Thermodynamics: Heat Exchange and Vapor Pressure

Problem of Mixing Two Phases

• Topic: Finding the final temperature when mixing ice and water.

• Key Concept: Heat exchange occurs between the ice (melting) and the warmer water, following the principles of thermodynamics.

Problem Statement

• Given:

  • Ice mass: 25.0 grams at 0°C

  • Water mass: 250 grams at 25°C

• Objective: Determine the final temperature of the liquid water after all the ice has melted and equilibrium has been reached.

Heat Exchange Process

• Heat Lost by Water: The heat lost by the warmer water as it cools down can be calculated using the formula:
  q{water} = m{water} \times c{water} \times (T{initial} - T_{final})
  Where:

  • (m_{water} = 250 \text{ grams})

  • (c_{water} = 4.184 \text{ joules/g°C})

  • (T_{initial} = 25°C)

  • (T_{final} = T) (unknown)
    This formula highlights that the energy transfer depends on the mass of the water, its specific heat capacity, and the temperature change.

• Heat Gained by Ice: The heating of the ice involves two significant processes:

  1. Melting of Ice (Heat of Fusion):
     The energy required for the ice to melt is given by the equation:
     [ q{fusion} = m{ice} \times \Delta H_{fusion} ]
     Where:

  • (m_{ice} = 25.0 \text{ grams})

  • (\Delta H_{fusion} = 334 \text{ joules/g})

  1. Heating the Melted Ice: After transformation into a liquid state, the newly formed water must gain additional heat to increase its temperature from 0°C to the final temperature T:
     [ q{ice \rightarrow water} = m{ice} \times c_{water} \times (T - 0) ]

  • This signifies that after melting, the ice, now water, continues to absorb heat until it reaches thermal equilibrium with the warmer water.

Equations Setup

• Heat Balance Equation: The foundational principle of conservation of energy leads us to equate the heat lost and gained:
  q{water} = q{ice}

• Substitute expressions for heat:

  • For water:
    q_{water} = 250 \times 4.184 \times (25 - T)

  • For ice:
    q_{ice} = 25 \times 334 + 25 \times 4.184 \times (T - 0)

Solve for T

1. Set the equations equal to each other:
   
   250 \times 4.184 \times (25 - T) = 25 \times 334 + 25 \times 4.184 \times T

2. Expand and simplify the expressions:

   • Left Side: 26150 - 1046T

   • Right Side: 8350 + 104.6T

3. Combine like terms to isolate T:
   
   26150 - 8350 = 1046T + 104.6T
   
   17800 = 1150.6T

4. Finally, solve for T by isolating it:
   
   T = \frac{17800}{1150.6} \approx 15.5°C

Conclusion

• The final temperature of the liquid water when the ice melts is approximately 15.5°C, which indicates that the system has reached thermal equilibrium. Understanding heat transfer is crucial for solving problems related to temperature changes during phase shifts.

• Common Mistake: Failing algebra skills can hinder solving the problem—always double-check calculations and ensure the correct application of physical principles.

Clausius-Clapeyron Equation

• Purpose: The Clausius-Clapeyron equation establishes the relationship between vapor pressure and temperature, providing insights into phase transitions of substances.

Equation:

• The equation can be expressed as: \text{ln}\left(\frac{P2}{P1}\right) = \frac{\Delta H{vaporization}}{R} \left( \frac{1}{T1} - \frac{1}{T_2} \right) Where:

  • (P1) and (P2) are the vapor pressures at temperatures (T1) and (T2) respectively.

  • (R = 8.31 \text{ joules/mol·K}) is the universal gas constant, a pivotal factor in calculating changes in heat and phase transitions.

  • (\Delta H_{vaporization}) refers to the heat required for a phase change from liquid to gas, critical for understanding boiling and evaporation processes.

Application of Equation

• Scenarios:

  1. Known (T1), (P1), and (\Delta H{vaporization}) -> Solve for (P2) using the equation.

  2. Known (P1), (P2), (T1), and (T2) -> Solve for (\Delta H_{vaporization}), which is valuable in thermodynamic applications and engineering.

  3. Known (T1), (P1), and one other (T) -> Solve for corresponding (P), an essential skill for predicting vapor behavior in various conditions.

Key Point

• It is pivotal to familiarize yourself with the use of scientific calculators especially for computing logarithmic functions, particularly natural logarithms (ln), to solve these equations effectively.

• Prepare for interactive problem-solving in the upcoming video sessions, which will further reinforce the concepts discussed here and enhance practical understanding of thermodynamic principles.