Raising the Temperature of Ice

Oct 13, 2025 by recoveryshake.com 99 views

How Many Btus Are Required to Change One Pound of Ice at $20^{\circ} F$ to Steam at $220^{\circ} F$ ?

Understanding the Problem

To solve this problem, we need to consider the latent heat of fusion and the latent heat of vaporization of water. The latent heat of fusion is the energy required to change a substance from a solid to a liquid, while the latent heat of vaporization is the energy required to change a substance from a liquid to a gas.

Latent Heat of Fusion

The latent heat of fusion of water is approximately 144 Btu/lb. This means that 144 Btu are required to change one pound of ice at $32^{\circ} F$ to water at $32^{\circ} F$ . However, in this problem, we are starting with ice at $20^{\circ} F$ , which is below the freezing point of water. To change the ice to water, we need to add energy to raise the temperature of the ice to $32^{\circ} F$ and then change it to water.

Raising the Temperature of Ice

To raise the temperature of one pound of ice from $20^{\circ} F$ to $32^{\circ} F$ , we need to add energy. The specific heat capacity of ice is approximately 0.5 Btu/lb- $^{\circ} F$ . Therefore, the energy required to raise the temperature of one pound of ice from $20^{\circ} F$ to $32^{\circ} F$ is:

Q = mc\Delta T = 1 \, \text{lb} \times 0.5 \, \text{Btu/lb-$^{\circ} F$} \times (32^{\circ} F - 20^{\circ} F) = 6 \, \text{Btu}

Changing Ice to Water

Now that we have raised the temperature of the ice to $32^{\circ} F$ , we can change it to water. The latent heat of fusion of water is approximately 144 Btu/lb. Therefore, the energy required to change one pound of ice at $32^{\circ} F$ to water at $32^{\circ} F$ is:

Q = mL = 1 \, \text{lb} \times 144 \, \text{Btu/lb} = 144 \, \text{Btu}

Latent Heat of Vaporization

Now that we have water at $32^{\circ} F$ , we need to change it to steam at $220^{\circ} F$ . The latent heat of vaporization of water is approximately 970 Btu/lb. Therefore, the energy required to change one pound of water at $32^{\circ} F$ to steam at $220^{\circ} F$ is:

Q = mL = 1 \, \text{lb} \times 970 \, \text{Btu/lb} = 970 \, \text{Btu}

Raising the Temperature of Steam

Finally, we need to raise the temperature of the steam from $212^{\circ} F$ to $220^{\circ} F$ . The specific heat capacity of steam is approximately 0.48 Btu/lb- $^{\circ} F$ . Therefore, the energy required to raise the temperature of one pound of steam from $212^{\circ} F$ to $220^{\circ} F$ is:

Q = mc\Delta T = 1 \, \text{lb} \times 0.48 \, \text{Btu/lb-$^{\circ} F$} \times (220^{\circ} F - 212^{\circ} F) = 6.4 \, \text{Btu}

Total Energy Required

The total energy required to change one pound of ice at $20^{\circ} F$ to steam at $220^{\circ} F$ is the sum of the energies required for each step:

Q_{\text{total}} = Q_{\text{raise temperature}} + Q_{\text{latent heat of fusion}} + Q_{\text{latent heat of vaporization}} + Q_{\text{raise temperature of steam}}

Q_{\text{total}} = 6 \, \text{Btu} + 144 \, \text{Btu} + 970 \, \text{Btu} + 6.4 \, \text{Btu} = 1126.4 \, \text{Btu}

However, this is not the correct answer. We need to consider the energy required to change the ice to water at $32^{\circ} F$ and then to steam at $220^{\circ} F$ . The correct answer is:

Q_{\text{total}} = Q_{\text{raise temperature}} + Q_{\text{latent heat of fusion}} + Q_{\text{latent heat of vaporization}} + Q_{\text{raise temperature of steam}}

Q_{\text{total}} = 6 \, \text{Btu} + 144 \, \text{Btu} + 970 \, \text{Btu} + 6.4 \, \text{Btu} + 6 \, \text{Btu} + 144 \, \text{Btu} + 970 \, \text{Btu} + 6.4 \, \text{Btu}

Q_{\text{total}} = 3208 \, \text{Btu}

The final answer is: $\boxed{3208}$
Q&A: How Many Btus Are Required to Change One Pound of Ice at $20^{\circ} F$ to Steam at $220^{\circ} F$ ?

Q: What is the latent heat of fusion of water?

A: The latent heat of fusion of water is approximately 144 Btu/lb. This means that 144 Btu are required to change one pound of ice at $32^{\circ} F$ to water at $32^{\circ} F$ .

Q: What is the specific heat capacity of ice?

A: The specific heat capacity of ice is approximately 0.5 Btu/lb- $^{\circ} F$ . This means that 0.5 Btu are required to raise the temperature of one pound of ice by one degree Fahrenheit.

Q: How much energy is required to raise the temperature of one pound of ice from $20^{\circ} F$ to $32^{\circ} F$ ?

A: The energy required to raise the temperature of one pound of ice from $20^{\circ} F$ to $32^{\circ} F$ is:

Q = mc\Delta T = 1 \, \text{lb} \times 0.5 \, \text{Btu/lb-$^{\circ} F$} \times (32^{\circ} F - 20^{\circ} F) = 6 \, \text{Btu}

Q: What is the latent heat of vaporization of water?

A: The latent heat of vaporization of water is approximately 970 Btu/lb. This means that 970 Btu are required to change one pound of water at $212^{\circ} F$ to steam at $212^{\circ} F$ .

Q: How much energy is required to raise the temperature of steam from $212^{\circ} F$ to $220^{\circ} F$ ?

A: The energy required to raise the temperature of one pound of steam from $212^{\circ} F$ to $220^{\circ} F$ is:

Q = mc\Delta T = 1 \, \text{lb} \times 0.48 \, \text{Btu/lb-$^{\circ} F$} \times (220^{\circ} F - 212^{\circ} F) = 6.4 \, \text{Btu}

Q: What is the total energy required to change one pound of ice at $20^{\circ} F$ to steam at $220^{\circ} F$ ?

A: The total energy required to change one pound of ice at $20^{\circ} F$ to steam at $220^{\circ} F$ is the sum of the energies required for each step:

Q_{\text{total}} = Q_{\text{raise temperature}} + Q_{\text{latent heat of fusion}} + Q_{\text{latent heat of vaporization}} + Q_{\text{raise temperature of steam}}

Q_{\text{total}} = 6 \, \text{Btu} + 144 \, \text{Btu} + 970 \, \text{Btu} + 6.4 \, \text{Btu} + 6 \, \text{Btu} + 144 \, \text{Btu} + 970 \, \text{Btu} + 6.4 \, \text{Btu}

Q_{\text{total}} = 3208 \, \text{Btu}

Q: What is the correct answer?

A: The correct answer is 3208 Btu.

Q: Why is the answer 3208 Btu and not 1126.4 Btu?

A: The answer 3208 Btu is correct because it takes into account the energy required to change the ice to water at $32^{\circ} F$ and then to steam at $220^{\circ} F$ . The answer 1126.4 Btu is incorrect because it only takes into account the energy required to change the ice to water at $32^{\circ} F$ and not the energy required to change the water to steam at $220^{\circ} F$ .