If 12.0 gram of ice are dropped in 150. ml of water at 45 degrees celcius, what is the final temperature of the water?

Answer 1

Note the correct spelling of celsius.
I assume the temperature of the ice is zero C.
heat gained by ice + heat gained by melted ice + heat lost by 45 degree water = 0.
[mass ice x heat fusion] + [mass melted ice x specific heat water x (Tfinal-Tinitial)] + [mass water x specific heat water x (Tfinal-Tinitial) = 0
Solve for Tfinal. I get something like 36 degrees but that's an approximation.

Answer 2

Filling in the equation, here is what I have. Still not coming up with 36 degrees. Please help:

[mass ice x heat fusion]
12.0g x 334j/g

[mass melted ice x specific heat water x (Tfinal-Tinitial)]
12.0g x 4.18j/g x 45 (Delta T is 45 because temp of ice from 0 degrees to water at 45 degrees)

[mass water x specific heat water x (Tfinal-Tinitial)
150g x 4.18j/g x (Tf - 45)

Also, how would I go about solving for Tfinal?

Answer 3

After some sleep, i was able to come up with the following. It's still not 36 degrees, but close. Can some check my calculation?

[mass ice x heat fusion] + [mass melted ice x specific heat water x (Tfinal-Tinitial)] + [mass water x specific heat water x (Tfinal-Tinitial) =0
12.0g x 334j/g = 4008j + 12.0g x 4.18j/gC x 45C =2257j + 150g x 4.18j/gC x (Tf – 45C) = 0
-6265j = 150g x 4.18j/gC x (Tf – 45C)
-6265 = 627C x (Tf – 45C)
-9.99 C = Tf – 45C
Tf = 35.01C

Answer 4

The middle term is the one that is not right. 12.0 g x 4.184 x (Tfinal-Tinitial). Tfinal is Tfinal no matter where you see it; therefore, Tfinal is what you solve for. You have one Tfinal in the middle term and another in the last term. As far as delta T goes for the middle terms, that is not 45 (45-0) but Tfinal-0). The melted ice starts at zero C and goes to Tfinal.

Answer 5

Makes perfect sense. Thank you.