Thermal Physics

This problem involves kinetic theory of gases and thermodynamics.

Given: - Air: average speed $v_{air} = 500$ m/s, molecular mass $m_{air} = 29$
- Hydrogen: molecular mass $m_H = 2.0$, same temperature means same average KE
- $m_{total} = 1.0$ g of hydrogen
a) Average speed of hydrogen molecule:

At same temperature, average KE is equal: $$\frac{1}{2}m_{air}v_{air}^2 = \frac{1}{2}m_H v_H^2$$

$$v_H = v_{air}\sqrt{\frac{m_{air}}{m_H}} = 500 \times \sqrt{\frac{29}{2.0}}$$

$$v_H = 500 \times \sqrt{14.5} = 500 \times 3.81$$

$$v_H = 1905 \text{ m/s} \approx 1900 \text{ m/s}$$

b) Average velocity of hydrogen molecules:

The molecules move in random directions. For every molecule moving in one direction, there's another moving in the opposite direction. The vector average is zero.

$$\vec{v}_{avg} = 0$$

c) Speed of sound comparison:

Speed of sound in a gas: $v_{sound} = \sqrt{\frac{\gamma k T}{m}}$

Where $\gamma = C_p/C_v$, $k$ is Boltzmann constant, and $m$ is molecular mass.

Since hydrogen molecules are much lighter ($m_H \ll m_{air}$), they move much faster at the same temperature.

Pressure variations are transmitted by molecular motion. With faster molecules, pressure variations propagate more quickly.

Therefore, speed of sound in hydrogen is much faster than in air at the same temperature.

d) Total kinetic energy of hydrogen:

$$KE_{total} = \frac{1}{2}m_{total}v_H^2$$

$$KE_{total} = \frac{1}{2}(1.0 \times 10^{-3})(1905)^2$$

$$KE_{total} = 0.5 \times 10^{-3} \times 3.63 \times 10^6$$

$$KE_{total} \approx 1810 \text{ J}$$

e) Comparison with air:

At the same temperature, all gases have the same average kinetic energy per molecule. This is a fundamental principle of thermodynamics.

Equal numbers of molecules at the same T would have identical total kinetic energy.

So air with the same number of molecules would also have approximately 1810 J.

f) Maximum height of hydrogen molecule:

Using energy conservation: $\frac{1}{2}mv^2 = mgh$

$$h = \frac{v^2}{2g} = \frac{(1905)^2}{2 \times 9.8}$$

$$h = \frac{3.63 \times 10^6}{19.6} = 185,000 \text{ m} = 185 \text{ km}$$

g) Comparison with air molecule:

For air molecule at same T: $$v_{air} = 500 \text{ m/s}$$

$$h_{air} = \frac{v_{air}^2}{2g} = \frac{500^2}{2 \times 9.8} = \frac{250,000}{19.6} \approx 12,755 \text{ m} \approx 13 \text{ km}$$

$$\frac{h_H}{h_{air}} = \frac{185}{13} \approx 14$$

The hydrogen molecule reaches much higher (about 14$\times$) than the air molecule.

h) How hydrogen escapes atmosphere:

The Maxwell-Boltzmann distribution describes molecular speeds in a gas. While there's an average speed, molecules actually have a distribution of speeds:
- Some molecules move faster than average
- Some move slower than average
The fastest hydrogen molecules in the tail of the distribution have:
- Higher KE than average
- Can reach greater heights
- May exceed escape velocity of Earth
This is how hydrogen gradually escapes from the atmosphere over geological time, even though the "average" molecule cannot escape.

This problem involves thermal expansion and pendulum clock accuracy.

a) Units of $\alpha$:

$$\alpha = \frac{\Delta \ell}{\ell} \cdot \frac{1}{\Delta \theta}$$

Units: $\frac{\text{m}}{\text{m}} \cdot \frac{1}{^\circ\text{C}} = ^\circ\text{C}^{-1}$

b) Oscillations in one day:

One day = 24 hours = $24 \times 3600$ seconds = 86,400 seconds

With period $T = 1$ second: $$N = \frac{86,400}{1} = 86,400 \text{ oscillations}$$

c) Oscillations at different temperatures:

At $15^\circ$C: clock runs 5 s fast per day (more oscillations) $$N_{15} = 86,400 + 5 = 86,405 \text{ oscillations}$$

At $30^\circ$C: clock runs 10 s slow per day (fewer oscillations) $$N_{30} = 86,400 - 10 = 86,390 \text{ oscillations}$$

d) Calculate periods:

$$T = \frac{\text{time}}{\text{oscillations}}$$

$$T_{15} = \frac{86,400}{86,405} = 0.999942 \text{ s}$$

$$T_{30} = \frac{86,400}{86,390} = 1.000116 \text{ s}$$

e) Calculate lengths:

