Heat Capacity and Calorimetry
Specific Heat Capacity ($ c = \frac{Q}{m \Delta T} $)
When heat is added to a substance, its temperature usually increases. The amount by which the temperature rises depends on the amount of heat added, the mass of the substance, and the nature of the substance itself. Different substances require different amounts of heat to raise the temperature of a given mass by one degree. This property is quantified by heat capacity and specific heat capacity.
Heat Capacity (C)
The heat capacity ($C$) of a body is defined as the amount of heat energy required to raise the temperature of the entire body by one degree (Celsius or Kelvin).
$ C = \frac{Q}{\Delta T} $
where $Q$ is the amount of heat added and $\Delta T$ is the resulting change in temperature. The units of heat capacity are Joules per degree Celsius (J/°C) or Joules per Kelvin (J/K). Heat capacity is an extensive property, meaning it depends on the size or mass of the body.
Specific Heat Capacity (c)
The specific heat capacity ($c$) of a substance is defined as the amount of heat energy required to raise the temperature of a unit mass of the substance by one degree (Celsius or Kelvin).
$ c = \frac{\text{Heat Energy}}{\text{Mass} \times \text{Change in Temperature}} = \frac{Q}{m \Delta T} $
where $m$ is the mass of the substance. The units of specific heat capacity are Joules per kilogram per degree Celsius (J/kg·°C) or Joules per kilogram per Kelvin (J/kg·K). Specific heat capacity is an intensive property, meaning it is a characteristic of the substance itself and does not depend on the amount of the substance.
From the definition, the heat required to change the temperature of a mass $m$ of a substance with specific heat capacity $c$ by $\Delta T$ is:
$ Q = m c \Delta T $
This is a fundamental formula for calculating heat transfer associated with temperature changes in a substance.
The heat capacity of a body is related to the specific heat capacity of the substance it is made of by $C = mc$.
Different substances have different specific heat capacities. Water has a high specific heat capacity (approx. 4186 J/kg·°C or 4186 J/kg·K). This means water requires a relatively large amount of heat to change its temperature, which is why it is used as a coolant and why coastal areas have moderate climates (large bodies of water store and release heat slowly).
Molar Heat Capacity ($C_m$)
For gases, it is often more convenient to use molar heat capacity ($C_m$), which is the heat capacity per mole of the substance. The units are Joules per mole per degree Kelvin (J/mol·K).
$ C_m = \frac{Q}{n \Delta T} $
where $n$ is the number of moles. The heat required for $n$ moles is $Q = n C_m \Delta T$.
For gases, the heat required to raise the temperature by a certain amount depends on whether the process occurs at constant volume or constant pressure, because work may be done by or on the gas during expansion or compression. Therefore, two specific molar heat capacities are defined for gases:
- Molar heat capacity at constant volume ($C_v$): Heat required per mole to raise temperature by 1 degree at constant volume.
- Molar heat capacity at constant pressure ($C_p$): Heat required per mole to raise temperature by 1 degree at constant pressure.
For an ideal gas, $C_p > C_v$ because at constant pressure, some of the heat added is used by the gas to do work during expansion, in addition to increasing its internal energy (which is directly related to temperature). At constant volume, all the heat added increases internal energy and temperature.
The relationship between $C_p$ and $C_v$ for an ideal gas is $C_p - C_v = R$, where $R$ is the ideal gas constant. The ratio $\gamma = C_p/C_v$ is important in thermodynamics.
Heat Required vs. Temperature Change
The formula $Q = mc\Delta T$ is applicable when there is no change of state (like melting or boiling) occurring. During a change of state, the temperature remains constant while heat is added or removed; this heat is called latent heat (discussed in the next section).
Example 1. How much heat energy is required to raise the temperature of 0.5 kg of water from 20°C to 80°C? (Specific heat capacity of water = 4186 J/kg·°C).
Answer:
Mass of water, $m = 0.5$ kg.
Initial temperature, $T_{initial} = 20^\circ\text{C}$.
Final temperature, $T_{final} = 80^\circ\text{C}$.
Change in temperature, $\Delta T = T_{final} - T_{initial} = 80^\circ\text{C} - 20^\circ\text{C} = 60^\circ\text{C}$.
Specific heat capacity of water, $c = 4186$ J/kg·°C.
The amount of heat energy required is given by $Q = mc\Delta T$.
$ Q = (0.5 \text{ kg}) \times (4186 \text{ J/kg} \cdot ^\circ\text{C}) \times (60 \, ^\circ\text{C}) $
$ Q = 0.5 \times 4186 \times 60 $ Joules
$ Q = 2093 \times 60 $ J
$ Q = 125580 $ J.
The heat energy required is 125,580 Joules, or 125.58 kJ.
Calorimetry
Calorimetry is the science or act of measuring the amount of heat transferred during a physical or chemical process. The key instrument used in calorimetry is a calorimeter, a device designed to measure heat exchange while isolating the system from its surroundings.
The principle of calorimetry is based on the conservation of energy, specifically the principle that in an isolated system, heat lost by hot bodies equals the heat gained by cold bodies.
