Specific heat capacity and specific latent heat

1. Overview

Thermal physics at the A-Level focuses on the transfer of energy and its effect on the internal energy of a system. When thermal energy is transferred to a substance, it results in an increase in the random distribution of kinetic and potential energies associated with its molecules. This energy transfer manifests in two distinct ways: a change in temperature (governed by specific heat capacity) or a change in phase (governed by specific latent heat). The fundamental principle is that energy is conserved, and the macroscopic changes we observe are direct consequences of microscopic energy redistributions.

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Key Definitions

• Specific Heat Capacity ($c$): The thermal energy per unit mass required to raise the temperature of a substance by one degree.
  • Note: This applies when there is no change of state.
• Specific Latent Heat ($L$): The thermal energy per unit mass required to change the state of a substance at constant temperature.
• Specific Latent Heat of Fusion ($L_f$): The thermal energy per unit mass required to change a substance from solid to liquid (or vice versa) at its melting point without a change in temperature.
• Specific Latent Heat of Vaporisation ($L_v$): The thermal energy per unit mass required to change a substance from liquid to gas (or vice versa) at its boiling point without a change in temperature.
• Internal Energy: The sum of the random distribution of kinetic and potential energies associated with the molecules of a system.

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Content

3.1 Specific Heat Capacity and Temperature Change

When a substance is heated and does not change state, the energy supplied increases the mean kinetic energy of the molecules. Since temperature is a macroscopic measure of the average random kinetic energy of the molecules, the temperature rises.

The energy required is calculated using: $$\mathbf{\Delta Q = mc\Delta \theta}$$

• $\Delta Q$: Change in thermal energy (J)
• $m$: Mass of the substance (kg)
• $c$: Specific heat capacity ($\text{J kg}^{-1} \text{K}^{-1}$ or $\text{J kg}^{-1} ,^\circ\text{C}^{-1}$)
• $\Delta \theta$: Change in temperature (K or $^\circ\text{C}$)

Microscopic Perspective: In solids, molecules vibrate more vigorously about fixed positions. In liquids and gases, the translational kinetic energy of the molecules increases. Because $\Delta \theta \propto \Delta E_k$, the temperature change is directly proportional to the energy absorbed per unit mass.

3.2 Specific Latent Heat and Phase Transitions

During a change of state, energy is supplied to the system, but the temperature remains constant. This energy is not increasing the kinetic energy of the molecules; instead, it is doing work against the intermolecular forces of attraction.

The energy required is calculated using: $$\mathbf{Q = mL}$$

• $Q$: Thermal energy transferred (J)
• $m$: Mass of the substance changing state (kg)
• $L$: Specific latent heat ($\text{J kg}^{-1}$)

Microscopic Perspective:

1. Melting (Fusion): Energy is used to break the regular, rigid lattice structure of the solid. The molecules move slightly further apart, increasing their potential energy.
2. Boiling (Vaporisation): Energy is used to completely overcome the attractive forces between molecules. The molecular separation increases significantly, leading to a massive increase in potential energy.

3.3 Comparing Fusion and Vaporisation

For any given substance, the specific latent heat of vaporisation ($L_v$) is significantly greater than the specific latent heat of fusion ($L_f$). There are two primary reasons for this:

1. Molecular Separation: In fusion, the increase in separation between molecules is small (the density of a liquid is usually similar to its solid). In vaporisation, the molecules must be moved very far apart to overcome all intermolecular forces, requiring much more work.
2. Work against the Atmosphere: When a liquid turns into a gas, it undergoes a massive increase in volume. The substance must do work to push back the surrounding atmosphere. According to the First Law of Thermodynamics ($\Delta U = q + w$), some of the energy supplied ($q$) is used for this expansion work ($w$), meaning more energy is required to achieve the phase change.

3.4 Experimental Determination of $c$ and $L$

In the laboratory, electrical heaters are typically used. The electrical energy supplied is $E = IVt$ (or $P \times t$).

