# How much does my mug cool my hot water?

As the tea lovers here no doubt know, green tea is typically brewed at some temperature less than boiling. Typically, the figure I hear is 180 F.

So last night I decide to make some green tea. I put water in the electric kettle, let it boil, then pour it into my mug and check the temperature with a digital kitchen thermometer. (Yeah, I’m a bit anal that way.) Much to my surprise, the temperature reads nothing near 212 F; instead, it was just a bit over 180 F and quickly falling.

I wanted my tea, so I didn’t waste too much time trying to figure this out.

Is it reasonable that the temperature would drop so, just pouring the water into a basic ceramic mug? For the mathematically inclined, I had approximately 12 oz. water; the mug weighs 17 oz. I would guess that the mug’s temperature was something like 65 degrees.

Years ago I’d have done this myself, but all the science has leaked out of my head; the space it took up is being filled with law. Thanks all!

The heat capacity of ceramics at ~100°c is in the range of 4 to 8 cal/g °C, so raising the temperature
12 oz of water is 350 grams.
A 17 oz mug masses 482 grams (seems awfully heavy)

The total heat capacity of the tea is thus 350 cal/°C
While the heat capacity of the mug is 482X4 = 1928 cal/°C

So heating the mug 1°C will cool the tea 1928/350 = 5.5°C.

That assumes that the heat is conducted through the mug efficiently though. I think one of the reasons we use ceramics for coffee mugs, is that it does not conduct heat very well. So you would not expect to heat the entire mug to the same temperature as the coffee inside very quickly.

Also, I think –100?C is a bit cool for tea, but if it’s the only number you have then I’m good because I don’t have any.

~That’s a tilde, not a dash, meaning ‘approximately 100°C’.

Not disagreeing with your method, and thanks for showing your work, but 4 to 8 calories per gram-degree centigrade? I’d figure aluminum oxide in the solid phase as having a heat capacity of 79.04 J/(mol K), or around 0.2 cal/g deg C.

You’re right, I’m wrong. I slipped a factor of 10 in there somehow.
Heating the mug 1°C would cool the tea about 0.5°C.

By the way, that really depends on the type and quality of tea. It should be around 140 F for Gyokuro. And the standard way of getting that temperature is to pour boiling water into a ceramic yuzamashi (cooling cup) and then pouring it into a pot or cup.

You should make sure that the boiling water was really at 212F to start with just before you pour it. Sometimes just 190F can appear to be fairly “boiling” to an observer.

Gotcha. Also I see you revised your figure from 5 ˚C to 0.5 ˚C which is much more agreeable intuitively.

I’m not sure how you’d get it hotter… Most of us prefer to drink our tea, not breathe it.

Some of us do like to breathe tea. Tell me you’ve never sat over a steaming cup of tea just inhaling? But it looks like Christopher was just confused and thought Squink said negative 100 degrees Celsius.

It seems to me the calculations of the mug must be a bit off, given that the handle doesn’t heat up all that much. About how much of the mug’s mass is composed of that part of the handle that doesn’t warm up much?

The thermal conductivity of ceramics is fairly low, the contact area between the handle and the cup is small, and the surface area to volume ratio of the handle is pretty high, so the handle probably loses heat at pretty much the same rate as is transmitted to it after it picks up a few degrees.

I’ve got formulas for the amount of radiant energy based on the surface area, temperatures, and emissivity, but throwing that in is going to make it pretty complicated, as you’ll have to allow for convection, evaporation, and the fact that the temperature of the tea and the cup are both constantly changing as they heat up the environment, and of course the heating of the environment by the tea and cup reduces the driving force available to transfer heat from the tea to that environment.

Just call it 0.5 degrees C and call it a day is my advice.

That’s allright Bill Door. What Squink provided us with is a maximum temperature loss, and as you suggested, it is the best we can do without getting complicated. Maybe there are a few engineers out there that are willing to get complicated, but as far as I’m concerned Squink nailed it.