New answers tagged ice
For the hypothetical case of a thermally perfectly insulated system, I'm sure you can work out yourself from the specific heat and the enthalpy of fusion for water. Given that the enthalpy of fusion (330 kJ/kg) and the specific heat of ice (2 kJ/kg-K) have a ratio of 165 K, and you need the entire ice bucket to stay below the melting point, and your 20:1 ...
I guess the puzzle is to transfer heat out to surrounding and, at the same time, do work to the surrounding. This doesn't violate the first law for sure as the energy is conserved by decreasing internal energy. And this doesn't violate the second law of thermodynamics as well.
Consider the water liquid at its freezing point. When water freezes its internal energy goes down - bonds are made. The decrease in internal energy is equal to the heat removed from the water and the work done by the water in expanding against its surroundings.
There is no inconsistency. The first law of thermodynamics tells you that during any process the internal energy $U$ of any simple compressible system will satisfy $\Delta U = W + Q$ where here $W$ is the heat received by the system and $Q$ the heat received by it. Upon changing from water to ice, the part of the system that undergoes the phase change does ...
It does work because it operates like a Heat Engine where the energy travels from something with higher energy (water) to something with less energy (air) to create the work done by the ice. The air has to be colder than the water and below 0C in normal conditions for the water to even freeze. It is just another Heat Engine-like working.
Top 50 recent answers are included