posted by Amir on .
An ice cube at 0.00 degree celsius with a mass of 23.5 g is placed into 550.0 g of water, initially at 28.0 degree celsius, in an insulated container. Assuming that no heat is lost to the surroundings, what is the temperature of the entire water sample after all the ice has melted?
heat to melt ice + heat to raise T of melted ice from zero to final T - heat lost by water at 28 in falling to final T.
[mass ice x heat fusion] + [mass melted ice x specific heat water x (Tfinal-Tinitial)] + [mass warm water x specific heat water x (Tfinal-Tinitial)] = 0
There is only one unknown in this equation. That is Tfinal. Solve for that. I think it is easier (far easier) to substitute the numbers and work it through that way rather than try to solve algebraically for Tfinal.
thanks for the help. im still a bit confused.,,
[23.5g x 6.02] + [? x 2.09 x (Tfinal- 0)] + [550.g x 4.184 x (Tfinal- 28.0)] = 0
is this right so far? How would you be able to get the mass of melted ice?
Close but not quite right. I see you have used kJ/mol for heat fusion but J/g for specific heat water. Units must be consistent. I think the heat fusion is 333 J/g but check me out on that. For the mass of the melted ice don't you think that 23.5 g solid ice will produce 23.5 g liquid water (and 23.5 g steam as far as that goes)?
I asked my professor and she said when the ice cube melts, it absorbs energy from the water. We can figure out how much energy using the heat of fusion. The water cools down; since we know how much energy it lost, how much mass it has (including the mass of the ice cube, now part of the water), and the specific heat of water we can calculate the temperature change using the heat equation from Ch 3 (q = mass x delta T x specific heat).
So what I did was 23.5 g x 1mol/18.02g x 6.02kj/1 mol =7.850 kj
Then Plugged into the equation of q=mCdeltat 7850j= 573.5 g x 4.184 j/g x (x-28) and from
That I solved for x = 31.27 but that doesn't make any sense since it should cool down... The answer in the book is 23.6...