Heat Balance in an Energy Conversion System: A Detailed Analysis
This article provides a comprehensive analysis of a basic thermodynamic problem involving the heat balance between steam and ice. We will delve into the specific energy requirements and heat exchange processes to determine the final temperature when steam at high temperature meets ice at a low temperature. This process is a key concept in the field of energy conversion.
Understanding the Problem
Suppose we have 20.0 grams of steam at 110°C mixed with 25.0 grams of ice at -40°C. The goal is to determine the final temperature of the mixture. To solve this problem, we will consider the specific heat capacities, latent heats, and energy changes involved in the process.
Key Properties and Constants
The following constants are essential for solving this problem:
Specific Heat Capacity: Water vapor: 1.970 J/gK Liquid water: 4.190 J/gK Ice: 2.100 J/gK Latent Heat of Condensation: 2260 J/g Latent Heat of Melting Ice: 334 J/gCalculation of Energy Balance
First, let's calculate the energy required to heat the ice from -40°C to 0°C:
Energy Required to Heat Ice from -40°C to 0°C
Using the specific heat capacity of ice (2.100 J/gK), the energy required can be calculated as:
Energy mass × specific heat capacity × temperature change
Energy 25.0g × 2.100 J/gK × 40K 2100 J
This value is not 25302.1 J as mentioned; thus, there might be a misinterpretation or an error in the given value. We will use the correct value of 2100 J for further calculations.
Energy Required to Melt Ice at 0°C
The latent heat of melting ice (334 J/g) is used to calculate the energy required to melt the ice:
Energy mass × latent heat of melting
Energy 25.0g × 334 J/g 8350 J
Energy Required to Heat Water from 0°C to 100°C
The specific heat capacity of water (4.190 J/gK) is used to calculate the energy required to heat the water from 0°C to 100°C:
Energy mass × specific heat capacity × temperature change
Energy 25.0g × 4.190 J/gK × 100K 10475 J
Heat Released by Cooling Steam to 100°C
The heat of vaporization of water (2260 J/g) is used to calculate the energy released by cooling the steam to 100°C:
Energy mass × latent heat of vaporization
Energy 20.0g × 2260 J/g / 2 985 J
Determining the Final Temperature
Now, let's set up the heat balance equation:
Heat lost by steam Heat gained by ice
25.0 × 1.970 × (110 - T) 25.0 × 2.100 × 40 25.0 × 334 25.0 × 4.190 × (100 - 0)
49.25 × (110 - T) 2100 8350 10475
49.25 × (110 - T) 21000
110 - T 21000 / 49.25 ≈ 425.28
T 110 - 425.28 ≈ -315.28°C
However, the correct interpretation should be in the context of the maximum possible temperature at which the ice can fully melt. The maximum temperature achievable in this scenario, without violating the phase transition properties, is:
T 100°C (as the ice will completely melt, and the system will reach a state of liquid water at 100°C)
Conclusion
In the context of the given problem, the system will ultimately reach a state where all the ice has been fully melted, and the steam has condensed, resulting in a final temperature of 100°C. This temperature is the maximum achievable in the system given the available heat and mass.
Understanding these principles is crucial in the realm of energy conversion systems, where such calculations are often performed to optimize efficiency and performance. Whether it is in the design of industrial processes, the improvement of thermal management systems, or even in everyday applications, the concepts of heat balance and energy conversion play a pivotal role.
For a more detailed exploration of these topics, consider consulting textbooks on thermodynamics or conducting further research in the field of heat transfer and energy systems.