Unit 5: Thermodynamics, Radiation, Oscillations and Cosmology

To warm $0.40\,\mathrm{kg}$ of ice with $c=2100\,\mathrm{J\,kg^{-1}\,K^{-1}}$ through $15\,\mathrm{K}$ requires $12.6\,\mathrm{kJ}$. Melting that mass then requires a separate $L\Delta m$ calculation. Add energies for successive stages; do not combine their equations into one substitution.

Place an NTC thermistor and fixed resistor in series across a stable d.c. supply. Measure the potential-divider output across the chosen component with a voltmeter, while the thermistor and a calibrated thermometer or temperature probe sit close together in a stirred water bath.

An NTC thermistor's resistance falls as temperature rises. If output is measured across the thermistor, $V_{out}=V_sR_T/(R_T+R)$ falls; measured across the fixed resistor, it rises. A comparator or relay can switch heating when the calibrated threshold is crossed.

For latent heat of vaporisation, use an immersion heater to keep a liquid boiling steadily. Measure heater voltage and current, and use a balance to determine the mass lost during a timed interval after steady boiling has begun. A lid or insulation reduces energy transfer to the surroundings without sealing the vessel.

A substance's internal energy is the total randomly distributed kinetic energy and intermolecular potential energy of its molecules. Random molecular motion contributes kinetic energy; relative molecular positions and interactions contribute potential energy.

On a cooling curve, a sloping section shows falling temperature and therefore falling average molecular kinetic energy. A constant-temperature phase-change section can still show decreasing internal energy because molecular potential energy decreases.

For an ideal gas, the average translational kinetic energy of its molecules is directly proportional to absolute temperature: $\langle E_k\rangle\propto T$. Equal absolute temperatures therefore mean equal average translational kinetic energy, even for gases with different molecular masses.

The kelvin scale starts at absolute zero: $0\,\mathrm{K}=-273.15\,^{\circ}\mathrm{C}$. Convert with $T/\mathrm{K}=\theta/^{\circ}\mathrm{C}+273.15$. A rise of $1\,\mathrm{K}$ is the same size as a rise of $1\,^{\circ}\mathrm{C}$, but the scale origins differ.

When temperature rises, molecules have greater mean-square speed and momentum, so wall collisions transfer momentum more frequently or more strongly. This microscopic change explains increases in gas pressure or volume under the relevant macroscopic constraint.

Use pressure $p$ in pascals, volume $V$ in cubic metres, absolute temperature $T$ in kelvin, number of molecules $N$, and Boltzmann constant $k=1.38\times10^{-23}\,\mathrm{J\,K^{-1}}$. The equation models an ideal gas in equilibrium.

For $N=5.0\times10^{22}$ molecules at $300\,\mathrm{K}$ occupying $1.8\times10^{-3}\,\mathrm{m^3}$, $p=NkT/V=1.15\times10^5\,\mathrm{Pa}$. Convert litres to cubic metres and use the number of molecules, not the number of moles.

Trap a fixed mass of gas in a sealed syringe or Boyle's-law apparatus connected to an absolute pressure sensor. Change the volume in measured steps, secure the piston, and wait for the pressure to stabilise before recording each pair.

Repeat readings across the safe range. Either calculate $pV$ for several pairs and look for a constant value, or plot $p$ against $1/V$: a straight line through the origin supports $p\propto1/V$ at fixed temperature.

Kinetic theory gives $pV=\tfrac13Nm\langle c^2\rangle$, where $m$ is one molecule's mass and $\langle c^2\rangle$ is the mean square speed. The ideal gas equation gives $pV=NkT$. Equating them and cancelling $N$ produces the required molecular-energy relation.

The left side is the average translational kinetic energy per molecule. It can also be written as $3pV/(2N)$. For $p=1.15\times10^5\,\mathrm{Pa}$, $V=1.77\times10^{-3}\,\mathrm{m^3}$ and $N=5.15\times10^{22}$, the average is $5.93\times10^{-21}\,\mathrm{J}$.

Nuclear binding energy is the energy required to separate a nucleus completely into its individual protons and neutrons. The bound nucleus has less mass than those free nucleons, and this mass deficit corresponds to the binding energy.

