Temperature, Heat, and Enthalpy Notes

Thermochemistry

• Thermochemistry studies energy changes in chemical and physical changes.
• Energy study is crucial for:
  • Food.
  • Climate temperature effects.
  • Fuel.
  • Home heat and power.

Energy

• Energy is the capacity to do work or cause change.
• Types:
  • Potential: Stored energy due to object position.
  • Kinetic: Energy due to object movement.
• Chemical energy is potential energy stored in chemical bonds.

Energy in Reactions

• Energy is a key reaction focus.
• Example: Burning gasoline for energy, not combustion products.
• Emphasis shift from matter to energy in processes.

Energy Units and Conversions

• Joule (J): SI unit of energy.
  • Energy to raise 1 g of water by 0.2930 C.
• Conversions:
  • $1 \text{ kJ} = 1000 \text{ J}$
  • $4.184 \text{ J} = 1 \text{ calorie}$
  • $1000 \text{ calories} = 1 \text{ Calorie}$
    • (Food calorie or kilocalorie)

Temperature (T)

• Measures average kinetic energy in particles.
• Measured using a thermometer.
• Higher temperature indicates faster particle movement.
• Units: Celsius (C) or Kelvin (K).

Heat (Q)

• Energy transferred between materials due to temperature difference.
• All objects above absolute zero have heat to transfer.
• Heat transfer amount depends on matter quantity.
• Example: Drop of boiling water vs. pot of boiling water.

Specific Heat (C_p)

• Heat to raise 1 gram of substance by 1 degree Celsius.
• Indicates how well a substance holds heat.
• Unique material property, especially for metals.
• Water's $C_p$: $4.184 \frac{\text{J}}{\text{g} \cdot \text{C}}$

Specific Heats of Common Materials

• Liquid water: $4.18 \frac{\text{J}}{\text{g} \cdot \text{°C}}$
• Solid water (ice): $2.11 \frac{\text{J}}{\text{g} \cdot \text{°C}}$
• Water vapor: $2.00 \frac{\text{J}}{\text{g} \cdot \text{°C}}$
• Dry air: $1.01 \frac{\text{J}}{\text{g} \cdot \text{°C}}$
• Basalt: $0.84 \frac{\text{J}}{\text{g} \cdot \text{°C}}$
• Granite: $0.79 \frac{\text{J}}{\text{g} \cdot \text{°C}}$
• Iron: $0.45 \frac{\text{J}}{\text{g} \cdot \text{°C}}$
• Copper: $0.38 \frac{\text{J}}{\text{g} \cdot \text{°C}}$
• Lead: $0.13 \frac{\text{J}}{\text{g} \cdot \text{°C}}$

Heat Equation

• Calculates heat transferred or lost/gained due to temperature change.
• Equation: $Q = m \cdot C_p \cdot \Delta T$
  • $Q$ = heat change (J).
  • $m$ = mass (g).
  • $C_p$ = specific heat $\frac{\text{J}}{\text{g} \cdot \text{°C}}$.
  • $\Delta T = T<em>f - T</em>i$ = final temp – initial temp.
  • $\Delta$ stands for “change in”.

Heat Equation Application

• Use only when an object changes temperature.
• Temperature can be in Celsius or Kelvin, as measuring change.
• $\Delta Q = m \cdot C_p \cdot \Delta T$
  • If Q is negative, exothermic (heat lost).
  • If Q is positive, endothermic (heat gained).

Heat Calculation Example

• Problem: Heat to raise 854 mL of water from 23.5 C to 85.0 C?
• Solution:
  • $Q = m \cdot C_p \cdot \Delta T = 854 \text{ g} \cdot 4.184 \frac{\text{J}}{\text{g} \cdot \text{°C}} \cdot (85.0 \text{°C} – 23.5 \text{°C}) = 2.20 \times 10^5 \text{ J}$
• Note:
  • $\Delta T$ is always final – initial.
  • Unit for heat is joules.
  • Water density is roughly 1 g/mL.

