Does cooling a pint glass down before putting beer in make any difference?

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.

• 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. Commented May 17, 2018 at 20:27
• @Chemomechanics That looks like an answer, not a comment Commented May 17, 2018 at 20:54
• 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. Commented May 17, 2018 at 21:12
• 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. Commented May 18, 2018 at 0:10

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 $$T_\mathrm{final}=\frac{m_\mathrm{glass}c_{P,\mathrm{glass}}T_\mathrm{glass,\,initial}+{m_\mathrm{beer}c_{P,\mathrm{beer}}T_\mathrm{beer,\,initial}}}{m_\mathrm{glass}c_{P,\mathrm{glass}}+m_\mathrm{beer}c_{P,\mathrm{beer}}}$$