Calculate the number of BTU's required to change 645 lbs of ice at 25 degrees to 645 lbs of steam at 295 degrees

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Answer to a math question Calculate the number of BTU's required to change 645 lbs of ice at 25 degrees to 645 lbs of steam at 295 degrees

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To calculate the total BTUs required for this change, we need to consider the following steps: 1. **Heating the ice from 25°F to 32°F (melting point):** * Specific heat of ice: 0.49 BTU/lb°F * Temperature change: 32°F - 25°F = 7°F * Heat required: 645 lbs * 0.49 BTU/lb°F * 7°F = 2221.05 BTU 2. **Melting the ice at 32°F:** * Latent heat of fusion of ice: 144 BTU/lb * Heat required: 645 lbs * 144 BTU/lb = 92880 BTU 3. **Heating the water from 32°F to 212°F (boiling point):** * Specific heat of water: 1 BTU/lb°F * Temperature change: 212°F - 32°F = 180°F * Heat required: 645 lbs * 1 BTU/lb°F * 180°F = 116100 BTU 4. **Vaporizing the water at 212°F:** * Latent heat of vaporization of water: 970 BTU/lb * Heat required: 645 lbs * 970 BTU/lb = 625650 BTU 5. **Heating the steam from 212°F to 295°F:** * Specific heat of steam: 0.48 BTU/lb°F (approximate value) * Temperature change: 295°F - 212°F = 83°F * Heat required: 645 lbs * 0.48 BTU/lb°F * 83°F = 25581.6 BTU **Total Heat Required:** Add up the heat required in each step: 2221.05 BTU + 92880 BTU + 116100 BTU + 625650 BTU + 25581.6 BTU = 862432.65 BTU **Therefore, approximately 862,433 BTUs are required to change 645 lbs of ice at 25°F to 645 lbs of steam at 295°F.**

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