Mixture of Steam and Ice-Water: Understanding the Thermodynamic Process

When dealing with the mixture of steam and ice-water, the thermodynamic principles of latent heat and specific heat are crucial in understanding the resulting mixture. This article explores the process when 10 grams of steam at 100 degrees Celsius is introduced to a mixture of 100 grams of ice at 0 degrees Celsius and 500 grams of water. Understanding this process can help in comprehending the concept of thermal equilibrium and the behavior of various substances under specific conditions.

Understanding Thermal Equilibrium and Latent Heat

Thermal equilibrium is a state where the temperatures of two substances in contact are equal. Latent heat, on the other hand, is the heat required to change the state of a substance without changing its temperature. For instance, the latent heat of vaporization (the amount of energy required to change a substance from liquid to gas) and the latent heat of fusion (the amount of energy required to change a substance from solid to liquid) are critical in the context of this problem.

Latent Heat of Vaporization and Fusion

The latent heat of vaporization for water is 540 calories per gram. This means that 540 calories are required to convert 1 gram of water into steam at 100 degrees Celsius. Similarly, the latent heat of fusion for water (the energy required to convert 1 gram of ice into liquid water at 0 degrees Celsius) is 80 calories per gram.

Thermal Energy Transfer in Mixture

Initially, we have 10 grams of steam at 100 degrees Celsius and a mixture of 100 grams of ice at 0 degrees Celsius and 500 grams of water at 0 degrees Celsius. The total mass is 610 grams. To understand the process, let's break down the energy transfer step by step.

When steam at 100 degrees Celsius is introduced to the ice and water mixture, the following processes may occur:

The steam will condense, releasing its latent heat of vaporization. The released latent heat will be used to melt the ice and warm the resulting water. Eventually, the system will reach a state of thermal equilibrium, where the temperature of the mixture will be the equilibrium temperature.

Calculating the Required Energy

First, we calculate the energy required to condense 10 grams of steam at 100 degrees Celsius to water at 0 degrees Celsius:

Energy required Latent heat of vaporization × mass of steam

Energy required 540 calories/gram × 10 grams 5400 calories

Next, we calculate the energy required to melt 80 grams of ice at 0 degrees Celsius to water at 0 degrees Celsius:

Energy required Latent heat of fusion × mass of ice

Energy required 80 calories/gram × 80 grams 6400 calories

The total energy required to both condense the steam and melt the ice is 6400 calories. Since the initial energy of the steam is 5400 calories, the remaining 1000 calories (6400 - 5400) must be used to warm the water in the mixture, without changing the ice-water mixture to water.

Final State of the Mixture

After the energy transfer, the ice will melt, and the temperature of the mixture will remain at 0 degrees Celsius. The mass of the final mixture will be the sum of the original mixture and the steam that has been condensed, which is 610 grams, including 10 grams of water from the condensed steam. Hence, the final state will be 620 grams of water at 0 degrees Celsius.

Conclusion

Understanding the thermodynamic principles of latent heat and specific heat is essential for predicting the final state of a mixture of steam and ice-water. The concept of thermal equilibrium helps in determining the temperature and mass of the final mixture. If you need help with such problems or homework, consider consulting your textbook or seeking assistance from classmates or teachers.

Note: For a more detailed analysis, refer to standard physics textbooks or online resources. Quorans and other sources can provide additional insights and explanations.

Keywords

Thermal equilibrium, Latent heat, Specific heat