# How does cooling scale with volume?

What equation would give me the answer to the question, "If i have a cup of water at a tempature of say boiling, how long would that cup of water take to cool off compared to say half that size of a cup of water." So the volume is in half. Its a general question I am just looking for where to start.

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Sean, google for "heat transfer", to learn about basics of a very complicated field, and that this is not a part of thermodynamics. –  Georg Dec 4 '11 at 19:29

An actual cup is slightly complicated because there are 2–3 distinct types of surfaces. Let's deal with free-floating cubes of water instead, both at an initial temperature Ti in an environment with temperature Te .

A cube with half the volume will have 50% the thermal mass C, but 63% of the surface area A. Newton's Law of Cooling implies $\frac{dT}{dt}=-h\frac{A}{C}(T-T_0\!)$ , where h is a property of the environment. So the smaller cube will cool 26% faster initially, when both cubes are at Ti .

If you want to know the temperature of a cube at any given time, the solution to the differential equation above is $\frac{T-T_e}{T_i-T_e}=\exp\left(-ht\frac{A}C\right)$. If you solve for t, it follows that when the small cube is at a given temperature, it will take the large cube 26% more time to reach it.

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This is not wrong, but utterly misleading, because vaporizing is 99 % of heat transfer in the case of boiling water (and well below 100°C still), as asked for! –  Georg Dec 5 '11 at 10:28
@Georg, vaporization and convection are both rolled up into the heat transfer coefficient. True, I assume that h is constant with temperature, but generally people aren't so anal about obvious back-of-the-envelope calculations. What's with the chip on your shoulder? –  rdhs Dec 5 '11 at 15:28
Vaporisation and convection (especially when not forced, but by Grashoff) are terribly nonlinear. To roll them up in a linear coefficient is misleading. (Especially to beginners!) –  Georg Dec 5 '11 at 15:34