In reply to spearing05:
> (In reply to Robert Durran)
> Boiling both together or both seperately still requires the same amount of heat. It is quicker to boil one cup than two but you have to do this twice (which is what I think you may be missing out?)
No, I am not missing this out.
I shall explain it again slightly differently:
Take your two cubes. Suppose each has a surface area A. If they are combined into a single lump, the surface area A' of this lump will be more than A (by a factor of 2^(2/3)=1.6 approx if they are moulded into a single larger cube).
Now, suppose the two cubes are heated separately, each taking time t. The heat loss from each will be proportional to A x t = At, and so the total heat loss from both will be propoptional to 2 x At = 2At (see, I have not missed this out!)
Now consider heating the combined lump. The time taken will be 2t (twice as much stuff to heat), so the heat loss will be proportional to A' x 2t = 2A't.
But, A' is greater than A, so the heat loss for the combined lump will be greater than the total for the two lumps separately.
In fact, the time taken to heat the combined lump will not be precisely twice that for a single cube, and an adjustment is needed taking into account the extra time needed to compensate for heat losses. My more detailed analysis using differential equations actually shows that this second order correction actually shows a greater relative inefficienccy using the combined lump than does the approximate argument above.
> In a perfect system with no losses it would take exactly twice as long to boil two cups as it would one.
Yes.
> The small difference in time that we see in the experimental data......
Note that my analysis does not take account of losses via steam; as I have said before, I accept that this is the dominant factor, more important than conduction, meaning that, in practice, the combined method is more efficient.