thermodynamics - 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.

Answer

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 T_i in an environment with temperature T_e .

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 T_i .

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.