BPhO R1 S2 · Section 2

Thermal Physics

3 questions — reveal each final answer.

2023 · Q3Thermal PhysicsFluidsThermal Physics & Fluids: Hot Air Balloons

This question is about hot air balloons.

A hot air balloon floats because the air sealed inside its envelope is less dense than the colder air outside, so the surrounding atmosphere pushes up on the balloon harder than gravity pulls it down. The problem begins with the simple buoyancy condition for a balloon to lift, recreates the pioneering 1783 Montgolfier and Charles flights using the ideal gas law, and then builds up the barometric (exponential atmosphere) model to follow the balloon's hydrogen as it expands and cools on the way up.

a) [3] A hot air balloon rises when the upthrust is greater than its total weight. Calculate the volume of hot air required to lift a balloon with a combined mass of $200\,\mathrm{kg}$ for the balloon and basket. You should include the mass of the hot air, which has a density of $0.95\,\mathrm{kg\,m^{-3}}$ in the total weight of the balloon. The density of the cold air surrounding the balloon is $1.23\,\mathrm{kg\,m^{-3}}$.

b) [4] In 1783, the Montgolfier brothers' hot air balloon freely ascended, but remained at a low height. The balloon had a volume of about $2000\,\mathrm{m^3}$, and a lifting capacity including its own weight of $830\,\mathrm{kg}$. The air inside the balloon was heated by a fire in a basket below. Estimate the temperature of the air in the balloon. Air density at ground level is $1.23\,\mathrm{kg\,m^{-3}}$. Surface atmospheric pressure is $1.01 \times 10^{5}\,\mathrm{Pa}$. The molar mass of air $0.029\,\mathrm{kg\,mol^{-1}}$.

figure

Figure 6: The Montgolfier brothers' hot air balloon of 1783, heated by a fire in the basket below.

c) [9] Charles' and the Robert brothers' balloon flew not long after but was filled with hydrogen. The balloon had a capacity of $380\,\mathrm{m^3}$ and was filled on the ground at a sea level pressure of $1.01 \times 10^{5}\,\mathrm{Pa}$, and temperature of $288\,\mathrm{K}$.

  1. (i) Calculate the mass of hydrogen in the balloon.
  2. (ii) The balloon was constructed of a single lamina of rubberised silk with a thickness $0.43\,\mathrm{mm}$. The silk cloth has a mass per unit area of $66.5\,\mathrm{g\,m^{-2}}$ and a density of $1.30\,\mathrm{g\,cm^{-3}}$. Rubber has a density of $1.34\,\mathrm{g\,cm^{-3}}$. Assuming the balloon is spherical, calculate the mass of the balloon to three significant figures.
  3. (iii) Taking the mass of the two passengers, gondola and rigging to be $270\,\mathrm{kg}$, calculate the initial acceleration of the balloon at sea level.
figure

Figure 7: Charles and the Robert brothers' hydrogen-filled balloon.

The density of air at sea level is $1.23\,\mathrm{kg\,m^{-3}}$.

d) [3] In an unrelated example, a balloon of mass $m_1$ with a small amount of ballast $m_2$ descends with constant velocity $v$ due to a drag force $F_\mathrm{d}$ acting on the balloon. The drag force is proportional to the speed of the balloon through the air. What is the mass of ballast $m_2$, expressed in terms of $F_\mathrm{d}$, that must be jettisoned from the balloon so that it rises with the same velocity $v$?

e) [6]

  1. (i) Atmospheric pressure is due to the weight of air above.

    figure

    Figure 9: A thin slab of atmosphere of thickness $\Delta h$, used to derive how pressure varies with altitude.

