Specific heat and temp - please check

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SUMMARY

This discussion centers on a calorimetry problem involving the mixing of water and ice, specifically calculating the final temperature (T_f) after the ice melts. Participants utilize the specific heat capacities of water (4190 J/(kg*K)) and ice (2100 J/(kg*K)), as well as the heat of fusion (334000 J/kg). The correct approach involves setting the heat gained by the ice equal to the heat lost by the water, leading to a final temperature calculation of approximately 71.78 degrees C. The discussion highlights common pitfalls, such as incorrect mass addition and sign errors in heat calculations.

PREREQUISITES
  • Understanding of specific heat capacity and its formula Q=mc(T_f-T_i)
  • Knowledge of heat of fusion and its application in phase changes
  • Ability to set up and solve energy balance equations in calorimetry
  • Familiarity with temperature scales and their consistency in calculations
NEXT STEPS
  • Review the principles of calorimetry and heat transfer in thermodynamics
  • Practice problems involving phase changes and specific heat calculations
  • Learn about the implications of energy conservation in closed systems
  • Explore advanced calorimetry techniques and their applications in real-world scenarios
USEFUL FOR

Students in chemistry or physics, educators teaching thermodynamics, and anyone interested in mastering calorimetry and heat transfer concepts.

jaded18
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In an insulated container, 0.50 kg of water at 80 degrees C is mixed with 0.050 kg of ice at -5.0 degrees C. After a while, all the ice melts, leaving only the water. Find the final temperature T_f of the water. The freezing point of water is 0 degrees C.
___________________________________________________
I know that the specific heat of water is 4190 J/(kg*K), the specific heat of ice is 2100 J(kg*K), and that the heat of fusion is 334000 J/kg
I also know that the expression for the amount of heat absorbed by ice is 0.05*2100(0-(-5))
The expression for the amount of heat released by water is (0.5)(4190)(T_f-80)
The expression for heat of transformation is (334000*0.05)

And then I set these Q's that I found to 0. And my answer that I get is 71.78 degrees C.

But I have a feeling I am doing something wrong. Any feedback will be awesome.
 
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no one can help...?
 
lol i did like a LOT of problems like these in ap chem last year.. but i forgot some of the formulas.. could u put up some formulas? please
Also, u should post this on the chemistry/bio/other science forum
 
oh by the way.. it says -->4190 J/(kg*K) convert it to kelvin
 
aq1q said:
lol i did like a LOT of problems like these in ap chem last year.. but i forgot some of the formulas.. could u put up some formulas? please
Also, u should post this on the chemistry/bio/other science forum

To solve a calorimetry problem like this one, Qnet = 0
In absence of phase changes, amount of heat Q required to change the temp of an object of mass m from Temp_i to Temp_f, we use Q=mc(T_f-T_i) where c=specific heat/
Since we are dealing with a phase change in this prob, we must also take into consideration the amount of heat involved here which is Q=mL where L is the heat of transformation.
 
ok i think i remember.. basically the heat released must equal to heat absorbed.. so what is the total heat absorbed? the heat of transformation + the heat it takes change the temperature of ice to 0.. right? so set your first + third equations = second
clear?
 
and it doesn't matter whether i use kelvin or celsius. as long as i am consistent throughout the prob, it doesn't matter...
 
thats true.. ok sorry.. but I'm pretty sure what i said earlier is correct. is that basically what u did?
 
jaded18 said:
In an insulated container, 0.50 kg of water at 80 degrees C is mixed with 0.050 kg of ice at -5.0 degrees C. After a while, all the ice melts, leaving only the water. Find the final temperature T_f of the water. The freezing point of water is 0 degrees C.
___________________________________________________
I know that the specific heat of water is 4190 J/(kg*K), the specific heat of ice is 2100 J(kg*K), and that the heat of fusion is 334000 J/kg
I also know that the expression for the amount of heat absorbed by ice is 0.05*2100(0-(-5))
The expression for the amount of heat released by water is (0.5)(4190)(T_f-80)
The expression for heat of transformation is (334000*0.05)

And then I set these Q's that I found to 0. And my answer that I get is 71.78 degrees C.

