Defining Specific Heat Capacity
- The specific heat capacity of substance is defined as:
The amount of thermal energy required to raise the temperature of $1 \mathrm{~kg}$ of a substance by $1{ }^{\circ} \mathrm{C}$ - This quantity determines the amount of energy needed to change the temperature of a substance
- The specific heat capacity is measured in units of Joules per kilogram per Kelvin $\left(\mathrm{Jg}^{-1} \mathrm{~K}^{-1}\right)$ or Joules per kilogram per Celsius $\left(\mathrm{J} \mathrm{kg}^{-10} \mathrm{C}^{-1}\right)$ and has the symbol $\boldsymbol{c}$
- Different substances have different specific heat capacities
- Specific heat capacity is mainly used in liquids and solids
- From the definition of specific heat capacity, it follows that:
- The heavier the material, the more thermal energy that will be required to raise its temperature
- The larger the change in temperature, the higher the thermal energy will be required to achieve this change
Calculating Specific Heat Capacity
The amount of thermal energy $Q$ needed to raise the temperature by $\Delta \theta$ for a mass $m$ with specific heat capacity $c$ is equal to:
$
\Delta \mathbf{Q}=\operatorname{mc} \Delta \boldsymbol{\theta}
$
- Where:
- $\Delta \mathrm{Q}=$ change in thermal energy (J)
- $\mathrm{m}=$ mass of the substance you are heating up $(\mathrm{kg})$
- $\mathrm{c}=$ specific heat capacity of the substance $\left(\mathrm{Jgg}^{-1} \mathrm{~K}^{-1}\right.$ or $\left.\mathrm{J} \mathrm{kg}^{-1} \mathrm{C}^{-1}\right)$
- $\Delta \theta=$ change in temperature $\left(\mathrm{K}\right.$ or $\left.{ }^{\circ} \mathrm{C}\right)$
- If a substance has a low specific heat capacity, it heats up and cools down quickly
- If a substance has a high specific heat capacity, it heats up and cools down slowly
- The specific heat capacity of different substances determines how useful they would be for a specific purpose eg. choosing the best material for kitchen appliances
- Good electrical conductors, such as copper and lead, are also excellent conductors of heat due to their low specific heat capacity
Worked example: Calculating specific heat capacity
A kettle is rated at $1.7 \mathrm{~kW}$. A mass of $650 \mathrm{~g}$ of a liquid at $25^{\circ} \mathrm{C}$ is poured into a kettle. When the kettle is switched on, it takes 3.5 minutes to start boiling. Calculate the specific heat capacity of the liquid.
Answer/Explanation
Step 1:
Calculate the Energy from the power and time
$
\begin{gathered}
\text { Energy }=\text { Power } \times \text { Time } \\
\text { Power }=1.7 \mathrm{~kW}=1.7 \times 10^3 \mathrm{~W} \\
\text { Time }=3.5 \text { minutes }=3.5 \times 60=210 \mathrm{~s} \\
\text { Energy }=1.7 \times 10^3 \times 210=3.57 \times 10^5 \mathrm{~J}
\end{gathered}
$
Step 2:
Thermal energy equation
$
\Delta Q=\operatorname{mc} \Delta \theta
$
Step 3:
Rearrange for specific heat capacity
$
c=\frac{\Delta Q}{m \Delta \theta}
$
Step 4:
Substitute in values
$
\begin{gathered}
m=650 \mathrm{~g}=650 \times 10^{-3} \mathrm{~kg} \\
\Delta \theta=100-25=75^{\circ} \mathrm{C}
\end{gathered}
$
$
c=\frac{3.57 \times 10^5}{650 \times 10^{-3} \times 75}=7323.07 \ldots=7300 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{C}^{-1}(2 \mathrm{~s} . \mathrm{f})
$
Exam Tip
The difference in temperature $\Delta \theta$ will be exactly the same whether the temperature is given in Celsius or Kelvin. Therefore, there is no need to convert between the two since the difference in temperature will be the same for both units.
Defining Latent Heat Capacity
- Energy is required to change the state of substance
- Examples of changes of state are:
- Melting $=$ solid to liquid
- Evaporation/vaporisation/boiling = liquid to gas
- Sublimation $=$ solid to gas
- Freezing $=$ liquid to solid
- Condensation $=$ gas to liquid
- When a substance changes state, there is no temperature change
- The energy supplied to change the state is called the latent heat and is defined as:
The thermal energy required to change the state of $1 \mathrm{~kg}$ of mass of a substance without any change of temperature - There are two types of latent heat:
- Specific latent heat of fusion (melting)
- Specific latent heat of vaporisation (boiling)
- The specific latent heat of fusion is defined as:
The thermal energy required to convert $1 \mathrm{~kg}$ of solid to liquid with no change in temperature
- This is used when melting a solid or freezing a liquid
- The specific latent heat of vaporisation is defined as:
- The thermal energy required to convert $1 \mathrm{~kg}$ of liquid to gas with no change in temperature This is used when vaporising a liquid or condensing a gas
Calculating Specific Latent Heat
- The amount of energy $Q$ required to melt or vaporise a mass of $m$ with latent heat $L$ is:
$
\mathbf{Q}=\mathbf{m L}
$
- Where:
- $Q=$ amount of thermal energy to change the state (J)
- $\mathrm{m}=$ mass of the substance changing state $(\mathrm{kg})$
- $L=$ latent heat of fusion or vaporisation $\left(\mathrm{Jgg}^{-1}\right)$
- The values of latent heat for water are:
- Specific latent heat of fusion $=330 \mathrm{~kJ} \mathrm{~kg}^{-1}$
- Specific latent heat of vaporisation $=2.26 \mathrm{MJ} \mathrm{kg}^{-1}$
- Therefore, evaporating $1 \mathrm{~kg}$ of water requires roughly seven times more energy than melting the same amount of ice to form water
- The reason for this is to do with intermolecular forces:
- When ice melts: energy is required to just increase the molecule separation until they can flow freely over each other
- When water boils: energy is required to completely separate the molecules until there are no longer forces of attraction between the molecules, hence this requires much more energy
Worked example: Specific latent heat
The energy needed to boil a mass of $530 \mathrm{~g}$ of a liquid is $0.6 \mathrm{MJ}$. Calculate the specific latent heat of the liquid and state whether it is the latent heat of vaporisation or fusion.
Answer/Explanation
Step 1: Write the thermal energy required to change state equation
$
\mathbf{Q}=\mathbf{m L}
$
Step 2:
Rearrange for latent heat
$
\mathrm{L}=\frar />Q=0.6 \mathrm{MJ}=0.6 \times 10^6 \mathrm{~J} \\
L=\frac{0.6 \times 10^6}{530 \times 10^{-3}}=1.132 \times 10^6 \mathrm{~J} \mathrm{~kg}^{-1}=1.1 \mathrm{MJ} \mathrm{kg} \text { (2 s.f.) }
\end{gathered}
$
$\mathbf{L}$ is the latent heat of vaporisation because the change in state is from liquid to gas (boiling)
Exam Tip
Use these reminders to help you remember which type of latent heat is being referred to:
- Latent heat of fusion = imagine ‘fusing’ the liquid molecules together to become a solid
- Latent heat of vaporisation = “water vapour” is steam, so imagine vaporising the liquid molecules into a gas
$
Step 3:
Substitute in values
$
\begin{gathered}
m=530 \mathrm{~g}=530 \times 10^{-3} \mathbf{k g} \\<b