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Calorimetry and Phase change question 
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#1
Jun1908, 07:58 AM

P: 9

The question is:
An insulated beaker with negligible mass contains liquid water with a mass of 0.225 kg and a temperature of 75.8 C. How much ice at a temperature of 23.6 C must be dropped into the water so that the final temperature of the system will be 31.0 C? Take the specific heat of liquid water to be 4190 J/kg*K, the specific heat of ice to be 2100 J/kg*k, and the heat of fusion for water to be 334 KJ/kg. mice=? This is what I have so far: mw= 0.225 kg Tw= 75.8 C Ti= 23.6 C mi= ? Tf= 31.0 C Cw= 4190 J/kg(K) Ci= 2100 J/kg(K) Lf= 334 KJ/kg = 334000J/kg (I would have to convert this right?) With the information given I come up with these: Qice warms= mi ci (0 (5)) Qice melts= mi Lf Qmix of waters= mi cw (310) Qwater cools= mw cw (31  75.8) Qwater cools = Qice warms + Qice melts + Q mix of waters so now we can solve the mass of ice. right? But when I calculate it the answer I is really weird. 


#2
Jun1908, 08:21 AM

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First if you calculate how much energy is lost due to the water cooling then you can find the mass of ice that would require that amount of energy to heat it to the equilibrium temperature.



#3
Jun1908, 08:33 AM

P: 9

okay, but isn't that what I did?
Qwater cools= mw cw (31  75.8) (energy lost from ice) and then i come up with 0 = Qice warms + Qice melts + Q mix of + Qwater cools 


#4
Jun1908, 08:37 AM

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Calorimetry and Phase change question
Yes that is what you did so how are you calculating the mass of ice? You will notice that you have three terms all containing the mass of the ice so you can easily take it out and rearrange to find the value.



#5
Jun1908, 08:43 AM

P: 9

haha okay. I'm doing
mi((2100*5)+(334000)+(4190*31) = 0.225*4190*44.8 mi = 0.0890 kg ... which is wrong. 


#6
Jun1908, 08:46 AM

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Why is there a 5 in there? The ice starts at 23.6.



#7
Jun1908, 06:50 PM

P: 9

ooo my goodness! thank you so much. i hate when i do that! thank you!



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