Specific Heat lecture

Lab Preparation

• Date of lab: Friday

• Important Requirement: Wear close-toed shoes.

Specific Heat Equation

• Formula: $q = m imes c imes \Delta t$

  • q: Energy in Joules

  • m: Mass of substance in grams

  • c: Specific heat capacity (can be found on reference sheet)

  • \Delta t: Change in temperature

  • Calculate as: Final Temperature - Initial Temperature

Common Mistake Warning

• Do NOT mix up initial (Ti) and final (Tf) temperatures; always subtract Initial from Final.

• Example: ( \Delta t = Tf - Ti )

Properties of Water

• Unique Attribute: Water has a very high specific heat of $4.186 ext{ J/(g°C)}$

• Comparison: Most substances range between $0.1 ext{ to } 0.4 ext{ J/(g°C)}$.

• Implication: Higher specific heat means greater resistance to temperature changes.

Real-World Implications of Water’s Specific Heat

• San Francisco Climate:

  • Moderate temperatures year-round due to Pacific Ocean's water specific heat.

  • Water's ability to heat slowly in summer and cool slowly in winter stabilizes regional climate.

• Alaska Climate:

  • Coastal areas experience milder winter temperatures due to the ocean's moderating effects.

  • In contrast, inland areas can reach much colder temperatures.

Reference Sheet Data

• Specific heats by phase of water:

  • Liquid Water: $4.186 ext{ J/(g°C)}$

  • Solid Water (Ice): $2.03 ext{ J/(g°C)}$

  • Water Vapor/Steam: $2.01 ext{ J/(g°C)}$

  • Note: Specific heat varies by phase.

Example Problem Using Specific Heat Equation

• Given: 250 g water from 25°C to 70°C

  • Calculate (\Delta t): $70°C - 25°C = 45°C$

  • Specific heat $c = 4.186$ J/(g°C)

  • Plug values into equation:

  • $q = 250 imes 4.186 imes 45$

  • Total energy absorbed: $q = 47192.5 ext{ J}$ (rounded to 47000 J based on significant figures)

Practice Problem

• Using specific heat, find unknown:

  • 10 Joules injected into 15.05 g of a substance, temp increases from 21°C to 40°C.

  • Calculate specific heat and identify the substance from the chart.

Changing For Final/Initial Temperature

• For problems solving for (Tf) or (Ti):

  • Rearranging using (\Delta t) method is effective:

  • Example: $\Delta t = T<em>f - T</em>i$

Heat of Fusion and Vaporization

• Distinction for phase changes:

  • For melting (solid to liquid): $q = m imes \Delta h_f$

  • Heat of fusion for water: 335 J/g.

  • For vaporization (liquid to gas): $q = m imes \Delta h_v$

  • Heat of vaporization for water: 2260 J/g.

Example Problem: Heat of Fusion

• Scenario: Calculate energy to melt 75 grams of ice.

  • Use: $q = m imes \Delta h_f$

  • Plug values: $q = 75 imes 335 = 25125 ext{ J}$

Example Problem: Heat of Vaporization

• Scenario: Calculate heat of vaporization given 86.1 J to vaporize 58.2 g.

  • Setup: $86.1 = 58.2 imes c_v$

  • Solve: $c_v = \frac{86.1}{58.2} = 1.48 ext{ J/g}$

Important Conversion Notes

• Kilojoules to Joules: 1 kJ = 1000 J.

• Use molar mass to convert from Joules per mole to Joules per gram.

Key Points Summary

• Understanding specific heat is crucial for temperature change calculations.

• The specific heat equation can be applied to phase changes by using heat of fusion or vaporization formulas.

• Always keep significant figures, units, and careful arithmetic in mind during calculations.