Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
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

2 Answers 2

up vote 1 down vote accepted

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.

share|improve this answer
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

It might help to start by looking at Newton's law of cooling.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.