Thermal Physics

AS UNIT 3

Oscillations and Nuclei

3.4 Thermal Physics 

Learners should be able to demonstrate and apply their knowledge and understanding of:
a) The idea that the internal energy of a system is the sum of the potential and kinetic energies of its molecules
b) Absolute zero being the temperature of a system when it has minimum internal energy
c) The internal energy of an ideal monatomic gas being wholly kinetic so it is given by [math]U = \frac{1}{2} nRT[/math]
d) The idea that heat enters or leaves a system through its boundary or container wall, according to whether the system’s temperature is lower or higher than that of its surroundings, so heat is energy in transit and not contained within the system
e) The idea that if no heat flows between systems in contact, then they are said to be in thermal equilibrium, and are at the same temperature
f) The idea that energy can enter or leave a system by means of work, so work is also energy in transit
g) The equation W = pΔV can be used to calculate the work done by a gas under constant pressure
h) The idea that even if p changes, W is given by the area under the p – V  graph
i) The use of the first law of thermodynamics, in the form ΔU= Q – W and know how to interpret negative values of ΔU, Q, and W
j) The idea that for a solid (or liquid), W is usually negligible, so Q = ΔU
k)

Q = mcΔθ for a solid or liquid, and this is the defining equation for specific heat capacity, c

Specified Practical Work

o   Estimation of absolute zero by use of the gas laws

o   Measurement of the specific heat capacity for a solid

  • a)    Internal Energy of a System

  • The internal energy (U) of a system is the total energy contained within the system due to the motion and interactions of its molecules. It is the sum of two components:
  • 1. Kinetic Energy:
  • – Due to the motion of molecules (translational, rotational, vibrational).
  • – In ideal gases, only translational kinetic energy contributes to internal energy.
  • 2. Potential Energy:
  • – Due to intermolecular forces (bonding, interactions).
  • – In ideal gases, we assume no intermolecular forces, so potential energy is zero.
  • Thus, the internal energy of an ideal monatomic gas is wholly kinetic.

  • b)   Absolute Zero and Minimum Internal Energy

  • Absolute zero (0K or – 273.15°C ) is the lowest possible temperature.
  • At absolute zero:
  • – Molecular motion completely stops (except for quantum effects).
  • – Kinetic energy is minimum.
  • – Internal energy is at its lowest possible value.
  • Since temperature is proportional to the average kinetic energy of molecules, a lower temperature means lower molecular movement, and at absolute zero, there is no thermal motion.
  • Figure 1 Ideal Gas Law

  • c)    Internal Energy of an Ideal Monatomic Gas

  • For an ideal monatomic gas, internal energy is only kinetic, and given by:
  • [math]U = \frac{3}{2} nRT[/math]
  • ⇒ Derivation of Internal Energy U
  • From kinetic theory:
  • [math]\text{Mean energy of one mole} = \frac{3}{2} kT[/math]
  • Total kinetic energy for N molecules:
  • [math]U = N \times \text{KE}_{\text{mean}} \\
    U = N \times \frac{3}{2} kT[/math]
  • Using ​[math]N =  n N_A \text{and} k = \frac{R}{N_A}[/math]:
  • [math]U = n N_A \times \frac{3}{2} \left( \frac{R}{N_A} \right) T[/math]
  • Thus, for an ideal monatomic gas, internal energy depends only on temperature and increases when heat is added.

  • d)   Heat as Energy in Transit

  • Heat (Q) is not contained within a system; it is energy in transit.
  • Heat enters or leaves a system due to a temperature difference between the system and its surroundings.
  • Heat is transferred by:
    1. Conduction – Direct molecular collisions (e.g., metal rod heating).
    2. Convection – Fluid motion transfers heat (e.g., boiling water).
    3. Radiation – Electromagnetic waves (e.g., sunlight heating Earth).
  • Since heat is energy in transit, once it enters a system, it transforms into internal energy or does work on the system.
  • Heat is not a substance but energy moving between systems.
  • A system does not contain heat—it contains internal energy.
  • The transfer of heat stops when thermal equilibrium (equal temperature) is reached.

