Condensing the Steam into Liquid Water

Oct 13, 2025 by miodeya.com 52 views

10.0 g of Steam at 120.0°C Converted into Ice at -20.0°C: A Comprehensive Analysis

In this article, we will explore the process of converting 10.0 g of steam at 120.0°C into ice at -20.0°C. This process involves a significant change in the state of matter, from gas to solid, and a substantial release of energy. We will create a graph to illustrate this change and calculate the total energy released during the conversion.

Step 1: Creating a Graph to Indicate the Change

To visualize the change in state, we can create a graph showing the temperature of the steam as it cools and condenses into liquid water, and then freezes into ice. The graph will have temperature on the y-axis and time on the x-axis.

**Graph: Temperature vs. Time**
---------------------------

| Temperature (°C) | Time (s) |
| --- | --- |
| 120.0 | 0 |
| 100.0 | 10 |
| 80.0 | 20 |
| 60.0 | 30 |
| 40.0 | 40 |
| 20.0 | 50 |
| 0.0 | 60 |
| -20.0 | 70 |

Step 2: Calculating the Total Energy Released

To calculate the total energy released during the conversion, we need to consider the energy required to cool the steam from 120.0°C to 100.0°C, condense it into liquid water, and then freeze it into ice at -20.0°C.

Heat of Vaporization

The heat of vaporization is the energy required to convert a substance from liquid to gas. For water, the heat of vaporization is approximately 2250 kJ/kg.

Heat of Fusion

The heat of fusion is the energy required to convert a substance from solid to liquid. For water, the heat of fusion is approximately 334 J/g.

Heat of Condensation

The heat of condensation is the energy released when a substance changes from gas to liquid. For water, the heat of condensation is approximately 2250 kJ/kg.

Calculations

To calculate the total energy released, we need to calculate the energy required for each step of the process.

Cooling the Steam from 120.0°C to 100.0°C

The energy required to cool the steam from 120.0°C to 100.0°C can be calculated using the formula:

Q = mcΔT

where Q is the energy required, m is the mass of the steam, c is the specific heat capacity of water, and ΔT is the change in temperature.

For water, the specific heat capacity is approximately 4.184 J/g°C. Therefore:

Q = 10.0 g x 4.184 J/g°C x (100.0°C - 120.0°C) = -10.0 g x 4.184 J/g°C x 20.0°C = -836.8 J

Condensing the Steam into Liquid Water

The energy required to condense the steam into liquid water can be calculated using the formula:

Q = mL

where Q is the energy required, m is the mass of the steam, and L is the heat of condensation.

For water, the heat of condensation is approximately 2250 kJ/kg. Therefore:

Q = 10.0 g x 2250 kJ/kg = 22500 J

Freezing the Liquid Water into Ice at -20.0°C

The energy required to freeze the liquid water into ice at -20.0°C can be calculated using the formula:

Q = mL

where Q is the energy required, m is the mass of the liquid water, and L is the heat of fusion.

For water, the heat of fusion is approximately 334 J/g. Therefore:

Q = 10.0 g x 334 J/g = 3340 J

Total Energy Released

The total energy released during the conversion can be calculated by adding the energy required for each step of the process:

Q_total = Q_cooling + Q_condensation + Q_freezing = -836.8 J + 22500 J + 3340 J = 24303.2 J

Therefore, the total energy released during the conversion of 10.0 g of steam at 120.0°C into ice at -20.0°C is approximately 24303.2 J.

In this article, we have explored the process of converting 10.0 g of steam at 120.0°C into ice at -20.0°C. We have created a graph to illustrate the change in state and calculated the total energy released during the conversion. The results show that the total energy released is approximately 24303.2 J. This process involves a significant release of energy, which can be harnessed and utilized in various applications.
Q&A: 10.0 g of Steam at 120.0°C Converted into Ice at -20.0°C

In our previous article, we explored the process of converting 10.0 g of steam at 120.0°C into ice at -20.0°C. We created a graph to illustrate the change in state and calculated the total energy released during the conversion. In this article, we will answer some frequently asked questions related to this process.

Q: What is the heat of vaporization of water?

A: The heat of vaporization of water is approximately 2250 kJ/kg. This is the energy required to convert a substance from liquid to gas.

Q: What is the heat of fusion of water?

A: The heat of fusion of water is approximately 334 J/g. This is the energy required to convert a substance from solid to liquid.

Q: What is the heat of condensation of water?

A: The heat of condensation of water is approximately 2250 kJ/kg. This is the energy released when a substance changes from gas to liquid.

Q: How much energy is required to cool the steam from 120.0°C to 100.0°C?

A: The energy required to cool the steam from 120.0°C to 100.0°C can be calculated using the formula:

Q = mcΔT

where Q is the energy required, m is the mass of the steam, c is the specific heat capacity of water, and ΔT is the change in temperature.

For water, the specific heat capacity is approximately 4.184 J/g°C. Therefore:

Q = 10.0 g x 4.184 J/g°C x (100.0°C - 120.0°C) = -10.0 g x 4.184 J/g°C x 20.0°C = -836.8 J

Q: How much energy is required to condense the steam into liquid water?

A: The energy required to condense the steam into liquid water can be calculated using the formula:

Q = mL

where Q is the energy required, m is the mass of the steam, and L is the heat of condensation.

For water, the heat of condensation is approximately 2250 kJ/kg. Therefore:

Q = 10.0 g x 2250 kJ/kg = 22500 J

Q: How much energy is required to freeze the liquid water into ice at -20.0°C?

A: The energy required to freeze the liquid water into ice at -20.0°C can be calculated using the formula:

Q = mL

where Q is the energy required, m is the mass of the liquid water, and L is the heat of fusion.

For water, the heat of fusion is approximately 334 J/g. Therefore:

Q = 10.0 g x 334 J/g = 3340 J

Q: What is the total energy released during the conversion of 10.0 g of steam at 120.0°C into ice at -20.0°C?

A: The total energy released during the conversion can be calculated by adding the energy required for each step of the process:

Q_total = Q_cooling + Q_condensation + Q_freezing = -836.8 J + 22500 J + 3340 J = 24303.2 J

In this article, we have answered some frequently asked questions related to the process of converting 10.0 g of steam at 120.0°C into ice at -20.0°C. We hope that this information has been helpful in understanding this process and its associated energy requirements.