Thermo warm-up

If someone points out to you that your pet theory of the universe is in disagreement with Maxwell's equations---then so much the worse for Maxwell's equations. If it is found to be contradicted by observation---well these experimentalists do bungle things sometimes. But if your theory is found to be against the second law of thermodynamics I can give you no hope; there is nothing for it but to collapse in deepest humiliation.
Sir Arthur Stanley Eddington

Problem 1: the cold of space

When designing a suit for open space, what should engineers be more careful of: heating or heat extraction?

solution

Despite popular belief that space is cold (temperature is less than 3 K), coldness, considered as a rate of cooling, could be treated from different perspectives. Since thermal conductance in vacuum is close to zero, hot body in open space will lose energy only due to radiation. Radiation power is proportional to 4--th power of temperature. For example, if astronaut stayed in open space (and far from nearest stars, so that heating from external sources was negligible) and was unable to return to a spaceship, he would not be covered by the crust of ice or suffered from ice death. His temperature, 310 K, is sufficient to remain in comfortable thermal conditions for some time (at least, until arrival of space rescue service). Assuming, that there is no energy release in astronaut's body and evaporation from his skin is eliminated (astronaut is in sealed suit without heat insulation (?)), his temperature would decrease by one dergee in 40 minutes, even if its suit is absolutely black and hence emitting radiation most efficiently. When temperature decreases, according to Stefan--Boltzmann law, cooling rate would decrease. In reality, astronauts are threatened not by cold, but by overheating, since the power of thermal heat release in human body is about 100 Wt; effective heat extraction is one of the main design problem in construction of space suits.

Problem 2: calculating photons

Estimate a number of photons in a gas oven at room temperature and at maximum heat.

solution

Typical gas oven has volume of $40~\mbox{l}$ and heats up to $520~K$. Assuming that walls and door of an oven are black bodies, we obtain the number of photons $$ N = V{\varsigma(3)\over \pi^2}\left(kT\over hc\right)^3 = 1.14\times10^{14}. $$ At room temperature this number decreases $(520/300)^3\approx 5$ times.

Problem 3: equilibrium

Estimate the temperature at the surface of the Sun, assuming that the Earth with mean temperature at its surface $15\,C^\circ$ is in thermal equilibrium with the Sun.

solution

Assuming, that the Sun and the Earth emit as black bodies, Stefan--Boltzmann law leads to a condition of thermal equilibrium at the sufrace of the Earth: $$ \sigma (4\pi R_\oplus^2) T_\oplus^4 = \left({R_\oplus \over 2D}\right)^2\sigma (4\pi R_\odot^2) T_\odot^4, $$ where $R_\oplus$ and $R_\odot$ are radii of the Earth and the Sun, while $T_\oplus$ and $T_\odot$ are their temperatures and $D$ is the distance between them. Hence, for Sun's temperature we obtain $$ T_\odot = \sqrt{2D\over R_\odot} T_\oplus\approx 10.73\times 288 \approx 5970~K. $$

Problem 4: entropy of gravity

What is the difference between entropy of gravitational degrees of freedom and ordinary entropy (e.g., entropy of ideal gas)?

Problem 5: the Sun as the source of low entropy

One of the most used classifications divide physical systems into open and isolated. The entropy in an isolated system can only increase, eventually reaching the maximum at thermal equilibrium. In contrast, in open systems the entropy can decrease due to external interactions, for example, through absorption of a component with low entropy. Explain why the Sun is a source of low entropy for the Earth.

solution

The Earth returns the same amount of energy to environment as it receives from the Sun. However, the photons coming from the Sun, that carry the same amount of energy, possess considerably less entropy: as the absorbed "yellow" photons are more energetic than the infrared photons of Earth's radiation, the energy from the Sun is transmitted by smaller number of photons and, hence, smaller number of degrees of freedom and phase volume. As a result, the Sun's photons have less entropy in comparison with the photons emitted by the Earth.
This chain could be continued both to the future and the past.
to the future: plants use the" low entropy energy" of the Sun in photosynthesis, thus reducing their entropy. We do the same when we eat plants or those, who eat plants.
to the past: the low entropy component is a consequence of gravitational collapse during the creation of the Sun.

Problem 6: infrared vs ultraviolet photons

Estimate how many infrared photons Earth radiates per one ultraviolet photon it absorbs from the Sub.

solution

The ratio of outcoming/incoming numbers of photons is equal to the ratio of the solar and Earth surface temperatures $ \approx 6000/300 \approx 20$

Problem 7: entropy gain of the Universe per person

A human during his lifetime increases the entropy of the Universe by converting chemical energy contained in the food to heat. Estimate the corresponding entropy gain.

solution

A simple estimate is based on the assumption that one person consumes about $2000\;kcal \approx 10^7 J$ in food per day for about $75\,yr$ and dissipates most of it as heat at $ \approx 300\,K$ Consequently, our contribution in the entropy production is \[\Delta S = \frac{\Delta Q}{T} \approx \frac{10^7 J \times 365 \times 75}{300\;K} \approx 10^9 \,J/K\]

Problem 8: Earth's entropy production

Estimate the contribution of Earth during its existence to the entropy production.

solution

This estimate is based on the incoming power $10^{3} \, W/m^{2} $ of solar radiation. The area of the Earth's surface is $S\approx 1.3\times 10^{14} m^{2} $ and the age of the Earth is $\approx 5\times 10^{9} yr$. Then the total received energy is about \[\Delta Q\approx 10^{3} Wm^{-2} \cdot 1.3\times 10^{14} m^{2} \cdot 5\times 10^{9} \cdot 3\times 10^{7} \sec \approx 2\times 10^{34} J\] The same amount of energy is released as heat at the Earth 's surface temperature of $T\approx 300\, K$, yielding the entropy increase \[\Delta S=\frac{\Delta Q}{T} \approx \frac{2\times 10^{34} J}{300\; K} \approx 10^{32} \, J/K\]

Problem 9: Sun and entropy production

Show that the Sun has contributed an entropy increase of about $10^{40} J/K$

solution

The solar power is $\approx 4\times 10^{26} W$and the age of the Sun $\approx 5\times 10^{9} yr$. Solar surface temperature is$\approx 6\times 10^{3} K$. This values yield the entropy increase \[\Delta S=\frac{\Delta Q}{T} \approx \frac{4\times 10^{26} W\cdot 5\times 10^{9} \cdot 3\times 10^{7} \sec }{6\times 10^{3} \; K} \approx 10^{40} \, J/K\] The mechanism of increase of entropy is emergence of new degrees of freedom, which came into being by producing about $7\times 10^{6} $photons emitted into space per nucleon subject to nucleosynthesis. Indeed, The number of created photons follows from the ratio of released nuclear energy per particle $\approx 7\, MeV$ to the energy of a visible photon $\approx 1\, eV$, corresponding to the solar surface temperature.