Tap water in the cold glass It is fairly long time I was in school and I was particularly bad at thermodynamics. This is an question I was trying to comprehend.
I have 500ml of tap water @ 20°C.
I pour it into heavy glass (730g) which is cooled to -18°C.
I would like to be able to calculate what would be the end temperature of this two if I disregard the room temperature. And the glass thickness is same all around (except on top) if this is any relevant data. As I said I don't even know how to approach this. 
I think here are some constants that would be useful:


*

*Specific heat (glass): 0.8 J/g/K 

*Thermal conductivity (glass): 0.8 W/m K 

*Specific heat (water - liquid): 4.18 J/g/K 

*Thermal conductivity(water): 0.598 W/m K


Thank you a lot. 
(P.S.: This might seem to be an school related problem but I assure you it's not)
 A: Since you only want to know what the final temperature will be, it's easy to do this by simply considering how much energy will flow between the water and the glass.  Knowing that the final temperatures of the two will be equal, and that energy is conserved (so any energy leaving the water must enter the glass), we can set up a couple of linear equations whose solution gives you the final state.
In symbols, let $E$ be the energy transferred, and let $T_\text{water}$ and $T_\text{glass}$ be the temperatures thought of as functions of $E$:
$$T_\text{water} = 20^\circ\text{C} - \frac{E}{500\;\text{g} \cdot 4.18\;\text{J/g/K}}\\
T_\text{glass} = -18^\circ\text{C} + \frac{E}{730\;\text{g} \cdot 0.8\;\text{J/g/K}}$$
Now you can set $T_\text{water} = T_\text{glass}$ and solve for $E$.  You can do the algebra yourself, but the answer is that the final temperature is 11.7°C and the energy transferred is 17.3 kJ.
Note that if you wanted to know how long it takes to come to equilibrium (within some tolerance), or you wanted to see the temperatures as a function of time, then the problem gets a lot more complicated. You would have to know something about the geometry of the glass, take account of the thermal conductivities and so on.
It would also be more complicated if the glass were chilled so much that the water could freeze. Then you would have to account for the energy transferred during the freezing process, as well as the different specific heats and conductivities of liquid water vs ice.