    1. i. From the hydrostatic equation, $p = \rho g h$ we can trivially write down an expression for $\Delta p$. This contains both $p$ and $\rho$ so you need another equation to eliminate $\rho$.
    2. ii. From the ideal gas equation, obtain a relation between $p$ and $\rho$, $R$, $T$ and $M$, the molar mass.
    3. iii. Now using the hydrostatic equation, obtain an expression for $\dfrac{\mathrm{d}p}{p}$ in terms of $M$, $g$, $h$, $R$ and $T$.
    4. iv. The temperature decreases linearly with height, in the form $T = T_0 - \alpha h$. Use this with the previous result to produce an expression for atmospheric pressure as a function of altitude $p(h)$, in terms of sea level pressure $p_0$, $g$, the molar mass of air $M$, the molar gas constant $R$, the air temperature at sea level $T_0$, the rate of decrease of temperature with altitude $\alpha$ and altitude $h$. Humidity should not be considered.
  2. (ii) In a second flight, Charles ascended to $3000\,\mathrm{m}$. Calculate the ambient temperature and pressure at this altitude. Take the molar mass of air to be $0.0290\,\mathrm{kg\,mol^{-1}}$, and the molar gas constant to be $8.31\,\mathrm{J\,mol^{-1}\,K^{-1}}$. The rate of decrease of temperature with height is $0.00976\,\mathrm{K\,m^{-1}}$, the reference sea level temperature $T_0 = 288\,\mathrm{K}$ and the reference sea level pressure $p_0 = 1.01 \times 10^{5}\,\mathrm{Pa}$.
  3. (iii) The balloon expands in volume by $30\,\%$ at a height of $3000\,\mathrm{m}$.
    1. i. Calculate the pressure inside the balloon, assuming no hydrogen has been lost.
    2. ii. Determine the stress in the skin of the balloon.
    3. iii. Estimate the Young's Modulus of the composite skin material.

[25 marks]

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Answers
  • (a)$\displaystyle V = 714\,\mathrm{m^3}$
  • (b)$\displaystyle T = 430\,\mathrm{K} = 160\,^\circ\mathrm{C}$
  • (c)(i)$\displaystyle m_{\mathrm{H_2}} = 32\,\mathrm{kg}$
  • (c)(ii)$\displaystyle m_\mathrm{balloon} = 146\,\mathrm{kg}$
  • (c)(iii)$\displaystyle a = 0.42\,\mathrm{m\,s^{-2}}$
  • (d)$\displaystyle m_2 = \frac{2F_\mathrm{d}}{g}$
  • (e)(i)$\displaystyle \Delta p = -\rho g\,\Delta h$$\displaystyle \rho = \frac{pM}{RT}$$\displaystyle \frac{\mathrm{d}p}{p} = -\frac{Mg}{RT}\,\mathrm{d}h$$\displaystyle p(h) = p_0\left(\frac{T_0 - \alpha h}{T_0}\right)^{\!\frac{Mg}{R\alpha}}$
  • (e)(ii)$\displaystyle T = 259\,\mathrm{K}, \qquad p(3000) = 7.0 \times 10^{4}\,\mathrm{Pa}$
  • (e)(iii)$\displaystyle p_2 = 70\,\mathrm{kPa}$$\displaystyle \sigma = 4 \times 10^{8}\,\mathrm{Pa}$$\displaystyle E = 4 \times 10^{9}\,\mathrm{Pa}$

Worked solutions are in the book.

2024 · Q2Thermal PhysicsElectricity & MagnetismThermal Physics: Solar Energy, Energy Losses and Capacitors

This question is about solar energy, energy losses and capacitors.

A simple sun-warmed can of water is used to estimate the intensity of solar radiation. The energy gained by the water is first treated as loss-free and then refined by modelling the heating and cooling of the can as a “leaky” capacitor circuit, before the same ideas are scaled up to estimate the power output and mass loss of the Sun itself.

a) [7] A simple experiment to estimate solar power output is conducted by a student on a clear sunny day in Spain. The setup is shown in Fig. 1, and consists of a thin aluminium walled matte black can filled with $300\,\mathrm{g}$ of water. A thermometer, fitted through a small hole in the black painted lid, measures the temperature of the water. The can, lid, water and thermometer are allowed to come to ambient temperature in the shade, before being placed on an insulating surface in full sunlight.

figure

Figure 1: A water filled can with a lid and thermometer, sitting on an insulated base in the sunshine.

After $20\,\mathrm{minutes}$, the temperature of the water in the can has risen from $27.0\,^\circ\mathrm{C}$ to $30.0\,^\circ\mathrm{C}$. The angle of the sun to the horizontal is $50^\circ$. The can has a diameter of $6.0\,\mathrm{cm}$ and a height of $10.0\,\mathrm{cm}$. Neglecting the heat capacity of the can, lid and thermometer, calculate a value for the intensity ($\mathrm{W\,m^{-2}}$) of the Sun's radiation. Ignore energy losses from the can. Assume that the can and lid absorb all of the incident radiation.

b) [6] (This part is about the capacitor circuit only, which is a model). A student believes that it might be possible to model the behaviour of the energy transfer above, by considering the can of water to be a capacitor being charged up with charge $Q$, and the energy losses to be a leak of current across the capacitor as shown in Fig. 2. We can determine the time constant for the circuit shown in terms of $r$, $R$ and $C$.

figure

Figure 2: Circuit with a “leaky” capacitor.

