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))

  • $riangle T = T<em>{final} - T</em>{initial}$

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: $Q<em>{Total} = Q</em>1 + Q_2 + …$

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?