Water and Ice mixture, Find original temperature of water

[Moon_tm]

Homework Statement

An 8 cm3 ice cube (temperature = 0.00 °C) is dropped into a glass with 3 dL juice which results in the ice cube being melted. By doing this, the juice drops to a temperature of 3.00 °C. What was the temperature of the juice before the ice cube was added? (Treat juice as water in this assignment. The density for ice is 920
kg/m3. Assume no energy is lost to the surroundings).

ice
m₁=(920)(8E-6) = 7.36E-3
T₁=0 C

juice=water
m₂=(1000)(0.3) = 0.3kg
T_0water=?

mixture
T_final=3°C

Homework Equations

[/B]
Q=mcΔT
H=mL_f

The Attempt at a Solution

Energy gained by ice = Energy lost by water

(Ice → Water_0C) + (Water_0C → Water_3C) = Water_initial → Water_3C

m₁L_f+m₁c(T₃-T₀) = m₂c(T_initial-T₃)
m₁[L_f+c(T₃-T₀)]/m₂c=T_initial-T₃

I am not sure whether in the case of water ΔT is (T_final-T_initial) or vice versa.
I get either a negative temperature, which is bs, or after swapping them around T_initial= 5°C which I do not quite believe, because who would want to cool down water by 2°C?!

Please help, I am on the verge of questioning my sanity after spending more than an hour on this...
(One year ago it would've been no problem for me, but after a year of little to no physics my brain went ..dumb :D)

 

[BvU]

There are a c and a L_f in this calculation; what are the numerical values you use ?
And yes, one cube for 3 dl isn't much...

 

[CWatters]

Method looks ok. Show your working.

 

[Moon_tm]

values from google or specific tables
c_water= 4180 J/KgK
Lf_ice= 334E3 J/Kg

Working? Like substituting in the numbers? I mentioned the results I got after having substituted. One was negative, close to -1 and the other was 5 after swapping the final and initial in
Can you please focus on the question whether in the case of water/juice ΔT is (Tfinal-Tinitial) or (Tinitial-Tfinal)

http://www4c.wolframalpha.com/Calculate/MSP/MSP28701i007f2a3h8b607b000022ab0bhbac85297h?MSPStoreType=image/gif&s=30 =5.03
http://www4c.wolframalpha.com/Calculate/MSP/MSP7451i161d717h0b1gbe00004g3egf9d7604ce93?MSPStoreType=image/gif&s=61 =-0.97
editable: http://goo.gl/dxxaum

 

[haruspex]

Tinitial= 5°C which I do not quite believe, because who would want to cool down water by 2°C?!

As BvU observes, that is not much ice for 300mL of drink.

(Tfinal-Tinitial) or (Tinitial-Tfinal)

Make a decision, which makes sense to you? Remember ΔQ_water=-ΔQ_ice.