Lecture 13 Notes: Heat Transfer and Specific Heat Capacity

Lecture 13 Notes

Date: February 20, 2026
Time: 9:12 AM
Topic: Heat Transfer and Specific Heat Capacity

Aktiv Warm Up

• Chemical Reaction Balancing

  • Balance the following chemical equation:

  • $<em>{CH</em>4} + <em>{O</em>2} \rightarrow <em>{CO</em>2} + <em>{H</em>2O}$

• Example Calculation:

  • When reacting 12 g of $CH<em>4$ with excess $O</em>2$, calculate the moles of $H_2O$ produced.

Heat Capacity and Specific Heat

• Heat Capacity (C)

  • Defined as the energy needed to raise the temperature of a substance by $1 °C$:

  • $C = \frac{q}{\Delta T}$, where $q$ is heat energy and $\Delta T$ is the change in temperature.

• Specific Heat Capacity (c)

  • Defined as heat capacity per gram or mol of a substance.

  • Formula:

  • $c = \frac{C}{m}$, where $m$ is the mass of the substance.

Comparison of Heats

• Heat capacities of various substances:

  • Iron:

  • Heat capacity $C = 20.8 ext{ J/°C}$, specific heat $c = 0.45 ext{ J/g°C}$

  • Water:

  • Heat capacity $C = 41.7 ext{ J/°C}$, specific heat $c = 4.18 ext{ J/g°C}$

• The energy required to raise the temperature varies by substance:

  • More mass (more "stuff") requires more energy to raise temperature (KE).

  • Higher heat capacity means harder to move, requiring more energy to raise temperature (KE).

Observational Patterns in Molar Specific Heats

• Grouping substances by molar specific heat, formula, or phases reveals patterns:

  • Noble Gases:

  • Helium ($He$): $c_m = 21 ext{ J/mol°C}$

  • Neon ($Ne$): $c_m = 21 ext{ J/mol°C}$

  • Solids:

  • Aluminum ($Al$): $c_m = 24 ext{ J/mol°C}$

  • Liquids:

  • Water ($H<em>2O$): $c</em>m = 50 ext{ J/mol°C}$

  • Ionic Solids:

  • Sodium Chloride ($NaCl$): $c_m = 50 ext{ J/mol°C}$

  • Potassium Bromide ($KBr$): $c_m = 52 ext{ J/mol°C}$

• Key Observations:

  • Metals exhibit similar molar specific heats.

  • Diatomic gases such as $H<em>2$ and $N</em>2$ require more energy to raise their temperature (higher specific heats).

  • Ionic solids have similar molar specific heats, larger molecules (like $C<em>8H</em>{18}$) require more energy due to their complexity.

Specific Heat Capacity in Different States

• The specific heat capacity in liquids and solids is influenced by bonding and intermolecular forces (IMFs).

  • When temperature increases, molecules move more vigorously (higher kinetic energy).

• In gases, specific heat capacity depends on "degrees of freedom", which are:

  • Rotation

  • Vibration

  • Translation

• Heat Energy Distribution Example:

  • For $9 of energy:$

  • $3$ on rotations, $3$ on vibrations, $3$ on translations in a gas.

  • In gases, if heat energy only increases translations, the final temperature is higher.

• Example:

  • $c_m (He) = 21 ext{ J/mol°C}$

Understanding Through Examples

• Test Your Understanding 1:

  • Given two gas samples at $20 °C$:

  • Sample A: $1 L ext{ of } CO_2$

  • Sample B: $1 L ext{ of } CH<em>3CH</em>3$

  • Energy Input: Adding $50 J$ to each sample, determine which sample will end up hotter.

Heat Transfer Calculations

• When heat is transferred, it's described by the principle:

  • $q_{ ext{input}} = C \Delta T$

  • where $C$ is heat capacity and $\Delta T$ is change in temperature.

  • Rearranged, this can be used to calculate heat:

  • $q = m c \Delta T$

• Illustrative Example:

  • Adding $250 J$ of heat to $10 ext{ g}$ of water at $10 °C$ where $c = 4.18 ext{ J/g°C}$ determines the final temperature using:

  • $250 = 10 x 4.18 x (T_{ ext{final}} - 10)$
    

  • Solving gives:

  • $T_{ ext{final}} = 15.98 °C$

Test Your Understanding 2

• Problem: Adding $250 J$ of heat to $10 g$ of isopropanol at $10 °C$, using $c = 2.60 ext{ J/g°C}$, find the final temperature:

  • $250 = 10 x 2.6 x (T_{ ext{final}} - 10)$

• Rearranged and solved gives final temperature of $T_{ ext{final}} = 19 °C$.

Nutrition Facts and Ratios

• Example of using nutrition facts for calorimetry from a cookie recipe with:

  • Serving Size: $54 g$

  • Calories: $171$

  • Components include Total Fat, Saturated Fat, Cholesterol, and Sodium.

• Ratios are crucial when calculating calories per batch (can scale recipe).

Calorimetry Principles

• Heat exchange between the system and the surroundings, using the equation:

  • $q<em>{ ext{surr}} = -q</em>{ ext{sys}}$

• The calorimeter measures heat exchanged:

  • $q<em>{ ext{water}} = m</em>{ ext{water}} C<em>{ ext{water}} \Delta T</em>{ ext{water}}$

  • $q<em>{ ext{metal}} = m</em>{ ext{metal}} C<em>{ ext{metal}} \Delta T</em>{ ext{metal}}$

  • Final temperatures should match to conserve energy in an isolated system.

• Example Calculation of Metal's Specific Heat:

  • Given 100 g of water at $27 °C$ and 100 g of metal at $77 °C$, with the water reaching $31.8 °C$:

  • $q_{ ext{water}} = 100 ext{ g} * 4.18 ext{ J/g°C} * (31.8°C - 27°C)$

  • Rearranging gives specific heat of metal as:

  • $C_{ ext{event}} = 0.44 ext{ J/g°C}$

Changes in Kinetic Energy

• Changes in heat transfer affect the kinetic energy (KE) of substances:

  • Changes are determined by both the amount of the substance and its specific heat capacity.

• Relationship to the change in temperature:

  • $\Delta KE = m c \Delta T$