  1. (i) Write down the potential around the circuit, in terms of $\varepsilon$, the current $I$ through $R$, and the potential across the capacitor from $C$ and its charge, $Q$.
  2. (ii) Now write down the current through $R$ (and the cell), $I$, in terms of the current through $r$ (using the potential on $C$) and the current flowing onto $C$, both in terms of $Q$.
  3. (iii) Hence write an equation for $\varepsilon$ in terms of $\dfrac{\mathrm{d}Q}{\mathrm{d}t}$ and $Q$.
  4. (iv) What is the expression for the time constant $\tau$ of this capacitor circuit? (You may read this off the differential equation without doing any integration).
  5. (v) By observation of the circuit diagram rather than an equation, what would be the steady state (equilibrium) charge, $Q_\mathrm{ss}$, on the capacitor after a long time, in terms of $R$, $r$, $\varepsilon$ and $C$?
  6. (vi) Sketch a graph of the charge on the capacitor against time, marking on $\varepsilon$ and $Q_\mathrm{ss}$.

c) [7] Since the power reaching the can of water is constant, in our model the cell in Fig. 2 is replaced with a constant current supply (the device adjusts the voltage output to maintain a specified current). The circuit becomes that of Fig. 3, in which the source current is $I_0$.

figure

Figure 3: Circuit with a “leaky” capacitor and constant current supply.

  1. (i) After a long time, the circuit of Fig. 2 will reach a steady state. By observation of the circuit diagram what will be the maximum charge on the capacitor, and its energy stored, in terms of $I_0$, $r$ and $C$?

The rate of flow of charge to the capacitor is analogous to the rate of increase in internal energy of the water in the can. The current through the resistor $r$ is analogous to the power lost to the surroundings, given by $P_\mathrm{leak} = kT$, where $T$ is the temperature difference between the water in the can and the surroundings and $k$ is a constant.

  1. (ii) Write down an equation for the power absorbed by the can, $P_\mathrm{can}$ in terms of the rate of temperature increase of the water, the specific heat capacity of the water $c$, the mass of water $m$, the temperature excess $T$ and $k$.
  2. (iii) Using your result from part (ii), what is the simple expression for the steady state result for $P_\mathrm{can}$?
  3. (iv) Calculate the value of $k$ given that over 40 minutes the temperature of the can falls from $34.2\,^\circ\mathrm{C}$ to $32.8\,^\circ\mathrm{C}$ in an ambient temperature of $31.0\,^\circ\mathrm{C}$.
  4. (v) Produce a new estimate for the power output of the Sun, this time accounting for energy losses from the can.

d) [5] The solar constant is the intensity of the Sun's radiation at the top of the atmosphere, normal to the radial line between the Earth and the Sun and has a value of $1360\,\mathrm{W\,m^{-2}}$. The Sun lies at a distance of $1.5 \times 10^{11}\,\mathrm{m}$ from the Earth and has a radius of $696\,000\,\mathrm{km}$.

  1. (i) What is the power produced on average per cubic metre within the Sun?
  2. (ii) Using the equation of mass-energy equivalence, $E = mc^2$, estimate how much mass the Sun loses each second through radiation.

[25 marks]

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Answers
  • (a)$\displaystyle I \approx 520\,\mathrm{W\,m^{-2}}$
  • (b)(i)$\displaystyle \varepsilon = V_R + V_C = IR + \frac{Q}{C}$
  • (b)(ii)$\displaystyle I = \frac{\mathrm{d}Q}{\mathrm{d}t} + \frac{Q}{rC}$
  • (b)(iii)$\displaystyle \varepsilon = \frac{\mathrm{d}Q}{\mathrm{d}t}\,R + \frac{Q}{C}\left(\frac{R}{r} + 1\right)$
  • (b)(iv)$\displaystyle \tau = \frac{RrC}{R+r}$
  • (b)(v)$\displaystyle Q_\mathrm{ss} = V_C\,C = \frac{\varepsilon r C}{R+r}$
  • (c)(i)$\displaystyle Q_\mathrm{ss} = C V_C = I_0 r C$$\displaystyle E = \tfrac{1}{2} I_0^2 r^2 C$
  • (c)(ii)$\displaystyle P_\mathrm{can} - kT = mc\,\frac{\mathrm{d}T}{\mathrm{d}t}$
  • (c)(iii)$\displaystyle P_\mathrm{can} = kT_\mathrm{ss}$
  • (c)(iv)$\displaystyle k = 0.293\,\mathrm{W\,^\circ C^{-1}}$
  • (c)(v)$I\approx599\,\mathrm{W\,m^{-2}}$ assuming ambient temperature $27^\circ\mathrm{C}$; the mean-excess approximation gives $593.4\,\mathrm{W\,m^{-2}}$ and is also accepted. Clarified — the printed ≈600 W/m² needs an ambient-temperature assumption absent from the question; see the interactive walkthrough.
  • (d)(i)$\displaystyle \frac{P}{V} = 0.27\,\mathrm{W\,m^{-3}}$
  • (d)(ii)$\displaystyle m = 4.3\times 10^{9}\,\mathrm{kg\,s^{-1}}$

Worked solutions are in the book.