  • e)    Thermal Equilibrium and Temperature

  • Thermal equilibrium occurs when two or more systems in contact exchange no net heat energy, meaning they are at the same temperature.
  • When two objects of different temperatures are placed in thermal contact, heat flows from the hotter object to the cooler one until they reach the same temperature.
  • Once no more heat flows, they are in thermal equilibrium.
  • ⇒ Example:
  • – A hot cup of coffee in a cold room will lose heat to the air until it reaches room temperature.
  • – A thermometer measures temperature because it reaches thermal equilibrium with the object it touches.
  • Figure 2 Thermal Equilibrium
  • ⇒ Zeroth Law of Thermodynamics:
  • If system A is in thermal equilibrium with system B, and system B is in thermal equilibrium with system C, then A is also in thermal equilibrium with C.
  • This law defines temperature as the property that determines whether heat will flow between systems.
  • Figure 3 Zeroth Law of thermodynamics

  • f)     Work as Energy in Transit

  • Energy can enter or leave a system through work (W), just like heat (Q).
  • – Heat: Energy transfer due to temperature difference.
  • – Work: Energy transfer due to forces acting on or by the system (e.g., expansion or compression of a gas).
  • Since work transfers energy, it is not stored within the system. Instead, it either:
  • – Increases the system’s internal energy (if work is done on the system).
  • – Decreases the internal energy (if the system does work on the surroundings).
  • ⇒ Example of Work in Thermodynamics:
  • – Compressing a gas (e.g., pushing a piston inward) increases its internal energy.
  • – Allowing a gas to expand (e.g., releasing a piston) decreases its internal energy.

  • g) Work Done by a Gas:

  • When a gas expands or contracts at constant pressure (p), the work done by the gas is given by:
  • [math]W = pΔV[/math]
  • Where:
  • – W = work done by the gas,
  • – p = pressure (assumed constant),
  • – ΔV = change in volume ([math]V_f – V_i[/math] ).
  • Figure 4 Work done by an ideal gas
  • ⇒ Understanding [math]W = pΔV[/math]
  • If V increases (gas expands), [math]ΔV > 0[/math] gas does positive work on surroundings.
  • If V decreases (gas compresses),[math]ΔV < 0[/math] work is done on the gas (internal energy increases).
  • ⇒ Example Calculation:
  • A gas expands from 2 L to 5 L at a constant pressure of 100 kPa.
  • [math]W = p \Delta V \\
    W = (100 \times 10^3 \text{ Pa}) \times (5 – 2) \times 10^{-3} \text{ m}^3 \\
    W = 100,000 \times 3 \times 10^{-3} \\
    W = 300 \text{ J}[/math]
  • The gas does 300 J of work on its surroundings.

  • h) Work Done When Pressure Varies: Area Under [math]p -V[/math]  Graph

  • If pressure is not constant, the work done is found by calculating the area under the pressure-volume ([math]p – V[/math] ) graph.
  • ⇒ Work the Area Under a [math]p – V[/math]  Graph:
  • Work is defined as force × displacement.
  • In a gas,
  • [math]force = pressure × area, \, and \, displacement = volume \, change[/math]
  • Figure 5 Work done equals the area under the curve on a [math]p – V[/math] diagram
  • Mathematically, for varying pressure:
  • [math]W = \int_{V_i}^{V_f} p \, dV[/math]
  • Which represents the area under the [math]p – V[/math]  curve.
  • ⇒  Cases for [math]p – V[/math]  Graphs:
  • 1. Constant Pressure Process (Isobaric)
  • – p is constant → Rectangle on the [math][/math]  graph.
  • – Work is simply [math]W = pΔV[/math] (area of rectangle).
  • 2. Changing Pressure (Non-Isobaric Process)
  • – Work is not directly [math]pΔV[/math]  but the total area under the curve.
  • – Found using integration or graphical methods (e.g., trapezoidal rule).
  • 3. Constant Volume Process (Isochoric, ΔV=0)
  • – No change in volume → No work done (W=0).
  • – Internal energy changes only due to heat transfer.
  • 4. Adiabatic Process (No Heat Exchange, Q=0)
  • – Work done changes internal energy (ΔU=W).

  • i)   The First Law of Thermodynamics: [math]ΔU = Q – W/math]

