Calculate the number of BTU’s required to change 100 lbs. of ice at 22°F to 100 lbs. of steam at 275°F

To calculate the total number of BTUs required to change 100 lbs of ice at 22°F to 100 lbs of steam at 275°F, we need to consider several stages:

1. **Heating the ice from 22°F to 32°F** (the melting point of ice).
2. **Melting the ice at 32°F** to water at 32°F.
3. **Heating the water from 32°F to 212°F** (the boiling point of water).
4. **Converting the water at 212°F to steam at 212°F**.
5. **Heating the steam from 212°F to 275°F**.

### Step 1: Heating the Ice from 22°F to 32°F

The specific heat of ice is about 0.5 BTU/lb/°F.

\[
\text{BTUs} = \text{mass} \times \text{specific heat of ice} \times \text{temperature change}
\]

\[
\text{BTUs} = 100 \, \text{lbs} \times 0.5 \, \text{BTU/lb/°F} \times (32°F - 22°F) = 100 \times 0.5 \times 10 = 500 \, \text{BTUs}
\]

### Step 2: Melting the Ice at 32°F

The latent heat of fusion (the energy required to convert ice at 32°F to water at 32°F) is 144 BTU/lb.

\[
\text{BTUs} = \text{mass} \times \text{latent heat of fusion} = 100 \, \text{lbs} \times 144 \, \text{BTU/lb} = 14,400 \, \text{BTUs}
\]

### Step 3: Heating the Water from 32°F to 212°F

The specific heat of water is 1 BTU/lb/°F.

\[
\text{BTUs} = \text{mass} \times \text{specific heat of water} \times \text{temperature change}
\]

\[
\text{BTUs} = 100 \, \text{lbs} \times 1 \, \text{BTU/lb/°F} \times (212°F - 32°F) = 100 \times 1 \times 180 = 18,000 \, \text{BTUs}
\]

### Step 4: Converting the Water at 212°F to Steam at 212°F

The latent heat of vaporization (the energy required to convert water at 212°F to steam at 212°F) is 970 BTU/lb.

\[
\text{BTUs} = \text{mass} \times \text{latent heat of vaporization} = 100 \, \text{lbs} \times 970 \, \text{BTU/lb} = 97,000 \, \text{BTUs}
\]

### Step 5: Heating the Steam from 212°F to 275°F

The specific heat of steam is about 0.48 BTU/lb/°F.

\[
\text{BTUs} = \text{mass} \times \text{specific heat of steam} \times \text{temperature change}
\]

\[
\text{BTUs} = 100 \, \text{lbs} \times 0.48 \, \text{BTU/lb/°F} \times (275°F - 212°F) = 100 \times 0.48 \times 63 = 3,024 \, \text{BTUs}
\]

### Total BTUs Required

Adding all the stages together:

\[
\text{Total BTUs} = 500 \, \text{BTUs} + 14,400 \, \text{BTUs} + 18,000 \, \text{BTUs} + 97,000 \, \text{BTUs} + 3,024 \, \text{BTUs} = 132,924 \, \text{BTUs}
\]

So, **132,924 BTUs** are required to change 100 lbs of ice at 22°F to 100 lbs of steam at 275°F.

Would you like further details or have any questions?

### Related Questions
1. How does the latent heat of fusion differ from the latent heat of vaporization?
2. Why is the specific heat of steam lower than that of water?
3. What is the effect of atmospheric pressure on the boiling point of water?
4. How does the specific heat of a substance influence the energy required to heat it?
5. How would the BTU calculation change if the initial temperature of the ice was below 22°F?

### Tip
Remember, the total energy required for a phase change includes both sensible heat (for temperature change) and latent heat (for phase change).

Calculate the number of BTU’s required to change 100 lbs. of ice at 22°F to 100 lbs. of steam at 275°F.

Explore the step-by-step calculation of BTUs required to convert 100 lbs of ice at 22°F to steam at 275°F. This thermodynamic problem covers heat transfer, phase changes, and specific heat calculations. Suitable for students in Grades 10-12.

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