2025 · Q5Thermal PhysicsModern & Nuclear PhysicsElectricity & MagnetismKinetic Theory: Molecular Size, Alpha Absorption and the Geiger-Müller Tube

This question is about ions and atoms.

A sequence of estimates builds up a microscopic picture of a gas: first the size and number density of air molecules are deduced by comparing the liquid and gaseous densities, then this molecular picture is used to model the absorption of alpha particles as a random sequence of ionising collisions, and finally the same ideas of ionisation energy and electric field are applied to the operation of a Geiger-Müller tube.

a) [6] If the density of liquid air is $870\,\mathrm{kg\,m^{-3}}$ and the density of gaseous air is $1.2\,\mathrm{kg\,m^{-3}}$, estimate

  1. (i) the radius of an air molecule, and
  2. (ii) the number of air molecules per cubic metre in the gaseous air.

Relative molecular masses of $\mathrm{N_2} = 28$ and $\mathrm{O_2} = 32$.

b) [6] An alpha particle can be considered extremely small compared to the size of a molecule of air.

A beam of alpha particles is stopped by collision (ionisation) with air molecules after a certain distance. This is illustrated in Fig. 11.

figure

Figure 11: A beam of alpha particles being absorbed by successive ionising collisions with air molecules.

The probability $P(x)$ of the alpha particle interacting with an air molecule, in a depth $\mathrm{d}x$, is given by $P(\mathrm{d}x) = n\sigma\,\mathrm{d}x$, where $n$ is the number of gas molecules per cubic metre and $\sigma$ is the area of cross-section of an air molecule.

  1. (i) Determine an expression for the probability, $P(x)$, of an alpha particle interacting within some distance $x$.
  2. (ii) Determine an expression for the mean distance travelled, $\mu$, between interactions by an alpha particle given that $$\mu = \int_{0}^{\infty} x\,P(x)\,\mathrm{d}x.$$

c) [4] If the ionisation energy of an air molecule is approximately $3\times 10^{-18}\,\mathrm{J}$, estimate the range of a $5\,\mathrm{MeV}$ alpha particle in air.

d) [9] Ionising radiation might be detected using a Geiger-Müller tube, illustrated schematically in Fig. 12.

figure

Figure 12: A schematic illustration of a Geiger-Müller tube showing the central wire and the surrounding conducting metal cylinder.

The electric field strength $E$, within the tube, is given by $E = \dfrac{b}{r}$ where $r$ is the distance from the centre of the central electrode and $b$ is a constant. The diameter of the central electrode is $0.2\,\mathrm{mm}$, and that of the outer metal tube is $15\,\mathrm{mm}$.

  1. (i) If the central electrode is at $+400\,\mathrm{V}$ with respect to the grounded outer metal tube, determine the constant $b$.
  2. (ii) An electron with negligible kinetic energy is liberated from the inner side of the outer metal tube. Accelerated towards the central electrode, what distance must the electron travel without collision in order to ionise an atom of argon with a first ionisation energy of $15.7\,\mathrm{eV}$?

[25 marks]

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Answers
  • (a)(i)$\displaystyle r \approx 2.36\times 10^{-10}\,\mathrm{m}$
  • (a)(ii)$\displaystyle N \approx 2.5\times 10^{25}\,\mathrm{m^{-3}}$
  • (b)(i)$\displaystyle P(x) = \frac{N_{\mathrm{col}}}{N_{0}} = 1 - e^{-n\sigma x}.$
  • (b)(ii)$\displaystyle \mu = \frac{1}{n\sigma}$
  • (c)$\displaystyle R \approx 0.06\,\mathrm{m} \approx 6\,\mathrm{cm}$
  • (d)(i)$\displaystyle b \approx 92.6\,\mathrm{V}$
  • (d)(ii)$\displaystyle d \approx 1.2\,\mathrm{mm}$

Worked solutions are in the book.