Heat Transfer Between Substances

Jun 13, 2024

Heat Transfer Between Substances in Thermal Contact

Key Concepts

  • Thermal Contact and Temperature Equalization

    • When two substances are in thermal contact, the higher temperature substance transfers heat to the lower temperature substance until they reach the same temperature (thermal equilibrium).
    • Example: Hot metal placed in cooler water results in metal cooling down and water heating up until both reach the same temperature.
    • The substance with a higher specific heat will change temperature less than the one with a lower specific heat.

  • Heat Transfer Principles

    • Heat lost by the higher temperature substance (q<sub>lost</sub>) is equal in magnitude and opposite in sign to heat gained by the lower temperature substance (negative q<sub>gain</sub>).
    • Equation: q<sub>lost</sub> = -q<sub>gain</sub>
    • Formula: q = mcΔT
      • m = mass
      • c = specific heat
      • ΔT = change in temperature
    • Relationship: m<sub>metal</sub>c<sub>metal</sub>ΔT<sub>metal</sub> = -m<sub>water</sub>c<sub>water</sub>ΔT<sub>water</sub>

Example Problem 1: Calculation of Specific Heat of Metal

  • Given Data

    • Mass of metal: 25 grams
    • Initial temperature of metal: 87.7°C
    • Mass of water: 37.4 grams
    • Initial temperature of water: 10.3°C
    • Final temperature of system: 15.4°C

  • Steps to Solve

    1. Calculate change in temperature (ΔT)
    • ΔT<sub>metal</sub> = 15.4 - 87.7 = -72.3°C
    • ΔT<sub>water</sub> = 15.4 - 10.3 = 5.1°C
    1. Rearrange the formula to isolate c<sub>metal</sub>
    • c<sub>metal</sub> = -(m<sub>water</sub>c<sub>water</sub>ΔT<sub>water</sub>) / (m<sub>metal</sub>ΔT<sub>metal</sub>)
    1. Substitute known values and solve
    • c<sub>water</sub> = 4.184 J/g°C
    • c<sub>metal</sub> = (37.4 × 4.184 × 5.1) / (25 × -72.3)
    • Result: c<sub>metal</sub> = 0.44 J/g°C (for iron)

Example Problem 2: Calculation of Specific Heat of Silver

  • Given Data

    • Mass of silver: 10.0 grams
    • Initial temperature of silver: 100°C
    • Mass of water: 20.0 grams
    • Initial temperature of water: 23.0°C
    • Final temperature of system: 25.0°C

  • Steps to Solve

    1. Calculate change in temperature (ΔT)
    • ΔT<sub>silver</sub> = 25.0 - 100 = -75.0°C
    • ΔT<sub>water</sub> = 25.0 - 23.0 = 2.0°C
    1. Rearrange the formula to isolate c<sub>silver</sub>
    • c<sub>silver</sub> = -(m<sub>water</sub>c<sub>water</sub>ΔT<sub>water</sub>) / (m<sub>silver</sub>ΔT<sub>silver</sub>)
    1. Substitute known values and solve
    • c<sub>water</sub> = 4.184 J/g°C
    • c<sub>silver</sub> = (20 × 4.184 × 2.0) / (10 × -75)
    • Result: c<sub>silver</sub> = 0.223 J/g°C (for silver)

Summary

  • Understanding heat transfer principles helps solve problems involving thermal contact between substances.
  • The methods involve using specific heat relations, mass, and temperature changes to determine unknown variables like specific heat.