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Water and Ice mixture, Find original temperature of water

  • Thread starter Moon_tm
  • Start date
1. The problem statement, all variables and given/known data
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
T0water=?

mixture
Tfinal=3°C


2. Relevant equations

Q=mcΔT
H=mLf


3. The attempt at a solution
Energy gained by ice = Energy lost by water

(Ice → Water0C) + (Water0C → Water3C) = Waterinitial → Water3C

m1Lf+m1c(T3-T0) = m2c(Tinitial-T3)
m1[Lf+c(T3-T0)]/m2c=Tinitial-T3

I am not sure whether in the case of water ΔT is (Tfinal-Tinitial) or vice versa.
I get either a negative temperature, which is bs, or after swapping them around Tinitial= 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

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

CWatters

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Method looks ok. Show your working.
 
values from google or specific tables
cwater= 4180 J/KgK
Lfice= 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 [Broken] =5.03
http://www4c.wolframalpha.com/Calculate/MSP/MSP7451i161d717h0b1gbe00004g3egf9d7604ce93?MSPStoreType=image/gif&s=61 [Broken] =-0.97
editable: http://goo.gl/dxxaum
 
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haruspex

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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 ΔQwater=-ΔQice.
 

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