Using $T = 2\pi\sqrt{\ell/g}$, so $\ell = \frac{gT^2}{4\pi^2}$:

$$\ell_{15} = \frac{9.8 \times (0.999942)^2}{4\pi^2}$$

$$\ell_{15} = \frac{9.8 \times 0.999884}{4\pi^2} = 0.248208 \text{ m}$$

$$\ell_{30} = \frac{9.8 \times (1.000116)^2}{4\pi^2}$$

$$\ell_{30} = \frac{9.8 \times 1.000232}{4\pi^2} = 0.248294 \text{ m}$$

f) Calculate $\alpha$:

$$\Delta \ell = \ell_{30} - \ell_{15} = 0.248294 - 0.248208 = 0.000086 \text{ m}$$

$$\Delta T = 30 - 15 = 15^\circ\text{C}$$

$$\alpha = \frac{\Delta \ell}{\ell} \times \frac{1}{\Delta T}$$

Using average length $\ell \approx 0.248$ m: $$\alpha = \frac{0.000086}{0.248} \times \frac{1}{15}$$

$$\alpha = 2.3 \times 10^{-5} \, ^\circ\text{C}^{-1}$$

This problem involves work, energy, and heat transfer when a wire is drawn through a die.

Given:
- Specific heat capacity: $c = 400 J/(kg\cdot^{\circ}C)$
- Mass per unit length: $\mu = 5$ g/m $= 0.005$ kg/m
- Force required: $F = 600$ N
- Wire speed: $v = 8.4$ m/s
- Water specific heat: $c_w = 4200 J/(kg\cdot^{\circ}C)$ (standard value)
a) Sketch:

The setup shows a wire of initial larger diameter being pulled through a small hole (die) in a metal plate. A force $F = 600$ N is applied to pull the wire through from left to right. The wire emerges with reduced diameter.

b) Work done on 1 metre of wire:

Work = Force $\times$ Distance (in direction of force)

For 1 metre of wire pulled through: $$W = F \times d = 600 \times 1 = 600 \text{ J}$$

c) Work done on 1 kilogram of wire:

The wire has mass per unit length $\mu = 0.005$ kg/m.

Mass of 1 metre of wire: $$m = \mu \times 1 = 0.005 \text{ kg}$$

Work per kilogram: $$W_{\text{per kg}} = \frac{W}{m} = \frac{600}{0.005} = 120,000 \text{ J/kg}$$

d) Temperature rise of wire:

Assuming all work converts to heat: $$Q = mc\Delta T$$

Where $Q$ is heat energy, $m$ is mass, $c$ is specific heat, and $\Delta T$ is temperature rise.

Per kilogram: $\frac{Q}{m} = c\Delta T$

$$\Delta T = \frac{W_{\text{per kg}}}{c} = \frac{120,000}{400} = 300^{\circ}\text{C}$$

e) Mass of wire produced per second:

Wire production rate: $$\text{Rate} = \mu \times v = 0.005 \times 8.4 = 0.042 \text{ kg/s}$$

f) Work done per second (Power):

$$P = W_{\text{per kg}} \times \text{Rate} = 120,000 \times 0.042 = 5040 \text{ W}$$

Alternatively: $P = Fv = 600 \times 8.4 = 5040$ W (same result)

g) Mass of water needed for $12^{\circ}$C temperature rise:

Heat energy absorbed by water per second: $$Q_w = P = 5040 \text{ J/s}$$

For water temperature rise of $\Delta T_w = 12^{\circ}$C: $$Q_w = m_w c_w \Delta T_w$$

$$m_w = \frac{Q_w}{c_w \Delta T_w} = \frac{5040}{4200 \times 12}$$

$$m_w = \frac{5040}{50,400} = 0.1 \text{ kg/s}$$

Therefore, 0.1 kg of water per second flowing past the wire would maintain constant wire temperature by carrying away the heat.

This problem involves inflating a car tire with an air pump and Boyle's Law.

