Lowest Final Temp: Which Substance to Add?

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The discussion centers on determining which substance, when added to 250g of water at 90°C, results in the lowest final temperature. Initial calculations suggest that adding 15g of water at 0°C (option C) yields a final temperature of 84.9°C, while adding 5g of ice at 0°C (option D) requires consideration of latent heat, leading to a final temperature of approximately 86.67°C. Participants debate the accuracy of their calculations, particularly regarding the heat required to melt the ice and the specific heat values used. Ultimately, there is consensus that option C produces the lowest final temperature, although some calculations indicate that option D could also be competitive. The importance of using consistent units and careful calculations is emphasized throughout the discussion.
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Homework Statement


An insulated container is filled with 250g of water at 90degrees Celsius. Which of the following, when added to the hot water, will result in the lowest final temperature of the mixture?
A. 10g of water at 0 degrees C
B. 125g of water at 75 degrees C
C. 15g of water at 0 degrees C
D. 5g of ice at 0 degrees C
E. 250g of water at 80 degrees C

Homework Equations


I used (CM[dT]) + (CM[dT]) = 0
where the first part is the container's heat and the second is the supposed amount of added substance at a certain temperature.

The Attempt at a Solution



I began plugging in numbers, starting from
A: the final temperature was 86.5
B. the final temperature was 85
C. The final temperature was 84.9
D. The final temperature was 88.23
E. The final temperature was 85.

It seems that C would be the answers, but it's just too close to B and E. Also, I think that my equation is wrong because ice has latent heat... so logically, I think that D would be the right one, because it takes heat to first melt the 5g of ice. But could anyone help me with how I should set up the equation due to the latent heat of ice?
 
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chiurox said:
It seems that C would be the answers, but it's just too close to B and E. Also, I think that my equation is wrong because ice has latent heat... so logically, I think that D would be the right one, because it takes heat to first melt the 5g of ice. But could anyone help me with how I should set up the equation due to the latent heat of ice?

If the final temp is t° C, then,

heat required to melt ice + heat required raise the temp of melted ice to t = heat given out by the water at 90° C to cool down to t.
 
You will have to consider latent heat for the part with ice.
 
For choice D, this is what I did:
mL + (ice's)cmdT = (water's)cmdT
(0.005*3.35*10^5) + (2000*0.005*(T-0)) = 4186*0.250*(90-T)
and I got T = 87.5 is that right?
 
What is the 2000 in that calculation?
 
I think it's the specific heat of ice.
 
chiurox said:
I think it's the specific heat of ice.

The ice has turned to water by then though.
 
oh heh...
so then the final T would be 86.6 which is still not the lowest. So the answer would be C then?
 
That should be correct.
 
  • #10
chiurox said:
oh heh...
so then the final T would be 86.6 which is still not the lowest. So the answer would be C then?

I am getting 80.38 °C for D. Recheck your calculation.
 
  • #11
I'm still getting 86.6
I did what I posted:
"(0.005*3.35*10^5) + (4186*0.005*(T-0)) = 4186*0.250*(90-T)"
I believe 80.38 is correct though. But what are you doing exactly then?
 
  • #12
Sorry...

chiurox said:
I'm still getting 86.6

You got it right. I'm sorry for my wrong calculations. The ans is 86.67 °C.

[Just one point though: use CGS for this type of problems. The numbers are so much easier to deal with, provided you don't make careless mistakes like me. E.g., the equation for (D) is:

5*80 + 5(t-0) = 250(90-t).]
 
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