1. Determining Specific Heat Capacity (Electrical Method): For a solid block:

• An immersion heater is placed in a hole in the block, and a thermometer in another.
• The block is insulated to minimize heat loss.
• $IVt = mc\Delta \theta + H$, where $H$ is the heat lost to the surroundings.
• To improve accuracy, a graph of temperature vs. time is plotted. The gradient $\frac{\Delta \theta}{\Delta t}$ is used: $$P = mc \left( \frac{\Delta \theta}{\Delta t} \right) \implies c = \frac{P}{m \times \text{gradient}}$$

2. The Continuous Flow Calorimeter (For Liquids): This is a sophisticated method used to eliminate the unknown heat loss ($h$) to the surroundings.

• Liquid flows at a constant rate through a tube containing an electrical heater.
• The temperature difference ($\Delta \theta$) between the inflow and outflow is measured.
• Two different flow rates ($m_1/t_1$ and $m_2/t_2$) and two different heater powers ($P_1$ and $P_2$) are used to achieve the same $\Delta \theta$.
• Equation 1: $P_1 = \frac{m_1}{t_1}c\Delta \theta + h$
• Equation 2: $P_2 = \frac{m_2}{t_2}c\Delta \theta + h$
• Subtracting the equations: $P_2 - P_1 = (\frac{m_2}{t_2} - \frac{m_1}{t_1})c\Delta \theta$.
• The heat loss $h$ cancels out, allowing for a highly accurate value of $c$.

3. Determining Specific Latent Heat of Vaporisation:

• A liquid is boiled using an electrical heater.
• The mass of liquid turned to steam is measured over time using a balance.
• $P = \frac{dm}{dt}L_v + h$.
• Again, using two different power settings ($P_1$ and $P_2$) to maintain the same boiling conditions allows $h$ to be eliminated.

Worked example 1 — Continuous Flow Calorimetry

A continuous flow calorimeter is used to measure the specific heat capacity of a liquid. In the first experiment, a power of $25.0,\text{W}$ produces a temperature rise of $12.0,\text{K}$ for a mass flow rate of $10.0,\text{g min}^{-1}$. In the second experiment, a power of $45.0,\text{W}$ produces the same temperature rise of $12.0,\text{K}$ for a mass flow rate of $22.0,\text{g min}^{-1}$. Calculate the specific heat capacity of the liquid.

Step 1: Convert units to SI

• $\Delta \theta = 12.0,\text{K}$
• Flow rate 1 ($\dot{m}_1$) = $0.010,\text{kg} / 60,\text{s} = 1.667 \times 10^{-4},\text{kg s}^{-1}$
• Flow rate 2 ($\dot{m}_2$) = $0.022,\text{kg} / 60,\text{s} = 3.667 \times 10^{-4},\text{kg s}^{-1}$

Step 2: Set up the simultaneous equations

• $P_1 = \dot{m}_1 c \Delta \theta + h \implies 25.0 = (1.667 \times 10^{-4}) \times c \times 12.0 + h$
• $P_2 = \dot{m}_2 c \Delta \theta + h \implies 45.0 = (3.667 \times 10^{-4}) \times c \times 12.0 + h$

Step 3: Subtract the equations to eliminate $h$

• $P_2 - P_1 = (\dot{m}_2 - \dot{m}_1) c \Delta \theta$
• $45.0 - 25.0 = (3.667 \times 10^{-4} - 1.667 \times 10^{-4}) \times c \times 12.0$
• $20.0 = (2.00 \times 10^{-4}) \times c \times 12.0$

Step 4: Solve for $c$

• $20.0 = 2.40 \times 10^{-3} \times c$
• $c = \frac{20.0}{2.40 \times 10^{-3}} = 8333.33,\text{J kg}^{-1} \text{K}^{-1}$

Answer: $8.3 \times 10^3,\text{J kg}^{-1} \text{K}^{-1}$ (to 2 s.f.)

Worked example 2 — Energy Balance with Phase Change

A $0.20,\text{kg}$ piece of copper at $150,^\circ\text{C}$ is dropped into an insulated beaker containing $0.50,\text{kg}$ of water and $0.10,\text{kg}$ of ice, all at $0,^\circ\text{C}$. Calculate the final temperature of the mixture. (Data: $c_{Cu} = 390,\text{J kg}^{-1} \text{K}^{-1}$, $c_{water} = 4200,\text{J kg}^{-1} \text{K}^{-1}$, $L_f = 3.3 \times 10^5,\text{J kg}^{-1}$)

Step 1: Calculate energy available from the copper cooling to $0,^\circ\text{C}$

• $Q_{lost} = mc\Delta \theta = 0.20 \times 390 \times (150 - 0) = 11,700,\text{J}$

Step 2: Calculate energy required to melt all the ice

• $Q_{melt} = mL_f = 0.10 \times 3.3 \times 10^5 = 33,000,\text{J}$

Step 3: Compare energies

• Since $Q_{lost} (11,700,\text{J}) < Q_{melt} (33,000,\text{J})$, the copper does not provide enough energy to melt all the ice.
• The system will reach thermal equilibrium while some ice is still melting.