But I have a feeling I am doing something wrong. Any feedback will be awesome.

I get 88.22 which is IMPOSSIBLE. how the heck can the temp rise? Oy.
 
  • #10
lol why did u quote your own statements? I'm getting it to be 72.5 did u get 88.22 using what i said?

The reason u got u got 88.22 is because u added 8.22 instead of subtracting.. you have to realize that u have to put a negative before 4190.. the reason is due to its being released. i mean this negative positive is all relative, when this is consistent.

The reason I am getting 72.5 is because I'm dividing by .55 instead of .5... because I am adding the mass of the ice..
(I think that's what u do, it makes sense right?)
 
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  • #11
I did 525+16700 = 2095(T_f-80) and got T=88.2

i quoted the problem so i could refer to it when i type my response so that i don't have to scroll up so far^^
 
  • #12
aq1q said:
Also, u should post this on the chemistry/bio/other science forum
Why? This is a typical introductory physics thermodynamics question.

Remember that the heat lost by the water is equal to the heat gained by the ice. Be careful of your signs.
 
  • #13
your answer doesn't make much sense tho... why would temperature only decrease by 1 .. what i did to get 71.77 the first time was make my second equation negative since heat is being released. but I'm not so sure now, you know?
 
  • #14
ya I'm sorry hage. Well i know that but i suggested that because he posted it around 5. but no one responded..
 
  • #15
jaded18 said:
your answer doesn't make much sense tho... why would temperature only decrease by 1 .. what i did to get 71.77 the first time was make my second equation negative since heat is being released. but I'm not so sure now, you know?

ya i was plugging the numbers off, i missed a zero. but the reason I'm getting it to be 72.5 is because I'm adding the mass of the ice.

--i edited my comment, read it again
 
  • #16
I have noOoOo clue how you are arriving at that number. can you spell it out for me? and I totally understand the posting this on chem forum thing because i remember having to do these kind of probs in chem way back when.
 
  • #17
0.05*2100(0-(-5)) + (334000*0.05)=-(.55)(4190)(tf-80) ... ok
i have .55 because i added the mass of the ice.. now i can be wrong, but i don't think so.. because it makes sense. I used to be very good in chemistry class as well lol. The negative is there because its being released..
 
  • #18
aq1q said:
ya i was plugging the numbers off, i missed a zero. but the reason I'm getting it to be 72.5 is because I'm adding the mass of the ice.

--i edited my comment, read it again

I don't understand this "because I am adding the mass of the ice" thing. You added the mass 0.05kg to my answer of 71.77? Now that doesn't make much sense.
 
  • #19
no that's not what I'm saying
 
  • #20
lol you keep posting just when i have finished responding...cancel what i said above for now
 
  • #21
haha
 
  • #22
the answer key says lol what?!
 
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  • #23
hahahah um hmm... did u check to see if u are missing a 0 or added extra 0s.. for 2100 and 334000 stuff..

I remember that sometimes these were really tricky cause it gave u answers in KJ instead of J. and things like that.. but huh.. I'm assuming that u gave the right information.
 
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  • #24
wait.. u deleted that 65 out of the post? lol was 65 not the answer?
 
  • #25
HAHA yes it was but I realized it's stupid to post answers like that. If anyone else gets how to do this problem correctly, enlighten me. By the way, all info presented in the first post is correct. (aq1q, we are totally spamming this thread...)
 
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  • #26
Don't add the masses together. This is becoming a mess.

You need to consider

The heat required to bring the ice from -5 degrees to 0 degrees
The heat required for the phase change of ice at 0 degrees to water at 0 degrees
The heat required to bring the 0.05 kg of water now at 0 degrees to the final temperature of the system.

These three terms will be equal to the heat lost by the 0.5 kg of water at 80 degrees going to the final temperature of the sytem.

So heat gained by melting ice to water at 0 degrees (then to final temp) = - heat lost by water initially at 80 degrees
 
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  • #27
chemist here,

listen to hage567, that is a good procedure
 
  • #28
ah, that makes sense. oh this was good review for me
 

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