  • The First Law of Thermodynamics states that the internal energy (U) of a system can change due to heat transfer (Q) and work done (W):
  • [math]ΔU = Q – W[/math]
  • Where:
  • – ΔU = Change in internal energy of the system.
  • – Q = Heat energy added to the system (+ve if heat enters, -ve if heat leaves).
  • – W = Work done by the system on its surroundings (+ve if system expands, -ve if work is done on the system).
  • This equation expresses the conservation of energy in thermodynamics:
  • – The internal energy of a system increases if heat is added or if work is done on the system.
  • – The internal energy decreases if heat is lost or if the system does work on the surroundings.
  • Figure 6 First law of thermodynamics
  • ⇒ Interpreting Positive and Negative Values in [math]ΔU = Q – W[/math]
Quantity Positive Value (+) Negative Value (−)
ΔU (Change in Internal Energy) Internal energy increases Internal energy decreases
Q (Heat Transfer) Heat added to system Heat removed from system
W (Work Done) Work done by the system (expansion) Work done on the system (compression)
  • ⇒  Examples:
  • 1. Gas expansion at constant heat input:
  • – Q is positive (heat added).
  • – W is positive (gas expands, doing work).
  • – ΔU may be small if most of the heat goes into work rather than raising internal energy.
  • 2. Gas compression by an external force:
  • – W is negative (work is done on the gas).
  • – If no heat is added (Q=0), then ΔU is positive (internal energy increases, temperature rises).
  • 3. Adiabatic expansion (Q=0):
  • – W is positive (gas expands).
  • – ΔU is negative (internal energy decreases, temperature drops).

  • j)   Work Done in Solids and Liquids: [math]W ≈ 0 \text{and} Q = ΔU[/math]

  • For solids and liquids, work done due to volume change is negligible because:
  • – Solids and liquids do not expand significantly under normal temperature changes.
  • – Unlike gases, their compressibility is very low.
  • Since [math]W ≈ 0[/math], the First Law simplifies to:
  • [math]ΔU = Q[/math]
  • This means that for solids and liquids, almost all the heat added goes into changing the internal energy, which usually manifests as a temperature change.

  • e)    Specific Heat Capacity and the Equation [math]Q = mcΔθ[/math]

  • For solids and liquids, the amount of heat energy required to change the temperature is given by:
  • [math]Q = mcΔθ[/math]
  • OR
  • [math]Q = mcΔT[/math]
  • Where:
  • – Q = Heat energy supplied (Joules, J).
  • – m = Mass of the substance (kg).
  • – c = Specific heat capacity ( [math]Jkg^{-1} k^{-1}[/math], i.e., Joules per kilogram per degree Kelvin).
  • – ∆θ = Change in temperature (K or °C).
  • This equation defines specific heat capacity (c), which is:
  • [math]c = \frac{Q}{m∆θ}[/math]
  • Specific heat capacity is the amount of heat required to raise the temperature of 1 kg of a substance by 1 K (or 1 °C).
  • ⇒  Understanding Specific Heat Capacity
  • Different materials absorb different amounts of heat for the same temperature change.
  • High ccc (e.g., water, 4184 J/kgK) → absorbs a lot of heat with a small temperature rise.
  • Low ccc (e.g., metals like copper, 385 J/kgK) → heats up quickly.
  • ⇒  Example Calculation:
  • How much heat is needed to raise the temperature of 2 kg of water from 20°C to 80°C?
  • Given:
  • m = 2 kg
  • c = 4184J/kgK(for water)
  • Δθ = 80 − 20 = 60 K
  • [math]Q = mc \Delta \theta \\
    Q = (2)(4184)(60) \\
    Q = 502,080 \text{ J} \\
    Q = 502 \text{ kJ}[/math]
  • So, 502 kJ of heat is required to heat 2 kg of water by 60°C.

