How many Btu’s must be removed from one pound of water at 200°F for it to end up as ice at 30°F?
a. 144 Btu
b. 828 Btu
c. 313 Btu
d. 526 Btu
The Correct Answer and Explanation is:
Correct Answer: c. 313 Btu
To determine how many Btu’s must be removed from one pound of water at 200°F for it to become ice at 30°F, we need to account for the following steps:
- Cooling the water from 200°F to 32°F (its freezing point): This step involves reducing the temperature of the water without changing its phase.
- Freezing the water at 32°F: This step involves the phase change from liquid water to ice, which releases latent heat.
- Cooling the ice from 32°F to 30°F: Finally, we cool the ice down by 2°F.
Step 1: Cooling water from 200°F to 32°F
The specific heat of liquid water is 1 Btu per pound per °F. To find the heat required to cool the water, we use the formula:
[
Q = mc\Delta T
]
where:
- ( Q ) is the heat removed in Btu,
- ( m ) is the mass in pounds (1 lb),
- ( c ) is the specific heat (1 Btu/lb°F for liquid water),
- ( \Delta T ) is the change in temperature (200°F – 32°F = 168°F).
[
Q = 1 \times 1 \times 168 = 168 \text{ Btu}
]
Step 2: Freezing the water at 32°F
The latent heat of fusion for water is approximately 144 Btu per pound. This is the amount of heat that must be removed for water to freeze.
[
Q = 144 \text{ Btu}
]
Step 3: Cooling the ice from 32°F to 30°F
The specific heat of ice is 0.5 Btu per pound per °F. The temperature change here is 2°F (32°F – 30°F). The heat removed is:
[
Q = 1 \times 0.5 \times 2 = 1 \text{ Btu}
]
Total Heat Removed
Now, we sum the heat removed in each step:
[
Q_{\text{total}} = 168 \text{ Btu (cooling water)} + 144 \text{ Btu (freezing)} + 1 \text{ Btu (cooling ice)} = 313 \text{ Btu}
]