Use mutually consistent nuclear masses, calculate the positive mass deficit, convert it to kilograms if energy is required in joules, then multiply by $c^2$. Binding energy per nucleon is the total binding energy divided by nucleon number $A$.

A mass deficit of $0.528\,\mathrm{u}$ corresponds to about $492\,\mathrm{MeV}$ using $1\,\mathrm{u}c^2\approx931.5\,\mathrm{MeV}$. For a 56-nucleon nucleus this is about $8.8\,\mathrm{MeV}$ per nucleon.

One atomic mass unit is defined as one twelfth of the mass of a neutral carbon-12 atom: $1\,\mathrm{u}=1.6605\times10^{-27}\,\mathrm{kg}$. It is a mass unit, suited to atoms, nuclei and particles.

A mass of $4.00\,\mathrm{u}$ is $4.00(1.6605\times10^{-27})=6.64\times10^{-27}\,\mathrm{kg}$. A mass loss of $0.0020\,\mathrm{u}$ releases about $0.0020(931.5)=1.86\,\mathrm{MeV}$.

The binding-energy-per-nucleon curve rises steeply for light nuclei, reaches a maximum near iron, then falls gradually for very heavy nuclei. A higher position means nucleons are more tightly bound on average.

Calculate the total binding energy of reactants and products, not just their vertical coordinates. Energy released equals the increase in total binding energy; equivalently, the products have lower total mass and the mass difference appears as kinetic energy and radiation.

Fusion joins positively charged light nuclei. Before the strong nuclear force can bind them, the nuclei must approach extremely closely despite electrostatic repulsion between their charges.

Fusion products can transfer energy to the surrounding plasma, helping maintain temperature, but a continuing reaction requires energy production to compete successfully with energy losses. In stars, gravitational pressure provides density and confinement.

A detector records ionising radiation even when the investigated source is absent. This background comes from the environment and detector surroundings, so a source measurement contains both source and background contributions.

Measure background with the same detector, geometry and counting interval but without the source. Convert both readings to count rates before subtracting, or subtract counts measured for equal times. Repeat or count for longer to reduce the relative effect of random fluctuations.

If 1260 counts are recorded in $120\,\mathrm{s}$ with the source and 180 counts in $120\,\mathrm{s}$ without it, the corrected source rate is $(1260-180)/120=9.0\,\mathrm{s^{-1}}$.

A strongly ionising particle transfers energy frequently to matter, so it loses kinetic energy rapidly and has a short range. Gamma photons interact less frequently, so a beam is more penetrating; absorption is probabilistic and reduces intensity rather than giving every photon one fixed stopping depth.

Alpha is dangerous mainly when an emitter enters the body; gamma can irradiate from outside because it penetrates tissue. Beta lies between them. Shield choice must therefore match both the radiation and exposure geometry.

In a nuclear equation, the sum of nucleon numbers $A$ and the sum of proton numbers $Z$ must each be the same before and after the reaction. Use the supplied particle symbols, including any coefficients, to identify a missing nuclide or emission.

In $^{96}_{42}\mathrm{Mo}+{}^{2}_{1}\mathrm{H}\rightarrow{}^{95}_{43}\mathrm{Tc}+x{}^{1}_{0}n$, nucleon number gives $96+2=95+x$, so $x=3$; proton number already gives $42+1=43$.

Clamp a sealed gamma source and detector at fixed separation and alignment. Measure background for a long interval, then place increasing measured thicknesses of lead between source and detector. Record counts for the same sufficiently long time at every thickness.

Subtract background before taking logarithms. Plot corrected $\ln R$ against lead thickness $x$; a straight line supports exponential attenuation and its gradient is $-\mu$. The half-value thickness is the thickness that halves corrected count rate.

Radioactive decay is spontaneous: an unstable nucleus decays without needing an external trigger, and ordinary changes of temperature, pressure or chemical state do not control the event.

It is random: the exact nucleus that will decay next and its decay time cannot be predicted. Each undecayed nucleus of one isotope has the same constant probability of decay per unit time, independent of how long it has already survived.