Understanding the situation:

- Tire volume when inflated: $V_{tire} = 20$ litres
- Flat tire still has: $V_{initial} = 16$ litres at atmospheric pressure
- Pump rate: $4$ litres/minute (at atmospheric pressure)
- Temperature remains constant (isothermal process)
a) Why it takes much longer than 1 minute:

Naive calculation: $$\text{Volume needed} = 20 - 16 = 4 \text{ litres}$$ $$\text{Time} = \frac{4 \text{ litres}}{4 \text{ litres/min}} = 1 \text{ minute}$$ Why this is wrong:

The pump moves 4 litres of air at atmospheric pressure, but as the tire fills:
- Pressure inside the tire increases
- Air from atmosphere gets compressed when entering tire
- 4 litres at 1 atm $\neq$ 4 litres at higher pressure!
Boyle's Law: $P_1V_1 = P_2V_2$ (at constant temperature)

As pressure increases, the same amount of air occupies less volume.

b) Pressure vs. time graph: Shape of graph:
- Starts at 1 atm (atmospheric pressure)
- Increases with time
- Curve is concave downward (rate of pressure increase slows)
Why concave downward? - Initially: Large volume of air needed, pressure builds slowly
- Later: Smaller volume needed for same pressure increase
- Eventually: Approaches final inflated pressure asymptotically
c) Excess pressure needed: Given:
- Car mass: $M = 1200$ kg
- Contact area per tire: $A = 130 \text{ cm}^2$
- Number of tires: 4
- Atmospheric pressure: $P_{atm} = 1.0 \times 10^5$ Pa
Weight supported by each tire: $$F = \frac{Mg}{4} = \frac{1200 \times 9.8}{4} = 2940 \text{ N}$$ Excess pressure per tire: $$P_{excess} = \frac{F}{A} = \frac{2940}{130 \times 10^{-4}}$$

$$P_{excess} = \frac{2940}{0.013} = 226,154 \text{ Pa}$$

$$P_{excess} \approx 2.26 \times 10^5 \text{ Pa} = 2.26 \text{ atm}$$

Total pressure in tire: $$P_{total} = P_{atm} + P_{excess} = 1.0 + 2.26 = 3.26 \text{ atm}$$ d) Volume of air from atmosphere for excess pressure: Using Boyle's Law:

The excess pressure (above atmospheric) is $2.26$ atm. $$P_1V_1 = P_2V_2$$

Where $P_1 = 2.26$ atm, $V_1$ is volume at atmospheric pressure, and $V_2 = 20$ litres.

$$2.26 \times V_1 = 1 \times 20$$

$$V_1 = \frac{20}{2.26} = 8.85 \text{ litres}$$

Wait, this needs to be corrected. The actual calculation should be: $$P_{atm} \times V_{atm} = P_{tire} \times V_{tire}$$

$$1 \times V_{atm} = 3.26 \times 20$$

$$V_{atm} = 65.2 \text{ litres}$$

This is the volume of air (at atmospheric pressure) needed to fill the tire to 3.26 atm.

e) Total inflation time: Air needed from atmosphere: $$V_{needed} = 65.2 \text{ litres}$$ Initial air in tire: Already has 16 litres at 1 atm, so we only need to pump the excess.

Actually, let me reconsider. The tire starts with 16 litres at 1 atm. We need to end with 20 litres at 3.26 atm.

Using Boyle's Law to find equivalent atmospheric volume: $$V_{total} = 3.26 \times 20 = 65.2 \text{ litres (at 1 atm)}$$

Initial air (at 1 atm equivalent): 16 litres Additional air needed: $65.2 - 16 = 49.2$ litres

Pumping time: $$t = \frac{49.2 \text{ litres}}{4 \text{ litres/min}} = 12.3 \text{ minutes}$$

Adding the initial minute for the first 4 litres: $$\text{Total time} \approx 13.3 \text{ minutes}$$

Summary:
- Takes much longer than 1 minute due to compression
- Final pressure: 3.26 atm total (2.26 atm excess)
- Requires 65 litres of atmospheric air
- Takes about 13 minutes to inflate

This problem involves thermal physics and astronaut survival in space.

The danger scenario:

An astronaut on a spacewalk:
- Body generates heat through metabolism and exertion
- No convection or conduction to remove heat (vacuum of space)
- Only radiation can carry away excess heat
- If core temperature rises too much, astronaut becomes incapacitated
Given data:
- Astronaut mass: $m = 75$ kg
- Heat generation rate: $P = 240$ W
- Specific heat capacity: $c = 4200$ J/kg/$^{\circ}$C
- Normal core temperature: $T_0 = 38.5^{\circ}$C
- Maximum safe temperature: $T_{max} = 40.5^{\circ}$C
a) Energy for $1^{\circ}$C temperature rise:

Heat equation: $$\Delta E = mc\Delta\theta$$

Where $\Delta\theta = 1^{\circ}$C:

$$\Delta E = 75 \times 4200 \times 1$$

$$\Delta E = 315,000 \text{ J} = 315 \text{ kJ}$$

It takes 315 kJ to raise the astronaut's body temperature by $1^{\circ}$C!

b) Rate of temperature increase: Power = energy per unit time: $$P = \frac{\Delta E}{\Delta t}$$ Temperature rate: $$\frac{\Delta\theta}{\Delta t} = \frac{P}{mc} = \frac{240}{315,000}$$

$$\frac{\Delta\theta}{\Delta t} = 7.62 \times 10^{-4} \text{ }^{\circ}\text{C/s}$$

This is approximately $0.00076^{\circ}$C per second.

