What is the final temperature of composition of mixture of equal amount of ice at10C and steam at 100C are mixed together

To determine the final temperature of a mixture consisting of equal amounts of ice at 0°C and steam at 100°C, we need to consider the principles of heat transfer and phase changes. When ice and steam are mixed, the heat from the steam will be used to melt the ice, and the resulting water will then reach thermal equilibrium with the remaining steam. Let's break this down step by step.

Understanding the Components

We have two components in this scenario:

• Ice: Initially at 0°C.
• Steam: Initially at 100°C.

Phase Changes and Heat Transfer

When steam condenses into water, it releases a significant amount of heat, known as the latent heat of vaporization. Conversely, when ice melts into water, it absorbs heat, known as the latent heat of fusion. The specific values for these latent heats are:

• Latent heat of fusion for ice: approximately 334 J/g.
• Latent heat of vaporization for steam: approximately 2260 J/g.

Calculating Heat Transfer

Let's assume we have 100 grams of ice and 100 grams of steam. The heat lost by the steam as it condenses will be equal to the heat gained by the ice as it melts and warms up to the final temperature.

Heat Lost by Steam

When steam condenses, it releases heat:

Heat lost by steam (Q_steam) = mass_steam × latent heat of vaporization

Q_steam = 100 g × 2260 J/g = 226,000 J

Heat Gained by Ice

First, the ice will absorb heat to melt:

Heat gained by ice to melt (Q_melt) = mass_ice × latent heat of fusion

Q_melt = 100 g × 334 J/g = 33,400 J

After melting, the water will then warm up to the final temperature (T_f). The heat gained by the melted ice (now water) can be calculated as:

Heat gained by melted ice (Q_warm) = mass_water × specific heat capacity × (T_f - 0°C)

Q_warm = 100 g × 4.18 J/g°C × (T_f - 0°C)

Setting Up the Equation

Now, we can set up the equation for heat balance:

Heat lost by steam = Heat gained by ice

226,000 J = 33,400 J + 100 g × 4.18 J/g°C × (T_f - 0°C)

Simplifying the Equation

Subtract the heat gained by the ice from both sides:

226,000 J - 33,400 J = 100 g × 4.18 J/g°C × (T_f - 0°C)

192,600 J = 418 J/°C × T_f

Solving for Final Temperature

Now, divide both sides by 418 J/°C:

T_f = 192,600 J / 418 J/°C ≈ 460.5°C

This temperature is not physically possible since it exceeds the boiling point of water. This indicates that all the steam has condensed and the resulting water has reached a maximum temperature of 100°C, while the ice has melted completely. The final temperature of the mixture will stabilize at 100°C, as there is no more steam left to condense.

Final Outcome

Thus, when equal amounts of ice at 0°C and steam at 100°C are mixed, the final temperature of the mixture will be 100°C, with all the ice melted and the steam condensed into water.