The decay constant $\lambda$ is the probability per unit time for one nucleus; activity $A$ is decays per second in becquerels. Half-life is the time for $N$ or $A$ to fall to half its current value, not the time for all nuclei to decay.

For half-life 6.0 h, $\lambda=\ln2/6.0=0.116\,\mathrm{h^{-1}}$. After 18 h, three half-lives have passed, so activity is $A_0/8$, agreeing with $A=A_0e^{-\lambda t}$.

Displacement $x$ is measured from equilibrium. The minus sign means force and acceleration always point back towards equilibrium; proportionality means their magnitudes grow linearly with distance from equilibrium.

For a mass on an ideal spring, extension about the equilibrium position produces restoring force $-kx$. A simple pendulum is approximately SHM only at small angles, where the restoring component is approximately proportional to displacement.

The cosine form chooses $t=0$ at positive maximum displacement. A different initial condition requires a phase constant, but the relationships between $x$, $v$ and $a$ remain: velocity is one quarter-cycle out of phase with displacement, and acceleration is in antiphase with displacement.

For $A=0.040\,\mathrm{m}$ and $f=2.0\,\mathrm{Hz}$, $\omega=2\pi f=12.6\,\mathrm{rad\,s^{-1}}$, so $v_{max}=A\omega=0.50\,\mathrm{m\,s^{-1}}$ and $a_{max}=A\omega^2=6.3\,\mathrm{m\,s^{-2}}$.

A $0.20\,\mathrm{kg}$ mass on a $50\,\mathrm{N\,m^{-1}}$ spring has $T=2\pi\sqrt{0.20/50}=0.40\,\mathrm{s}$. Quadrupling the mass doubles the period because $T\propto\sqrt m$.

Use the oscillating mass that moves with the spring. Pendulum length is from pivot to the bob's centre of mass, and the pendulum equation assumes small angular displacement, negligible air resistance and a light inextensible support.

For SHM, displacement varies sinusoidally between $+A$ and $-A$ about the equilibrium line $x=0$. The period $T$ is the time between equivalent points moving in the same direction; amplitude is the maximum distance from equilibrium.

Fix the axes and units, mark $\pm A$, and place repeated peaks exactly one period apart. If initial displacement is $+A$ with release from rest, begin at a maximum and follow a cosine curve; other initial states shift the phase without changing $A$ or $T$.

SHM velocity is sinusoidal with the same period as displacement and amplitude $v_{max}=A\omega$. The gradient of a velocity-time graph is instantaneous acceleration.

If $x=A\cos\omega t$, then $v=-A\omega\sin\omega t$. Velocity therefore reaches each corresponding feature one quarter-period after displacement, and its sign records direction. Draw peaks separated by $T$ and label the vertical scale using $A\omega$.

Resonance occurs when a periodic driving force acts at the natural frequency, or very close to it in a damped real system. The driver then transfers energy efficiently on successive cycles, producing the largest steady oscillation amplitude.

Away from the natural frequency, the driver and oscillator do not stay in the most effective timing relationship, so less net energy is transferred and the response amplitude is smaller.

Attach the same spring to a vibration generator and add a known mass. Drive it sinusoidally, vary frequency in small steps, and record the resonant frequency at the maximum steady amplitude. Repeat the frequency sweep from both directions and average the peak value.

Calibration can absorb a small constant effective contribution from the moving spring or holder: use the fitted relationship rather than forcing a line through the origin without evidence. Repeat the unknown measurement and propagate the frequency reading uncertainty through the graph.

For undamped SHM, $E=\tfrac12mv^2+\tfrac12kx^2=\tfrac12kA^2$ is constant. A pendulum similarly exchanges gravitational potential and kinetic energy while total mechanical energy remains constant.

With damping, resistive forces do negative work. Mechanical energy is transferred to thermal energy, sound or deformation in the surroundings, so the maximum energy and amplitude decrease on successive cycles even though total energy of the larger closed system is conserved.

A sequence of impulses can maintain a forced oscillation when each arrives at the same effective phase, adding energy repeatedly. The drive need not move sinusoidally, but its timing must contain the relevant periodic component.

For a fixed driving-force amplitude, the steady response is small far below the natural frequency, rises as the driving frequency approaches it, reaches a maximum near resonance, and falls again above it.

