Thermodynamics: Mixing ice and water

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SUMMARY

The discussion centers on the thermodynamic calculations involved in mixing 10 g of ice at -20°C with 100 g of water at +5°C. The specific heat capacities of water (4.186 x 10^3 J/kg°C) and ice (2.108 x 10^3 J/kg°C), along with the latent heat of fusion for ice (333.55 x 10^3 J/kg), are critical to the calculations. The initial calculations were incorrect due to arithmetic errors and misapplication of units, specifically using 333,000 J/kg instead of 333 J/g for the latent heat of fusion. The correct conclusion is that 5 grams of ice will melt when the system reaches thermal equilibrium.

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iAlexN
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10 g of ice at -20C and 100g of water at +5C is mixed together. How much water and ice will there be once the system reaches equilibrium (assuming no heat is lost)?

Specific heat of water and ice: 4.186*10^3, 2.108*10^3
Latent heat of fusion of ice: 333.55*10^3

All in units (kg^{-1} * C^{-1})

My try:

Energy_{water} = (0.1 * 5 * 4.186 * 10^3) = 2093 J
Energy_{water} - Heating_{ice} = 2093 - (0.01 * 2.108 * 10^3 * 20) = 2071.92J

Melting the ice with the remaning energy:

2071.92 = 333.55*10^3 * x
x ≈ 0.062 grams

But the right answer is supposed to be 5 grams of ice melted.

Where did I go wrong? I thought it was a bit to simple to be correct.

Edit: +5 not -5 for water.

Thank you!
 
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Are you sure about that -5C for the water?
 
Chestermiller said:
Are you sure about that -5C for the water?

Thank you. It is supposed to be +5 C for the water.
 
The following is incorrect arithmetic: 2093−(0.01∗2.108∗103∗20)=2071.92J

Also, you should be dividing by 333 J/gm, not 333000.

Chet
 
Chestermiller said:
The following is incorrect arithmetic: 2093−(0.01∗2.108∗103∗20)=2071.92J

Also, you should be dividing by 333 J/gm, not 333000.

Chet

Oh, I see. Thanks!
 

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