Things I do for coffee (A.K.A., fun with thermodynamics)

I needed hot water at around 80-85 Celsius to use with
At the time, I didn’t have a thermometer that goes higher than 50C.
I know the room temperature and the temperature of near-boiling water (close to 100C ).
The result: a calculator, for a cool (pun intended), practical use of thermodynamics.

The Question

How many grams of ~20C water are needed to lower the temperature of 500g of ~100C water to ~85C?


Here is a calculator to do that for us. Go ahead and change the values
for desired temperature and room/ambient temperature to get your answer.

Desired temperature (Celsius)

Hot water weight (grams)

Hot Water temperature (Celsius)

Cold Water temperature (Celsius)


(Note: desired temperature should be between Cold and Hot water temperatures)

Detailed Solution

Via convection, heat transfers from the hotter liquid to the colder.


We can calculate heat transfer quantity using


Putting those two together we get

Q1 + Q2 = 0

M1•C•(T – T1) + M2•C•(T- T2) = 0

500•C•(85 – 100) + M2•C•(85 – 20) = 0

500•C•(-15) + M2•C•(65) = 0

M2•C•(65) = 500•C•(15)

And, solving for M2, we get




I need to add 115.3g of room-temperature water.

Published by pgk


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