 \text{ BTU}

]

Next, we need to raise the temperature of the resulting water from 32°F (the melting point of ice) to 45°F. The specific heat capacity of water is about 1 BTU per pound per degree Fahrenheit. Thus, the temperature increase here is:

[

45°F - 32°F = 13°F

]

The heat required to raise the temperature of the water is calculated as follows:

[

20 \text{ pounds} \times 1 \text{ BTU/pound/°F} \times 13°F = 260 \text{ BTU}

]

Now, we add the heat needed

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