<p>I would like to estimate the cooling timescale for an interstellar dust grain, starting at 200K, down to 100 K.</p>
<p>The equation I have come up with is:</p>
<p><span class="math-container">$\displaystyle t_{cooling} = \frac{mC\Delta T}{Q_{rad}P}$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$m$</span> = grain mass (<span class="math-container">$=1.3\times 10^{-14}$</span> kg)</li>
<li><span class="math-container">$C$</span> = specific heat capacity (<span class="math-container">$=2000$</span> J kg<span class="math-container">$^{-1}$</span> K <span class="math-container">$^{-1}$</span>)</li>
<li><span class="math-container">$T$</span> = change in temperature (<span class="math-container">$= 200-100 = 100 \text{ K}$</span>)</li>
<li><span class="math-container">$Q_{rad}$</span> = emission efficiency</li>
<li><span class="math-container">$P$</span> = radiative power of dust grain</li>
</ul>
<p><strong>My question is: How do I determine the emission efficiency <span class="math-container">$Q_{rad}$</span>?</strong></p>
<p>From <em>'The Physics and Chemistry of the Interstellar Medium (Tielens 2005)'</em>, for grains with sizes less than 1<span class="math-container">$\mu$</span>m, the Planck mean efficiency, &lt;<span class="math-container">$Q_P(T_d)$</span>&gt;, is given by:</p>
<p>&lt;<span class="math-container">$Q_P(T_d)$</span>&gt; <span class="math-container">$=$</span> <span class="math-container">$\displaystyle 1.25 \times 10^{-5} T_d^2 \frac{a}{1\mu m} \;\;\;\;\;\;\;\;\text{(eq. 5.35)}$</span></p>
<p>Is this analogous to the emission efficiency? Using my values I get &lt;<span class="math-container">$Q_P(T_d)$</span>&gt; <span class="math-container">$=0.5$</span></p>

CC BY-SA 4.0
2022-09-06T10:53:57.617Z
Mysterious!  I would have guessed that "emission efficiency" meant ratio of emissivity compared with a perfect black-body surface.   But what does "radiative power" mean?   Also,  perhaps this lecture will be helpful:    https://w.astro.berkeley.edu/~ay216/08/NOTES/Lecture06-08.pdf
Carl Witthoft
CC BY-SA 4.0
2022-09-06T10:55:35.120Z
Maybe also https://academic.oup.com/mnras/article/145/3/249/2602791  -- clickable link to the PDF there
Carl Witthoft
<temperature><radiation><interstellar-medium><dust><black-body>
Cooling timescale for an interstellar dust grain
lucas
# Cooling timescale for an interstellar dust grain

(Asked by: username_0 on 2022-09-05)

I would like to estimate the cooling timescale for an interstellar dust grain, starting at 200K, down to 100 K.

The equation I have come up with is:

$\displaystyle t\_{cooling} = \frac{mC\Delta T}{Q\_{rad}P}$

where:

* $m$ = grain mass ($=1.3\times 10^{-14}$ kg)
* $C$ = specific heat capacity ($=2000$ J kg$^{-1}$ K $^{-1}$)
* $T$ = change in temperature ($= 200-100 = 100 \text{ K}$)
* $Q\_{rad}$ = emission efficiency
* $P$ = radiative power of dust grain

**My question is: How do I determine the emission efficiency $Q\_{rad}$?**

From *'The Physics and Chemistry of the Interstellar Medium (Tielens 2005)'*, for grains with sizes less than 1$\mu$m, the Planck mean efficiency, <$Q\_P(T\_d)$>, is given by:

<$Q\_P(T\_d)$> $=$ $\displaystyle 1.25 \times 10^{-5} T\_d^2 \frac{a}{1\mu m} \;\;\;\;\;\;\;\;\text{(eq. 5.35)}$

Is this analogous to the emission efficiency? Using my values I get <$Q\_P(T\_d)$> $=0.5$

**Tags:** temperature, radiation, interstellar-medium, dust, black-body

> **Comments:**
> **username_1**: Mysterious!  I would have guessed that "emission efficiency" meant ratio of emissivity compared with a perfect black-body surface.   But what does "radiative power" mean?   Also,  perhaps this lecture will be helpful:    https://w.astro.berkeley.edu/~ay216/08/NOTES/Lecture06-08.pdf  
> **username_1**: Maybe also https://academic.oup.com/mnras/article/145/3/249/2602791  -- clickable link to the PDF there  

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