Difference between revisions of "Thermo warm-up"

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.

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.

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 $$ T_\odot = \sqrt{2D\over R_\odot} T_\oplus\approx 10.73\times 288 \approx 5970~K. $$

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.

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$

A simple estimate is based on the assumption that a 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 \[\Delta S = \frac{\Delta Q}{T} \approx \frac{10^7 J \times 365 \times 75}{300\;K} \approx 10^9 \,J/K\]

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 $\approx 5\times 10^{9} yr$. So that 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 \[\Delta S=\frac{\Delta Q}{T} \approx \frac{2\times 10^{34} J}{300\; K} \approx 10^{32} \, J/K\]