In more practical units: $$\frac{\Delta\theta}{\Delta t} = 7.62 \times 10^{-4} \times 60 = 0.046^{\circ}\text{C/min}$$

About $0.05^{\circ}$C per minute - slow but steady!

c) Time until incapacitation: Temperature rise needed: $$\Delta T = T_{max} - T_0 = 40.5 - 38.5 = 2^{\circ}\text{C}$$ Energy needed: $$E_{needed} = mc\Delta T = 315,000 \times 2 = 630,000 \text{ J}$$ Time calculation: $$t = \frac{E_{needed}}{P} = \frac{630,000}{240}$$

$$t = 2,625 \text{ seconds}$$

Convert to minutes: $$t = \frac{2,625}{60} = 43.75 \text{ minutes}$$ Summary:

The astronaut has approximately 44 minutes before becoming incapacitated from overheating if no cooling system removes the internally generated heat!

This highlights why spacesuit thermal management is critical - astronauts need active cooling (like the liquid cooling garment in real EMU suits) to survive extended spacewalks.

This problem involves heat transfer and cooling of the Earth.

Understanding Earth's thermal history:

Earth formed as a molten ball at $\approx 4000$ K and has been cooling ever since. We're analyzing the cooling process using a simplified model.

a) Heat transfer mechanisms: From deep within Earth to surface:

1. Convection: Hot mantle material rises toward crust
- Mantle convection cells transport heat
- Material is heated from below (core-mantle boundary)
- Less dense hot rock rises, transfers heat, cools, sinks
2. Conduction: Heat flows through cooler crust
- Solid crust cannot convect
- Heat conducts through rock via lattice vibrations
- Rate depends on thermal conductivity
From surface to space:

Only radiation is possible!

- Space is vacuum - no medium for convection or conduction
- Earth radiates as a blackbody: $P = \sigma A T^4$
- Stefan-Boltzmann constant: $\sigma = 5.67 \times 10^{-8}$ W/m$^2$K$^4$
b) Cooling formula analysis:

$$t_{cool} = \frac{Nk}{\sigma A}\left(\frac{1}{T_f^3} - \frac{1}{T_i^3}\right)$$

c) Why $T_i$ doesn't matter much:

The dependence is on $1/T^3$, which varies extremely rapidly:
- If $T_i = 4000$ K: $1/T_i^3 = 1.6 \times 10^{-11}$ K$^{-3}$
- If $T_f = 300$ K: $1/T_f^3 = 3.7 \times 10^{-8}$ K$^{-3}$
The final temperature term is $\approx 2300$ times larger!

$$\frac{1}{T_f^3} - \frac{1}{T_i^3} \approx \frac{1}{T_f^3}$$

Approximate formula: $$t_{cool} \approx \frac{Nk}{\sigma A}\left(\frac{1}{T_f^3}\right)$$ c) iii) Calculating number of atoms $N$: Mass of Earth: $$M = \rho V = \rho \times \frac{4}{3}\pi R^3$$

$$M = 5500 \times \frac{4}{3}\pi (6.4 \times 10^6)^3$$

$$M \approx 6.0 \times 10^{24} \text{ kg}$$

Number of atoms: $$N = \frac{M}{m_{atom}} = \frac{6.0 \times 10^{24}}{6 \times 10^{-26}}$$

$$N = 1.0 \times 10^{50} \text{ atoms}$$

c) iv) Calculating cooling time:

$$t_{cool} = \frac{(10^{50})(1.38 \times 10^{-23})}{(5.67 \times 10^{-8})(5.1 \times 10^{14})}\left(\frac{1}{300^3}\right)$$

$$t_{cool} = \frac{1.38 \times 10^{27}}{2.89 \times 10^{7}} \times \frac{1}{2.7 \times 10^7}$$

$$t_{cool} = 1.77 \times 10^{12} \text{ seconds}$$

Convert to years: $$t_{cool} = \frac{1.77 \times 10^{12}}{365.25 \times 24 \times 3600}$$

$$t_{cool} \approx 56,000 \text{ years}$$

c) v) Comment on result:

Actual Earth age: $\approx 4$ billion years

Model prediction: 56,000 years

The model is wildly wrong! This is because: 1. Radioactive heating in mantle (ignored in model) 2. Convection enhances cooling (model assumes only radiation) 3. Earth's formation was more complex than simple cooling 4. Model assumes uniform temperature throughout

Lord Kelvin used this model and got an age of 20-40 million years, which conflicted with geological evidence. The discovery of radioactivity resolved the paradox!