In a simple calorimetry experiment, a calorimeter (usually a well-insulated container, often made of copper or aluminium due to their high thermal conductivity, with a stirrer and thermometer) is used. A known mass of a substance at a known initial temperature is placed inside the calorimeter, which contains a known mass of a calorimeter liquid (usually water) at a known temperature. Heat is exchanged between the substance, the liquid, and the calorimeter itself until they all reach a common final temperature.
Principle of Method of Mixtures
The most common calorimetry technique is the method of mixtures. When substances at different temperatures are mixed in an insulated calorimeter, heat flows from the hotter substances to the colder substances until thermal equilibrium is reached. Assuming no heat is lost to the surroundings (due to insulation) or the calorimeter itself (if its heat capacity is negligible or accounted for), the total heat lost by the hot components is equal to the total heat gained by the cold components.
Heat Lost = Heat Gained
Consider mixing a hot substance A (mass $m_A$, specific heat capacity $c_A$, initial temperature $T_A$) with a cold substance B (mass $m_B$, specific heat capacity $c_B$, initial temperature $T_B$) in a calorimeter containing liquid C (mass $m_C$, specific heat capacity $c_C$, initial temperature $T_C$). Assume $T_A > T_C = T_B$. When they reach thermal equilibrium, let the final temperature be $T_f$.
Substance A loses heat: $Q_{lost, A} = m_A c_A (T_A - T_f)$ (since $T_A > T_f$)
Substance B gains heat: $Q_{gained, B} = m_B c_B (T_f - T_B)$ (since $T_f > T_B$)
Liquid C gains heat: $Q_{gained, C} = m_C c_C (T_f - T_C)$ (since $T_f > T_C$)
By the principle of calorimetry:
$ Q_{lost, A} = Q_{gained, B} + Q_{gained, C} $
$ m_A c_A (T_A - T_f) = m_B c_B (T_f - T_B) + m_C c_C (T_f - T_C) $
If the heat capacity of the calorimeter is considered ($C_{cal}$), the term $m_C c_C$ is replaced by $C_{cal}$. If the calorimeter liquid and the calorimeter itself are initially at the same temperature, the equation becomes:
$ m_A c_A (T_A - T_f) = (m_B c_B + C_{cal}) (T_f - T_B) $
If the specific heat capacity of the calorimeter material is known ($c_{cal}$) and its mass is $m_{cal}$, then $C_{cal} = m_{cal} c_{cal}$. Some problems use the concept of water equivalent of the calorimeter, which is the mass of water that has the same heat capacity as the calorimeter. If the water equivalent is $W_{equiv}$, then $C_{cal} = W_{equiv} c_{water}$.
Applications of Calorimetry
- Determination of specific heat capacities: The method of mixtures is commonly used to experimentally determine the specific heat capacity of unknown solids or liquids.
- Measurement of heat involved in reactions: Calorimetry is used in chemistry to measure the heat released or absorbed during chemical reactions (e.g., enthalpy changes). Bomb calorimeters are used for reactions at constant volume.
- Measurement of latent heats: Calorimetry experiments can be designed to measure the latent heat of fusion (melting) or vaporisation (boiling) of substances.
Example 2. A piece of metal of mass 0.2 kg at 150°C is dropped into 0.5 kg of water at 25°C. The final temperature of the mixture is 30°C. If the specific heat capacity of water is 4200 J/kg·°C, find the specific heat capacity of the metal. Assume the calorimeter is perfectly insulated and its heat capacity is negligible.
Answer:
Mass of metal, $m_{metal} = 0.2$ kg.
Initial temperature of metal, $T_{metal, initial} = 150^\circ\text{C}$.
Mass of water, $m_{water} = 0.5$ kg.
Initial temperature of water, $T_{water, initial} = 25^\circ\text{C}$.
Final temperature of mixture, $T_f = 30^\circ\text{C}$.
Specific heat capacity of water, $c_{water} = 4200$ J/kg·°C.
Let the specific heat capacity of the metal be $c_{metal}$.
Heat lost by the hot metal: $Q_{lost, metal} = m_{metal} c_{metal} (T_{metal, initial} - T_f)$.
$ Q_{lost, metal} = (0.2 \text{ kg}) \times c_{metal} \times (150^\circ\text{C} - 30^\circ\text{C}) = 0.2 c_{metal} \times 120 $ J
$ Q_{lost, metal} = 24 c_{metal} $ J.
Heat gained by the cold water: $Q_{gained, water} = m_{water} c_{water} (T_f - T_{water, initial})$.
$ Q_{gained, water} = (0.5 \text{ kg}) \times (4200 \text{ J/kg} \cdot ^\circ\text{C}) \times (30^\circ\text{C} - 25^\circ\text{C}) = 0.5 \times 4200 \times 5 $ J
$ Q_{gained, water} = 2100 \times 5 = 10500 $ J.
By the principle of calorimetry (Heat Lost = Heat Gained, neglecting calorimeter heat capacity):
$ Q_{lost, metal} = Q_{gained, water} $
$ 24 c_{metal} = 10500 $
$ c_{metal} = \frac{10500}{24} $ J/kg·°C.
$ c_{metal} = \frac{5250}{12} = \frac{2625}{6} = \frac{875}{2} = 437.5 $ J/kg·°C.
The specific heat capacity of the metal is 437.5 J/kg·°C.