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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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
Dexter
4.7114 Answers
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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