chemistry
posted by treasa on .
what is the final temperature of amixture containning 25 grams of ice at 20 degree celciusadded to water at room temperature 22oC mass of water is 50g
followed the steps given by you
but not getting the answer (0degree celsius)

That's because I assumed, incorrectly, that all of the ice would melt. It won't.
How many J does it take to move T of ice at 20C to OC? That's
q = (mass ice x specific heat ice x (TfTi) = [25g x 2.01 J/g x (20)] = 1005 J.
How much will that lower the T of the water?
1005 = mass H2O x specific heat H2O x delta T and solve for delta T. I obtained about 4.8 C so T H2O now 224.8 = about 17.2.
So how many J do you have in the H2O @ 17.2 C to lose? That's [mass H2O x specific heat H2O x (TfTi)] = 50*4.184 x (17.2) = 3598 if the H2O moves T all the way to zero C. Now, how much ice will that melt at zero C? That's
3698 = mass ice x heat fusion
3698 = mass ice x 334
mass ice = 3698/334 = approx 11g ice which will leave 25  about 11 = about 14g unmelted.
So you will have water at zero with 14 g ice floating.
I didn't realize all of the ice didn't melt until I started on the problem an came out with a negative value for Tf.