Find Ice Mass to Make Water Temp 29°C

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SUMMARY

The discussion revolves around calculating the mass of ice required to achieve a final temperature of 29°C when added to 0.275 kg of water at 77.7°C. The specific heats of water (4190 J/kg°C) and ice (2100 J/kg°C), along with the heat of fusion (334,000 J/kg), are utilized in the calculations. The initial calculation by the user, Magicarp, yielded 0.114 kg of ice, which was later corrected to 0.11277 kg after adjusting for significant figures. The consensus is that precision in significant figures is crucial for accurate results in such thermodynamic problems.

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Magicarp
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Homework Statement



An insulated beaker with negligible mass contains liquid water with a mass of 0.275kg and a temperature of 77.7.

How much ice at a temperature of -17.4 must be dropped into the water so that the final temperature of the system will be 29.0?
Take the specific heat of liquid water to be 4190 , the specific heat of ice to be 2100 , and the heat of fusion for water to be 334 x 10^3 .

Homework Equations


Q = mc\DeltaT
Q_tot = q_1 + q_2 ..etc
Fusion - Phase change

The Attempt at a Solution



I am almost certain I have done this correctly, but masteringphysics will not admit it! :/
I will leave off some units as I have made sure they all match up.

q1: Ice to 0C
q2: Ice fusion mw= mass water
q3: Ice to 29C mi = mass ice
q4: H20 to 29C((mw)(4190)(29C-77.7C) + (mi)(2100)(0C+17.4C)+(mi)(334 X 10^3)+(mi)(4190)(29C-0C))= 0

I solve for mi and get 0.114kg ice. MP no likey. Anyone care to check this?

Thanks in advance,
Magicarp
 
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Welcome to PF.

If it's any consolation I get a different answer from you.
Perhaps you should recalculate a little more carefully?

I initially dropped a term but it was inconsequential, and still different from your answer.

Edit: but quite close to yours. Maybe carry more precision?
 
Thanks for the speedy response! Are you sure there is nothing to fix in the setup? I could extend the answer to more sig figs, but this is one of those questions where MP conveniently doesn't tell you how many sig figs to include. It tells me that my answer is close and I may have rounded or used the incorrect number of sig figs. But usually when it says this, the answer is different by quite a lot.
 
Magicarp said:
Thanks for the speedy response! Are you sure there is nothing to fix in the setup? I could extend the answer to more sig figs, but this is one of those questions where MP conveniently doesn't tell you how many sig figs to include. It tells me that my answer is close and I may have rounded or used the incorrect number of sig figs. But usually when it says this, the answer is different by quite a lot.

I'd say your method is apparently OK.

I got .11277 kg when I equated what heats up to what cools down. In quickly checking your terms it seems your signs are OK.
 
Thanks LowlyPion, I simply had to extend the sig figs by 1 digit.

Cheers,
Magicarp
 
Magicarp said:
Thanks LowlyPion, I simply had to extend the sig figs by 1 digit.
I would have to back you up.
You are given the data to 3sig figs, more than 3sig figures in the answer is wrong. That's why I hate these online test things
 

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