How many Btus are required to change five pounds of ice at 20°F to steam at 220°F?

Question

Quizplus · Accepted Answer

To solve this, we need to calculate the energy required for each phase change and temperature change for the ice to become steam at 220°F. The steps are as follows:1. **Raising the temperature of ice from 20°F to 32°F (melting point):**    - The specific heat of ice is approximately 0.5 Btu/(lb·°F).    - $Q = m \cdot c \cdot \Delta T = 5 \, \text{lb} \cdot 0.5 \, \frac{\text{Btu}}{\text{lb}·°\text{F}} \cdot (32°F - 20°F) = 30 \, \text{Btu}$.2. **Melting the ice to water at 32°F:**    - The heat of fusion of ice is about 144 Btu/lb.    - $Q = m \cdot \text{latent heat} = 5 \, \text{lb} \cdot 144 \, \frac{\text{Btu}}{\text{lb}} = 720 \, \text{Btu}$.3. **Raising the temperature of water from 32°F to 212°F (boiling point):**    - The specific heat of water is approximately 1 Btu/(lb·°F).    - $Q = m \cdot c \cdot \Delta T = 5 \, \text{lb} \cdot 1 \, \frac{\text{Btu}}{\text{lb}·°\text{F}} \cdot (212°F - 32°F) = 900 \, \text{Btu}$.4. **Turning the water into steam at 212°F:**    - The heat of vaporization of water is about 970 Btu/lb.    - $Q = m \cdot \text{latent heat} = 5 \, \text{lb} \cdot 970 \, \frac{\text{Btu}}{\text{lb}} = 4850 \, \text{Btu}$.5. **Raising the temperature of steam from 212°F to 220°F:**    - The specific heat of steam is approximately 0.48 Btu/(lb·°F).    - $Q = m \cdot c \cdot \Delta T = 5 \, \text{lb} \cdot 0.48 \, \frac{\text{Btu}}{\text{lb}·°\text{F}} \cdot (220°F - 212°F) = 20 \, \text{Btu}$.Adding all these energies together: $30 \, \text{Btu} + 720 \, \text{Btu} + 900 \, \text{Btu} + 4850 \, \text{Btu} + 20 \, \text{Btu} = 6520 \, \text{Btu}$.

How Many Btus Are Required to Change Five Pounds of Ice