Calorimetry

Overview

Calorimetry is the measurement of heat transfer during chemical reactions and physical changes. A calorimeter is an insulated device that measures these heat changes.

Fundamental Equations

Heat Equation

$$q = mc\Delta T$$

Where:

• $q$ = heat (J or kJ)
• $m$ = mass (g)
• $c$ = specific heat capacity (J/g·°C)
• $\Delta T$ = temperature change (°C or K)

Using Molar Heat Capacity

$$q = nC\Delta T$$

Where:

• $n$ = moles
• $C$ = molar heat capacity (J/mol·°C)

Specific Heat Values

Substance	c (J/g·°C)
Water (liquid)	4.184
Water (ice)	2.09
Water (steam)	2.01
Ethanol	2.44
Aluminum	0.897
Copper	0.385
Iron	0.449
Glass	0.84

Types of Calorimeters

Coffee Cup Calorimeter (Constant Pressure)

• Simple, inexpensive
• Used for aqueous reactions
• Measures $q_p = \Delta H$
• Assumes no heat loss to surroundings

Bomb Calorimeter (Constant Volume)

• Used for combustion reactions
• Measures $q_v = \Delta E$ (internal energy)
• More accurate, accounts for calorimeter heat capacity

Coffee Cup Calorimetry

Basic Principle

$$q_{\text{reaction}} = -q_{\text{solution}}$$

Heat lost by reaction = Heat gained by solution (and vice versa)

For Aqueous Solutions

$$q_{\text{solution}} = m \times c \times \Delta T$$

Usually assume:

• $c \approx 4.184$ J/g·°C (like water)
• $m$ = total mass of solution

Example 1: Dissolution

When 5.0 g NaOH dissolves in 100 mL water, temperature rises from 22.0°C to 35.0°C.

$$q_{\text{solution}} = (105 \text{ g})(4.184 \text{ J/g·°C})(35.0 - 22.0)$$ $$q_{\text{solution}} = 5711 \text{ J} = 5.71 \text{ kJ}$$ $$q_{\text{reaction}} = -5.71 \text{ kJ}$$ $$\text{mol NaOH} = \frac{5.0 \text{ g}}{40.0 \text{ g/mol}} = 0.125 \text{ mol}$$ $$\Delta H_{\text{dissolution}} = \frac{-5.71 \text{ kJ}}{0.125 \text{ mol}} = -45.7 \text{ kJ/mol}$$

Example 2: Neutralization

50.0 mL of 1.0 M HCl + 50.0 mL of 1.0 M NaOH Temperature rises from 25.0°C to 31.6°C.

$$q_{\text{solution}} = (100 \text{ g})(4.184)(31.6 - 25.0) = 2762 \text{ J} = 2.76 \text{ kJ}$$ $$\text{mol HCl} = \text{mol NaOH} = 0.050 \text{ L} \times 1.0 \text{ M} = 0.050 \text{ mol}$$ $$\Delta H_{\text{neutralization}} = \frac{-2.76 \text{ kJ}}{0.050 \text{ mol}} = -55.2 \text{ kJ/mol}$$

Bomb Calorimetry

Heat Balance

$$q_{\text{reaction}} = -(q_{\text{water}} + q_{\text{calorimeter}})$$ $$q_{\text{reaction}} = -(m_{\text{water}} \times c_{\text{water}} \times \Delta T + C_{\text{calorimeter}} \times \Delta T)$$ $$q_{\text{reaction}} = -(m_{\text{water}} \times c_{\text{water}} + C_{\text{calorimeter}}) \times \Delta T$$

Where $C_{\text{calorimeter}}$ = heat capacity of calorimeter (J/°C)

Example: Combustion

1.50 g of glucose burns in a bomb calorimeter. The temperature rises from 22.0°C to 28.3°C.

• Mass of water: 2.00 kg
• Calorimeter constant: 850 J/°C

$$\Delta T = 28.3 - 22.0 = 6.3°C$$ $$q_{\text{water}} = (2000 \text{ g})(4.184 \text{ J/g·°C})(6.3°C) = 52,719 \text{ J}$$ $$q_{\text{calorimeter}} = (850 \text{ J/°C})(6.3°C) = 5,355 \text{ J}$$ $$q_{\text{total}} = 58,074 \text{ J} = 58.1 \text{ kJ}$$ $$q_{\text{reaction}} = -58.1 \text{ kJ}$$ $$\text{mol glucose} = \frac{1.50 \text{ g}}{180.16 \text{ g/mol}} = 0.00833 \text{ mol}$$ $$\Delta H_{\text{combustion}} = \frac{-58.1 \text{ kJ}}{0.00833 \text{ mol}} = -6,976 \text{ kJ/mol}$$

Determining Specific Heat

Heat Transfer Method

Mix hot substance with cold water and measure final temperature.

$$q_{\text{hot}} = -q_{\text{cold}}$$ $$m_{\text{hot}} \times c_{\text{hot}} \times (T_f - T_{\text{hot}}) = -m_{\text{water}} \times c_{\text{water}} \times (T_f - T_{\text{water}})$$

Example

A 50.0 g metal at 100.0°C is placed in 100.0 g water at 25.0°C. Final temperature is 28.0°C.

$$(50.0)(c_{\text{metal}})(28.0 - 100.0) = -(100.0)(4.184)(28.0 - 25.0)$$ $$(50.0)(c_{\text{metal}})(-72.0) = -(100.0)(4.184)(3.0)$$ $$-3600 c_{\text{metal}} = -1255.2$$ $$c_{\text{metal}} = 0.349 \text{ J/g·°C}$$

Enthalpy of Fusion and Vaporization

Melting Ice

Total heat = Heat to melt + Heat to warm water

$$q_{\text{total}} = n\Delta H_{fus} + mc\Delta T$$

Example

Calculate heat to convert 50.0 g ice at -10°C to water at 20°C.

$$q_1 = (50.0)(2.09)(0 - (-10)) = 1,045 \text{ J} \quad \text{(warm ice)}$$ $$q_2 = \frac{50.0}{18.0} \times 6,010 = 16,694 \text{ J} \quad \text{(melt ice)}$$ $$q_3 = (50.0)(4.184)(20 - 0) = 4,184 \text{ J} \quad \text{(warm water)}$$ $$q_{\text{total}} = 1,045 + 16,694 + 4,184 = 21,923 \text{ J} = 21.9 \text{ kJ}$$

Key Relationships

Enthalpy	Value (H₂O)
$\Delta H_{fus}$	6.01 kJ/mol or 334 J/g
$\Delta H_{vap}$	40.7 kJ/mol or 2260 J/g

Heating Curve Calculations

For complete phase changes:

$$q_{\text{total}} = q_{\text{solid warming}} + q_{\text{melting}} + q_{\text{liquid warming}} + q_{\text{vaporization}} + q_{\text{gas warming}}$$

Sources of Error

1. Heat loss to surroundings
2. Heat absorbed by calorimeter
3. Incomplete reactions
4. Temperature measurement errors
5. Impure reagents
6. Heat of stirring