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:
{CH4} + {O2} \rightarrow {CO2} + {H2O}
Example Calculation:
When reacting 12 g of CH4 with excess O2 , 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 ( H2O ): cm = 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 H2 and N2 require more energy to raise their temperature (higher specific heats).
Ionic solids have similar molar specific heats, larger molecules (like C8H{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 } CH3CH3
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{ ext{surr}} = -q{ ext{sys}}
The calorimeter measures heat exchanged:
q{ ext{water}} = m{ ext{water}} C{ ext{water}} \Delta T{ ext{water}}
q{ ext{metal}} = m{ ext{metal}} C{ ext{metal}} \Delta T{ 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