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

specific heat of ice =  05  

specific heat of water  = 1.0  

specific heat of steam = 0.5  

latent heat of melting = 144  

latent heat of boiling = 970  

To calculate the total number of BTUs required to change 100 lbs of steam at 275°F to 100 lbs of ice at 22°F, we need to account for the following steps:

1. **Cool steam from 275°F to 212°F** (steam cooling down to the condensation point).
2. **Condense the steam at 212°F to water**.
3. **Cool water from 212°F to 32°F** (cooling the water to the freezing point).
4. **Freeze the water at 32°F to ice**.
5. **Cool the ice from 32°F to 22°F** (cooling the ice).

### Step-by-Step Calculation:

#### 1. Cool the steam from 275°F to 212°F (steam cooling):
The formula for heat transfer is:
\[
Q = m \cdot C_s \cdot \Delta T
\]
Where:
- \(m = 100 \, \text{lbs}\) (mass)
- \(C_s = 0.5 \, \text{BTU/lb·°F}\) (specific heat of steam)
- \(\Delta T = 275°F - 212°F = 63°F\)

\[
Q_1 = 100 \cdot 0.5 \cdot 63 = 3,150 \, \text{BTUs}
\]

#### 2. Condense the steam at 212°F to water (latent heat of condensation):
The latent heat of condensation (same as latent heat of boiling) is 970 BTU/lb.
\[
Q_2 = m \cdot \text{latent heat} = 100 \cdot 970 = 97,000 \, \text{BTUs}
\]

#### 3. Cool the water from 212°F to 32°F (water cooling):
\[
Q = m \cdot C_w \cdot \Delta T
\]
Where:
- \(C_w = 1.0 \, \text{BTU/lb·°F}\) (specific heat of water)
- \(\Delta T = 212°F - 32°F = 180°F\)

\[
Q_3 = 100 \cdot 1.0 \cdot 180 = 18,000 \, \text{BTUs}
\]

#### 4. Freeze the water at 32°F to ice (latent heat of fusion):
The latent heat of melting (fusion) is 144 BTU/lb.
\[
Q_4 = m \cdot \text{latent heat} = 100 \cdot 144 = 14,400 \, \text{BTUs}
\]

#### 5. Cool the ice from 32°F to 22°F (ice cooling):
\[
Q = m \cdot C_i \cdot \Delta T
\]
Where:
- \(C_i = 0.5 \, \text{BTU/lb·°F}\) (specific heat of ice)
- \(\Delta T = 32°F - 22°F = 10°F\)

\[
Q_5 = 100 \cdot 0.5 \cdot 10 = 500 \, \text{BTUs}
\]

### Total Heat Transfer:
The total number of BTUs required is the sum of all the steps:
\[
Q_{\text{total}} = Q_1 + Q_2 + Q_3 + Q_4 + Q_5
\]
\[
Q_{\text{total}} = 3,150 + 97,000 + 18,000 + 14,400 + 500 = 133,050 \, \text{BTUs}
\]

So, **133,050 BTUs** are required to change 100 lbs of steam at 275°F to 100 lbs of ice at 22°F.

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Would you like more details on any part of the solution?

Here are 5 related questions for further practice:

1. How much heat is required to melt 50 lbs of ice at 32°F?
2. How many BTUs are required to vaporize 200 lbs of water at 212°F?
3. Calculate the energy needed to cool 150 lbs of steam from 300°F to 100°F.
4. How much heat is required to heat 80 lbs of ice at -10°F to water at 50°F?
5. What is the total energy required to convert 100 lbs of ice at -20°F to steam at 300°F?

**Tip:** Always pay attention to the different stages (phase changes and temperature changes) when calculating heat transfer.

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

Discover how to calculate the total BTUs required to change 100 lbs of steam at 275°F into ice at 22°F. This detailed thermodynamics problem involves phase changes and heat transfer, applying specific heat and latent heat formulas. Perfect for college-level students studying heat transfer and thermodynamics.

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