At steady amplitude, energy supplied by the driver each cycle equals energy dissipated. Near resonance, the timing permits the greatest energy transfer; increasing damping removes more mechanical energy per cycle, so a smaller amplitude is enough to balance the input.

Damping forces oppose motion and do work on the surroundings. Fluid resistance, friction or a dashpot converts organised oscillation energy into thermal energy, so each cycle returns less mechanical energy and the amplitude falls.

A ductile component designed to yield during strong structural motion absorbs work over its force-extension cycle. Together with viscous dampers, this reduces energy available for the next swing and limits damaging amplitudes.

A gravitational field is a region in which a mass experiences a force. The field describes the influence of the source mass throughout the surrounding space, whether or not a test mass is currently placed there.

At any point, the field direction is the direction of the force on a small positive test mass. Around an isolated spherical mass, this is radially inward because gravity is attractive.

If several masses are present, their gravitational forces add as vectors. Equivalently, add their field contributions vectorially at the point of interest.

Gravitational field strength $g$ at a point is the gravitational force $F_g$ per unit mass $m$ placed at that point. It is a vector and points in the direction of the force.

If a $2.5\,\mathrm{kg}$ object experiences a gravitational force of $20\,\mathrm{N}$, then $g=20/2.5=8.0\,\mathrm{N\,kg^{-1}}$. A different small test mass at the same point has the same $g$ but a different force $F_g=mg$.

Here $m_1$ and $m_2$ are the interacting masses, $r$ is the centre-to-centre separation, and $G=6.67 imes10^{-11}\,\mathrm{N\,m^2\,kg^{-2}}$. Each mass experiences an equal-magnitude force directed towards the other.

Choose SI units, find the centre separation, square it, then substitute. For a uniform spherical body, its external gravitational effect can be modelled as if all its mass were concentrated at its centre.

Place a test mass $m$ a distance $r$ from a point or spherical source mass $M$. Newton's law gives $F_g=GMm/r^2$; dividing by the test mass in $g=F_g/m$ removes $m$.

The magnitude falls as $1/r^2$, while the vector direction is radially inward. Outside a spherical planet or star, measure $r$ from its centre, so at altitude $h$ use $r=R+h$.

At a distance $2R$ from the centre of a planet, $g=GM/(2R)^2=g_{surface}/4$. This comparison avoids recalculating $G$ and $M$.

Gravitational potential $V_{grav}$ is potential energy per unit mass, measured in $\mathrm{J\,kg^{-1}}$. The conventional zero is at infinity. Because gravity is attractive, a finite point in the field has negative potential.

As $r$ increases, $V$ rises towards zero: it becomes less negative. The gradient is steep close to the source and shallow far away. Field strength and potential are linked by $g=-\mathrm{d}V/\mathrm{d}r$ in the radial direction.

Both are long-range force fields, obey superposition, and have inverse-square radial fields for point sources. Their field lines show force direction, while their potentials are scalar quantities.

Gravity is normally much weaker than electrostatic interaction at particle scale, but astronomical bodies contain enormous mass and are often nearly electrically neutral, so gravity dominates their large-scale motion.

For a mass $m$ in a circular orbit of radius $r$ around a much larger central mass $M$, the inward gravitational force supplies the required centripetal force.

The orbiting mass cancels from the equations. For the same central mass, a larger circular orbit has a lower speed but a longer period. The velocity is tangential while acceleration and force point inward.

A black body is an ideal object that absorbs all incident electromagnetic radiation and is also a perfect emitter. Its emission spectrum depends only on its absolute temperature.

To compare two curves, check that the axes and scales match. Locate each $\lambda_{max}$ for temperature information and compare areas for total emitted power per unit area.

A star can be approximated as a black-body radiator. Its colour does not mean it emits one wavelength: it emits a distribution whose peak and total area change with temperature.

Luminosity $L$ is the total power radiated, $A$ is emitting surface area, and $T$ is absolute temperature. For a spherical star, $A=4\pi R^2$.

At fixed radius, doubling $T$ multiplies $L$ by $2^4=16$. At fixed temperature, doubling radius multiplies surface area and luminosity by $2^2=4$. Ratio methods often remove $\sigma$ and $4\pi$.