  • Specified Practical Work

  • 1. Estimation of Absolute Zero by Use of the Gas Laws
  • ⇒ Principle:
  • At constant volume, the pressure of a fixed amount of gas is directly proportional to its absolute temperature (in kelvins). This relationship is expressed by the ideal gas law in the form:
  • [math]\frac{p}{T} = \text{constant} \quad \text{(at constant volume)}[/math]
  • Thus, if you measure the pressure of a gas at different temperatures (for example, by cooling and heating it) and plot pressure versus temperature (using a temperature scale in °C), you obtain a straight line. Extrapolating this line to the point where the pressure would be zero gives an estimate of absolute zero (which should be about -273.15 °C).
  • Figure 7 The ideal gas Law
  • ⇒  Experimental Setup:
  • 1. Apparatus:
  • – A rigid, sealed container (often a fixed-volume syringe or a glass vessel) with a pressure sensor.
  • – A thermometer (or a temperature probe) to measure the temperature of the gas inside the container.
  • – Means of varying the temperature (for example, an ice bath, water bath, and a heating source).
  • – Data logger (optional) to record pressure and temperature readings.
  • 2. Procedure:
  • – Step 1: Ensure the gas inside the sealed container is at thermal equilibrium at a known starting temperature (e.g., room temperature). Record the pressure ​ and the temperature ​[math]T_1[/math] (in °C).
  • – Step 2: Change the temperature of the container gradually (for example, by placing it in an ice bath and then a warm water bath) and record corresponding pressures p at various temperatures T (in °C).
  • – Step 3: Plot a graph of pressure p versus temperature T (using the Celsius scale). The data should lie approximately along a straight line.
  • – Step 4: Extrapolate the linear fit of the data to the point where the pressure would be zero. The temperature at this intercept gives an estimate of absolute zero in °C.
  • – Step 5: If desired, convert to Kelvin by adding 273.15.
  • ⇒  Data Analysis and Calculation:
  • For example, suppose you obtain a set of points and the best-fit line is given by:
  • [math]p = mT + b[/math]
  • Where m is the slope and b is the intercept (with p measured in kPa, for instance).
  • Set p=0 to find the critical temperature [math]T_c[/math]:
  • [math]0 = mT_c + b \\
    T_c = -\frac{b}{m}[/math]
  • If your calculated ​[math]T_c[/math] is around -273 °C, then your experiment has successfully estimated absolute zero.
  • ⇒  Points of Consideration:
  • Accuracy: The container must be rigid so that volume remains constant.
  • Calibration: The pressure sensor and thermometer must be properly calibrated.
  • Error Sources: Temperature gradients, sensor inaccuracies, and small leaks can introduce errors.
  • 2. Measurement of the Specific Heat Capacity for a Solid
  • ⇒  Principle:
  • The specific heat capacity ccc of a material is defined as the amount of energy Q required to raise the temperature of a unit mass mmm of the substance by one degree Celsius (or one Kelvin):
  • [math]Q = mcΔT[/math]
  • By supplying a known amount of energy to a solid and measuring its temperature rise, you can determine its specific heat capacity.
  • ⇒  Experimental Setup:
  • 1. Apparatus:
  • – A sample of the solid with known mass m
  • – An electrical heater (resistive element) attached to or embedded in the solid.
  • – A power supply with a means of measuring current I and voltage V.
  • – A thermometer or thermocouple to record the temperature change ΔT.
  • – An insulating container to minimize heat losses.
  • Figure 8 To determine specific heat capacity
  • 2. Procedure:
  • – Step 1: Weigh the solid sample accurately.
  • – Step 2: Set up the heater in contact with the solid; ensure good thermal contact.
  • – Step 3: Measure the initial temperature [math]T_o[/math]​ of the solid.
  • – Step 4: Turn on the heater and supply a known electrical energy for a specific time t. The electrical energy delivered is given by:
  • [math]Q = IVt[/math]
  • Where I is the current and V is the voltage.
  • – Step 5: Measure the final temperature [math]T_f[/math] after heating.
  • – Step 6: Calculate the temperature change:
  • [math]∆T = T_f – T_0[/math]
  • Step 7: Calculate the specific heat capacity ccc using:
  • [math]c = \frac{Q}{m∆T}[/math]
  • ⇒  Data Analysis and Calculation:
  • Example Calculation:
  • – Suppose the mass m = 0.2 kg of the sample.
  • – The heater operates at V = 12 V and I = 1 A for t = 300 s. Thus:
  • [math]Q = IVt \\
    Q = (1)(12)(300) \\
    Q = 3600 \text{ J}[/math]
  • – If the temperature increases from [math]T_0 = 20℃[/math] to [math]T_f = 40℃[/math], then:
  • [math]\Delta T = T_f – T_0 \\
    \Delta T = 40 – 20 \\
    \Delta T = 20^\circ \text{C}[/math]
  • ⇒  Points of Consideration:
  • Heat Losses:
  • To improve accuracy, the experiment should be conducted with good insulation or corrections made for heat losses.
  • Measurement Accuracy:
  • Accurate measurement of current, voltage, time, and temperature is crucial.
  • Uniform Heating:
  • Ensure the entire solid reaches a uniform temperature before taking the final reading.
  • ⇒  Conclusion:
  • Estimation of Absolute Zero:
  • By measuring the pressure of a gas at various temperatures and extrapolating the linear relationship (at constant volume), you can estimate absolute zero. The ideal gas law p/T = constant underpins this method.
  • Measurement of Specific Heat Capacity:
  • Using an electrical heating method, where the energy supplied is known (via [math]Q = IVt[/math]) and the temperature change of a known mass of solid is measured, the specific heat capacity is calculated as
  • [math]c = \frac{Q}{mΔT}[/math]
  • ​Both experiments offer practical insights into fundamental thermodynamic properties and illustrate how careful measurement and controlled conditions can reveal intrinsic physical constants and material properties.
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