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

Question

followed the steps given by you
but not getting the answer (0degree celsius)

Answer 1

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 (Tf-Ti) = [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 22-4.8 = 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 (Tf-Ti)] = 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.