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As the title says, how much heat would you actually 'save' by cooling a glass down before putting a pint of beer in it?

I know it has to do with specific heat capacity, but I'm unsure of how to make the calculation.

My gut instinct tells me that a glass at room temperature, filled with cold beer will barely get much warmer than a glass that's at room temperature.

Thanks in advance.

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    $\begingroup$ You can calculate the final temperature of such a binary system by equating the final temperature $T_\mathrm{final}$ and applying conservation of energy $U$. Since $\Delta U=C\Delta T$ for each system, where $C$ is the heat capacity (i.e., the specific heat capacity multiplied by the mass), we have $T_\mathrm{final}=(C_\mathrm{glass}T_\mathrm{glass,\,initial}+C_\mathrm{beer}T_\mathrm{beer,\,initial})/(C_\mathrm{glass}+C_\mathrm{beer})$. Now plug in some numbers. $\endgroup$ Commented May 17, 2018 at 20:27
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    $\begingroup$ @Chemomechanics That looks like an answer, not a comment $\endgroup$
    – David Z
    Commented May 17, 2018 at 20:54
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    $\begingroup$ A chilled mug will have a much greater effect on the beer than a chilled glass, because the mug will have a greater thermal mass. And, of course the temperature difference between beer and glass or mug is crucial. You could try chilling a bottle, then pour a small amount of beer into it and swirl it around thoroughly. Measure the temperature of the beer before and after. This will give you a "hands-on" feel for the relative heat capacities of beer and glass. $\endgroup$
    – S. McGrew
    Commented May 17, 2018 at 21:12
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    $\begingroup$ For the sake of science, I'm doing an experiment to answer this. It started about 25 years ago and I expect it to end soon. I'll write up my results when finished. $\endgroup$
    – Natsfan
    Commented May 18, 2018 at 0:10

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Prompted by @David_Z, I'll expand my comment into an answer. Let's assume that the beer and glass reach thermal equilibrium before much energy exchange with the environment. Then we can assume that there's a single final temperature $T_\mathrm{final}$, and we can apply conservation of energy $U$ to say that the same amount of energy lost by the glass or beer is gained by the other. Finally, we have the equation of state for each component $\Delta U=mc_P\Delta T$, where $m$ is the mass and $c_P$ is the constant-pressure specific heat capacity, which we'll assume is constant for simplicity.

From $\Delta U_\mathrm{glass}+\Delta U_\mathrm{beer}=0$, we have $$T_\mathrm{final}=\frac{m_\mathrm{glass}c_{P,\mathrm{glass}}T_\mathrm{glass,\,initial}+{m_\mathrm{beer}c_{P,\mathrm{beer}}T_\mathrm{beer,\,initial}}}{m_\mathrm{glass}c_{P,\mathrm{glass}}+m_\mathrm{beer}c_{P,\mathrm{beer}}}$$

Let's say the pint glass (specific heat 800 J/kg K) and the beer (specific heat 4200 J/kg K, temperature 5°C) each weigh 0.5 kg. If the glass starts at room temperature (20°C), then the beer and glass end up at 7.4°C. If the glass comes out of the freezer at -5°C, then they end up 3.4°C, or 4°C cooler. Whether this will substantially affect your drinking experience is a question only you can answer—I'll drink beer at any temperature.

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  • $\begingroup$ where do your specific heat numbers come from? $\endgroup$
    – pentane
    Commented May 18, 2018 at 3:54
  • $\begingroup$ Round numbers for soda lime glass and water. $\endgroup$ Commented May 18, 2018 at 4:45
  • $\begingroup$ what is your source though $\endgroup$
    – pentane
    Commented May 18, 2018 at 22:28
  • $\begingroup$ I don't remember; the calculation strategy was back of the envelope, and I found broad consensus (to within 25%, say) from various sites and textbooks accessed through an online search. The numbers are largely placeholders meant to demonstrate the calculation method; I don't really know how cold my fridge and freezer are, for example. $\endgroup$ Commented May 18, 2018 at 22:58
  • $\begingroup$ Thanks very much for the answer and comments. Now comes the more interesting part of the experiment. The taste test!! $\endgroup$ Commented May 20, 2018 at 14:58

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