For the purposes of this blog, I'm going to assume that I made a cup of coffee and then proceeded to forget about it, which is unfortunately a common occurrence. Here are some basics:

-The average mug I own is 16 ounces, and I always fill that to the highest level I can while still leaving room for milk, which is about 12 ounces (354.8 grams).

-My Keurig brews coffee at a temperature of 192 ºF, or 88.8 ºC.

-My refrigerator is ~36ºF on a good day, so 2.2ºC

-For a 12 ounce cup of coffee, I probably add about 2 ounces of milk (61.5 grams) at the previously mentioned 2.2ºC.

-c(milk) = 3130 J/kg, c(water, also coffee) = 4186 J/kg

-The thermostat in my apartment is set to 68ºF (20 ºC) and no one is allowed to touch it

-I'd like my coffee to stay above 120ºF (48.8ºC)

So, here we go with some math. My coffee starts out at 88.8ºC, and then I add the milk, so let's find our starting temp:

My mug has an opening that is 7.32×10

^{-3 }m2
Since it's getting late and I only have so much time to do this blog post, I'm going to pretend heat is only escaping through the opening in the top. This seems most like convection cus the air is interacting with the open mug, and some guy at MIT on the internet says the Kc of coffee is ~10 W/m2 x ºC, so I'm gonna use that.

So starting at 78.9ºC, in a 20ºC room, let's see how long it will take my coffee to reach 48.8ºC:

Now let's see how many Joules it will take to reach 48.8ºC, assuming the specific heat is not changed by the small amount of milk added:

So, 52453/4.31 is 12170 seconds, which is 202 minutes, aka 3.4 hours. Considering that the coffee is also losing heat through the mug itself, it must cool down quicker, but I still have more time than I thought I did.

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