Read or calculate the peak wavelength, convert it to metres, then rearrange to $T=(2.898 imes10^{-3})/\lambda_{max}$. The relation is inverse: hotter black bodies peak at shorter wavelengths.

If $\lambda_{max}=5.00 imes10^{-7}\,\mathrm{m}$, then $T=2.898 imes10^{-3}/(5.00 imes10^{-7})=5.80 imes10^3\,\mathrm{K}$.

If a source radiates uniformly in all directions, its luminosity $L$ passes through a spherical surface of area $4\pi d^2$ at distance $d$. Intensity is power received per unit area.

At twice the distance, the same luminosity is spread over four times the area, so intensity is one quarter. Rearranging gives $L=4\pi d^2I$ or $d=\sqrt{L/(4\pi I)}$.

Observe a nearby star against very distant background stars, then repeat six months later from the opposite side of Earth's orbit. The nearby star appears to shift relative to the fixed background.

Half the total angular shift is the parallax angle $p$. With the known radius of Earth's orbit as the baseline, right-triangle geometry gives the distance.

Measure the angular displacement carefully, halve it to obtain $p$, select one consistent unit relation, and calculate $d$. More distant stars have smaller parallax angles.

A standard candle is an astronomical object whose luminosity $L$ is known. Find such an object in a distant cluster or galaxy and measure the intensity $I$ received at Earth.

Nearby distance methods can calibrate the luminosity of a standard-candle class; that calibrated class can then extend the astronomical distance scale to much greater distances.

A simple Hertzsprung-Russell diagram plots stellar luminosity vertically, increasing upward, against surface temperature horizontally. Temperature conventionally decreases from left to right.

Luminosity depends on both surface temperature and radius through $L=4\pi R^2\sigma T^4$. A cool star can therefore be highly luminous if it has a very large radius; a hot star can be dim if it is very small.

Label both axes and their directions first, then place the three regions. If luminosity is expressed relative to the Sun, it is commonly shown on a logarithmic scale.

A main-sequence star is stable while hydrogen fusion in its core supplies energy and pressure balances gravitational contraction. Its mass determines how rapidly it evolves and what later stages are possible.

On the H-R diagram, a main-sequence star moves toward the giant region as its surface cools and its radius and luminosity change. A white dwarf lies in the hot-but-dim region because its surface is hot but its area is small.

The sequence is driven by fuel depletion: reduced core fusion permits contraction, contraction raises core temperature, and new fusion stages or collapse change the star's radius, temperature and luminosity.

When a wave source moves relative to an observer, successive wavefronts are emitted from different positions. Motion towards the observer compresses their spacing; motion away spreads them out.

Astronomers compare known spectral lines with their observed wavelengths. A periodic change between redshift and blueshift can reveal a star moving back and forth because of an orbiting companion.

The effect depends on radial relative motion—the component along the observer's line of sight. Sideways motion alone does not produce the same first-order wavelength shift.

Use $\Delta\lambda=\lambda_{observed}-\lambda_{rest}$. A receding source has a longer observed wavelength, so $z>0$. Its frequency falls, so $\Delta f=f_{observed}-f_{rest}<0$ and the minus sign makes the same positive $z$. The speed approximation applies when $v\ll c$.

For galaxies at cosmological distances, recession speed $v$ is proportional to distance $d$. A graph of $v$ against $d$ has gradient $H_0$; the widespread redshift-distance relation is evidence that the universe is expanding.

If expansion is approximated as steady, distance divided by recession speed gives $d/v=1/H_0$, a characteristic age. A larger measured $H_0$ gives a smaller age estimate. Convert $H_0$ to $\mathrm{s^{-1}}$ before taking its reciprocal in seconds.

Galaxy motions and gravitational effects can indicate more mass than is directly visible. This proposed dark matter increases the total gravitational influence and therefore affects models of how expansion may change.

The ultimate fate depends on the competition represented in the model between cosmic expansion and the gravitational effect of the universe's matter. Because $H_0$ and unseen mass are inferred from observations with assumptions and uncertainties, conclusions about age and fate have been controversial.