Question about thermal physics -- Ice cubes melting in water

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Homework Statement:

Ice cubes of total mass 150 g are placed in a vacuum flask holding 350 g of water at a temperature of 80 degrees Celsius. Assuming the vacuum flask loses no heat and has negligible heat capacity, what will be the final temperature of the water after all the ice has melted.

Relevant Equations:

Q = mLf,
Q = mcΔT
First, I calculated the heat required for the ice to melt:

Q=mLf

Q=0.150×330

Q=49.5 J

Then, I calculated the final temperature of the water by forming the following equation:

Q=mcΔT

−49.5=(0.15+0.35)×4200×(Tf −80)

Tf=80.0 degrees Celcius

But the answer says 32 degrees Celsius.
 

Answers and Replies

  • #2
haruspex
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Homework Statement:: Ice cubes of total mass 150 g are placed in a vacuum flask holding 350 g of water at a temperature of 80 degrees Celsius. Assuming the vacuum flask loses no heat and has negligible heat capacity, what will be the final temperature of the water after all the ice has melted.
What is the initial temperature of the ice?
−49.5=(0.15+0.35)×4200×(Tf −80)
Having melted the ice, what masses of water at what temperatures do you have?
 
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What is the initial temperature of the ice?

Having melted the ice, what masses of water at what temperatures do you have?
It didn't say anything about it. So I'm assuming it to be 0 degrees Celsius.
 
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haruspex
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It didn't say anything about it. So I'm assuming it to be 0 degrees Celsius.
Ok, but what about my second question?
 
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Welcome, ianc1339 :cool:
In real life, could the temperature of a mass of water remain the same (80 degrees Celsius in this case) if you add about half the mass of ice to it?

Your mass of water has more thermal energy than your mass of ice.
If you put both masses in direct contact, naturally a flow of energy from hotter to cooler will occur.

Once the thermal equilibrium is reached (there is no more energy flow), the original mass of frozen water will have a higher temperature and the original mass of water will have a lower temperature, the value of both being the same.

The calculation would be easier if you originally had two masses of water at different temperatures: the final temperature would be proportional to the masses.
In the case of this problem, some of the original energy of the mass of water must be used to melt the mass of ice, while its temperature remains at 0 degrees Celsius (heat of phase change).
From that point on, you have two masses of water at different temperatures and an ongoing flow of energy.

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/phase2.html#c1

:cool:
 
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  • #6
mjc123
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Q=mLf

Q=0.150×330

Q=49.5 J
What are the units attached to the numbers 0.150 and 330? Is your calculation consistent?

Q=mcΔT

−49.5=(0.15+0.35)×4200×(Tf −80)
Does all the water cool down from 80°C? Doesn't some of it have to warm up from 0°?
 
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  • #7
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Calculating latent heat required to melt ice:
##
Q=mL_f\\

Q=0.150kg\times 330Jkg^-1\\

Q=49.5 J
##

Calculating change in temperature of water:
##
Q=mc\Delta T\\

−49.5J=0.35kg\times 4200Jkg^{-1}K^{-1}\times (T_f −80^o C)\\

T_f=79.966^oC
##

Calculating temperature when the two bodies of water reach thermal equilibrium:
##
0.15kg\times 4200Jkg^{-1}K^{-1}\times (T_f - 0^oC) = 0.35kg\times 4200Jkg^{-1}K^{-1}\times (79.966^oC - T_f)\\

T_f = 56.0^oC
##

I still got the wrong answer... Where did I make an error in my working?
 
  • #8
haruspex
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## 330Jkg^{-1}##
Check that number, paying attention to units.
Didn't it seem wrong that the water was only losing 0.034oC?
(For the LaTeX, you need to put {} around the -1.)
 
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  • #9
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Check that number, paying attention to units.
Didn't it seem wrong that the water was only losing 0.034oC?
(For the LaTeX, you need to put {} around the -1.)
My book said the specific latent heat of ice was 330 Jkg^-1. But with a quick google search, it turns out that it is actually 330000 Jkg^-1.

And yes, I ended up getting 32 degrees Celsius. I don't know why the book wrote down the constant wrong which is quite annoying.

Anyways, thanks for the heads up!
 

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