Measuring Heat Change

• Always measure heat change in a system.
• Heat flows from high to low temperature until equilibrium.

Enthalpy (H or ∆H)

• Internal energy: total kinetic and potential energy in a system.
• Transferred as heat during processes.
• Enthalpy changes involve energy changes for:
  • Chemical reactions.
  • Solvation.
  • Changes in state

Enthalpy vs. Heat

• Enthalpy is the system's energy.
• Heat is energy transfer observation.
• Cannot measure enthalpy directly, but can measure its transfer as heat.

Enthalpy Example

• Boiling water gallon vs. teaspoon: same temperature, different enthalpy due to mass.

Law of Conservation of Energy

• Total energy in the universe is constant; cannot be created or destroyed; can change form.
• Basis of calorimetry.

Calorimetry

• Study of energy changes during chemical or physical changes.
• Based on heat transfer into a known mass of material.

Calorimeter

• Insulated chamber using water mass to monitor energy changes.
• Based on water temperature changes.

Calorimetry Principle

• Heat lost by one system equals heat gained by another.
• $-Q<em>{\text{lost}} = Q</em>{\text{gained}}$

Calorimetry Considerations

• Heated substance in contact with cooler substance leads to temperature equalization.

Importance of Isolation

• Essential to have an isolated system to prevent energy escape or entry.

Calorimetry Example 1

• Iron nail at 752 C added to 250.0 g of water.
• Water warms from 23.5 C to 26.3 C.
• Find the mass of the iron nail.
• Solution:
  • $Q<em>{\text{Fe}} = Q</em>{\text{H2O}}$
  • $Q<em>{\text{H2O}} = m</em>{\text{H2O}} \cdot C<em>{p\text{H2O}} (T</em>f - T_{i\text{H2O}}) = 250.0 \text{ g} \cdot 4.184 \frac{\text{J}}{\text{g} \cdot \text{°C}} \cdot (26.3 \text{°C} - 23.5 \text{°C}) = 2930 \text{ J}$
  • $Q<em>{\text{Fe}} = m</em>{\text{Fe}} \cdot C<em>{p\text{Fe}} (T</em>f - T_{i\text{Fe}})$
  • $2930 \text{ J} = m<em>{\text{Fe}} \cdot C</em>{p\text{Fe}} (T<em>f - T</em>{i\text{Fe}})$
  • $m_{Fe} = 8.95 \text{ g}$

Energy Balance Problems

• Think of heat lost and gained as separate.
• Sometimes, equate them in one large equation to find a variable.
• $-Q<em>{\text{lost}} = Q</em>{\text{gained}}$

Calorimetry Example 2

• 3.90 g aluminum at 99.3 C dropped into 10.0 mL water at 22.6 C.
• Find the final temperature of the system.
• Solution:
  • $-Q<em>{\text{lostAl}} = Q</em>{\text{gainedH2O}}$
  • $-m<em>{\text{Al}} \cdot C</em>{p\text{Al}} \cdot \Delta T<em>{\text{Al}} = m</em>{\text{H2O}} \cdot C<em>{p\text{H2O}} \cdot \Delta T</em>{\text{H2O}}$
  • $-(3.90 \text{ g})(0.897 \frac{\text{J}}{\text{g} \cdot \text{°C}})(T<em>f – 99.3 \text{°C}) = (10.0\text{g})(4.184 \frac{\text{J}}{\text{g} \cdot \text{°C}})(T</em>f - 22.6 \text{°C})$
  • $-3.498T<em>f + 347.38 = 41.84T</em>f - 945.58$
  • $45.34T_f = 1292.97$
  • $T_f = 28.5 \text{°C}$
• Final temperature for water and aluminum is the same.
• Heat transfers until temperatures equalize.

Solving for Variables

• Isolating the variable is challenging due to multiple variables and need to combine like terms.