Specific Heat Capacity and Heat Transfer Concepts

Specific Heat Capacity

• Physical property of materials pertaining to their ability to absorb heat.

Heat Flow

• Definition: The flow of thermal energy between two objects.

• Principle: Heat naturally flows from warmer to cooler objects.

  • Example: Ice warms while your hand cools down.

  • Example: A cup of warm coffee cools while your hand warms.

Specific Heat Capacity (C)

• Definition: The quantity of heat required to raise the temperature of 1 gram of a substance by one degree Celsius (°C) or Kelvin (K).

• Unit: Joules per gram per degree Celsius (J/g(°C)).

Units of Energy

• Joule (J): SI unit of energy.

  • Formula: 1 J = 1 kg•m^2/s^2

• KiloJoule (kJ): Commonly used unit, where 1000 J = 1 kJ.

• Calorie (cal): Alternate energy unit.

  • Relations: 1 cal = 4.184 J

  • 1000 cal = 1 kcal = 1 Cal

Specific Heat Capacities of Common Substances

• Water: 4.18 J/g(°C)

• Copper: 0.385 J/g(°C)

• Iron: 0.449 J/g(°C)

Heat Transfer Equation

• Equation: q = mC riangle T

  • Variables:

  • q = heat transferred (J)

  • m = mass of substance (g)

  • C = specific heat capacity (J/g(°C))

  • riangle T = T{final} - T{initial}

Example Calculations

Problem 1: Specific Heat of Silver

• Given:

  • Mass of silver: m = 15.4 g

  • Initial temperature: T_i = 20.0°C

  • Final temperature: T_f = 31.2°C

  • Heat added: q = 40.5 J

• From the equation:

  • 40.5 = 15.4(C)(31.2 - 20) \ $

  • solved to find C:

  • C = 0.235 J/g(°C)

Problem 2: Final Temperature of Silver

• Given:

  • Mass of silver: m = 5.8 g

  • Initial temperature: T_i = 30.0°C

  • Heat added: q = 40.5 J

  • Specific heat: C = 0.235 J/g(°C)

• From the equation:

  • 40.5 = 5.8(0.235)(T_f - 30.0)

  • Solving gives: T_f ext{ (Final temperature)} o 60.0°C

Problem 3: Heat Required for Water Heating

• Given:

  • Volume: 100 mL (Density = 1 g/mL, hence mass = 100 g)

  • Initial temperature: T_i = 45.6°C

  • Final temperature: T_f = 52.8°C

- Specific heat: C = 4.184 J/g(°C)

• Using the heat transfer equation:

  • q = 100 g imes 4.184 J/g(°C) imes (52.8°C - 45.6°C)

  • Result: q ext{ (Heat) } = 3012.48 J \ $q ext{ (rounded) } = 3010 J$

Heat of Phase Changes

• Heat of Fusion: Energy needed to change a solid to liquid at constant temperature.

  • Formula: Q = m(H_f)

• Heat of Vaporization: Energy needed to vaporize a liquid at constant temperature.

  • Formula: Q = m(H_v)

Example: Heat of Vaporization Calculation

• Given:

  • Total heat absorbed: Q = 2.27 imes 10^5 J

  • Mass: m = 115 g

• Using formula for Hv:

  • 2.27 imes 10^5 = 115(H_v) \ $

  • Solving yields: H_v o 1973.91 J/g

Example: Heat of Fusion Calculation

• Heat of fusion for ice: H_f = 3.4 imes 10^2 J/g

• Given:

  • Mass of ice: m = 75 g

• Calculate:

  • Q = 75 imes H_f = Q o 2.55 imes 10^4 J \ $Q ext{ (rounded) } = 2.6 imes 10^4 J$

Energy in Chemical Reactions

• Chemical reactions involve energy changes:

  • Endothermic Reaction: Heat enters the reaction.

  • Exothermic Reaction: Heat is released in the reaction.

Total Heat Calculation

• Formula: Q{Total} = Q1 + Q_2 + …

Practice Problem: Heat Removal from Water

• Given:

  • Mass: 283 g

  • Initial temperature: 23.5°C

  • Final temperature: -12.4°C

• Problem: How much energy is required to remove heat from water to reach this temperature?