Step 4: Determine final temperature

• As long as ice and water coexist in equilibrium at standard pressure, the temperature must be the melting point.

Answer: $0,^\circ\text{C}$

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Key Equations

Concept	Equation	Symbols & Units	Data Sheet?
Specific Heat Capacity	$\mathbf{\Delta Q = mc\Delta \theta}$	$m$: mass (kg), $c$: spec. heat capacity ($\text{J kg}^{-1}\text{K}^{-1}$), $\Delta \theta$: temp change (K)	Yes
Specific Latent Heat	$\mathbf{Q = mL}$	$m$: mass (kg), $L$: spec. latent heat ($\text{J kg}^{-1}$)	Yes
Electrical Power	$\mathbf{P = IV}$	$P$: power (W), $I$: current (A), $V$: voltage (V)	Yes
Energy Transfer	$\mathbf{Q = Pt}$	$Q$: energy (J), $P$: power (W), $t$: time (s)	No
Internal Energy Change	$\mathbf{\Delta U = q + w}$	$q$: heat added, $w$: work done on system	Yes

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Common Mistakes to Avoid

• ❌ Confusing $t$ (time) and $T$ or $\theta$ (temperature): In equations like $Pt = mc\Delta \theta$, students often mix up the symbols.
  • ✓ Right: Use $t$ for seconds and $\Delta \theta$ or $\Delta T$ for Kelvin/Celsius.
• ❌ Incorrect Mass Units: Using grams because the question provides "200g".
  • ✓ Right: Standardize all masses to kg immediately. $Q = mc\Delta \theta$ requires kg to match the units of $c$ ($\text{J kg}^{-1} \text{K}^{-1}$).
• ❌ Assuming Temperature Change during Phase Change: Calculating $mc\Delta \theta$ for a substance while it is melting.
  • ✓ Right: If the substance is at its melting or boiling point and energy is being added, use $Q = mL$. The temperature does not change until the phase transition is complete.
• ❌ Ignoring the Container: In calorimetry "mixing" problems, forgetting that the beaker/calorimeter also absorbs energy.
  • ✓ Right: $Q_{lost} = Q_{gained_by_liquid} + Q_{gained_by_container}$.
• ❌ Sign Errors in Thermodynamics: Forgetting that work is done by a gas during boiling.
  • ✓ Right: In $\Delta U = q + w$, if a gas expands, $w$ is negative because work is done by the system on the surroundings.

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Exam Tips

1. Definition Precision: When defining Specific Latent Heat, you must include the phrase "at constant temperature". Without this, you will likely lose the mark.
2. The "Why $L_v > L_f$" Question: This is a common 3-mark descriptive question. Structure your answer:
   • Point 1: Mention that $L_v$ involves a much larger increase in molecular separation than $L_f$.
   • Point 2: State that more work is done against intermolecular forces to completely break bonds.
   • Point 3: Mention that work must be done against the atmospheric pressure due to the large volume increase.
3. Graph Gradients: If you are given a graph of Temperature vs. Time, the gradient is $\frac{d\theta}{dt}$. From $P = mc\frac{d\theta}{dt}$, the specific heat capacity $c = \frac{P}{m \times \text{gradient}}$. If the gradient decreases over time, it indicates heat loss to the surroundings is increasing.
4. Thermal Equilibrium: In any "mixing" problem, the final temperature is the same for all components. Set up your equation as: $\sum (\text{Energy Gains}) = \sum (\text{Energy Losses})$.
5. Significant Figures: Cambridge 9702 is strict. If the data is given to 2 s.f., provide your answer to 2 s.f. (or 3 s.f. as a safe margin), but never 1 s.f. or 5 s.f.