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IOAA 2019, Теория — Задача 5. Cosmic Microwave Background Oven

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Theory problem sheet · 10 т.

Условие

Since the human body is made mostly of water, it is very efficient at absorbing microwave photons. Assume that an astronaut's body is a perfect spherical absorber with mass of $m = 60\ \mathrm{kg}$ , and its average density and heat capacity are the same as for pure water, i.e. $\rho = 1000\ \mathrm{kg\ m^{-3}}$ and $C = 4200\ \mathrm{J\ kg^{-1}\ K^{-1}}$ .

a) What is the approximate rate, in watts, at which an astronaut in intergalactic space would absorb radiative energy from the Cosmic Microwave Background (CMB)? The spectral energy distribution of CMB can be approximated by blackbody radiation of temperature $T_{\mathrm{CMB}} = 2.728\ \mathrm{K}$ . [5 т.]

b) Approximately how many CMB photons per second would the astronaut absorb? [3 т.]

c) Ignoring other energy inputs and outputs, how long would it take for the CMB to raise the astronaut's temperature by $\Delta T = 1\ \mathrm{K}$ ? [2 т.]

Решение

a)

If the body is spherical, its radius $r$ can be calculated as:

V = \frac{m}{\varrho} = \frac{4}{3}r^3\pi \rightarrow r = \left(\frac{3m}{4\pi\varrho}\right)^{1/3} \quad (5.1)

The area of the body:

A = 4r^2\pi = 4\pi\left(\frac{3m}{4\pi\varrho}\right)^{2/3} = (4\pi)^{1/3}\left(\frac{3m}{\varrho}\right)^{2/3} \quad (5.2) \quad (1\ \mathrm{p})

According to the Stefan-Boltzmann law, the total energy radiated per unit surface area across all wavelengths per unit time:

j = \sigma T_{\mathrm{CMB}}^4 \quad (5.3) \quad (1\ \mathrm{p})

The CMB is isotropic, which means the absorbed radiation by unit surface area of the spherical body is the same independent of the surface's normal direction. (1 p)
The total absorbed energy by the spherical body per unit time:

P = jA = \sigma T_{\mathrm{CMB}}^4 \times (4\pi)^{1/3}\left(\frac{3m}{\varrho}\right)^{2/3} \quad (5.4) \quad (1\ \mathrm{p})

Its numerical value:

P \approx [2.3 \times 10^{-6}\ \mathrm{W}] \quad (1\ \mathrm{p})

b)

The mean energy of CMB photons:

\langle\varepsilon\rangle = 3kT_{\mathrm{CMB}} = 1.1 \times 10^{-22}\ \mathrm{J} \quad (5.5) \quad (1\ \mathrm{p})

The number of CMB photons the body would absorb per second:

n = \frac{P}{\langle\varepsilon\rangle} \quad (5.6) \quad (1\ \mathrm{p})

Its numerical value:

n = [2.1 \times 10^{16}\ \mathrm{s}^{-1}] \quad (1\ \mathrm{p})

c)

Denote the time necessary for raising the temperature by $\Delta t$ , so:

P\Delta t = Cm\Delta T \rightarrow \Delta t = \frac{Cm\Delta T}{P} \quad (5.7) \quad (1\ \mathrm{p})

Its numerical value:

\Delta t \approx [1.1 \times 10^{11}\ \mathrm{s}] \approx 1\,253\,200\ \mathrm{d} \approx [3430\ \mathrm{yr}] \quad (1\ \mathrm{p})

Оригинал в Архива: TH-problem-final.pdf · официални решения: TH-solution-final.pdf · допълнителни данни: TH-anssheet.pdf