# Thermodynamics: Mixing ice and water

#### iAlexN

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!

Last edited:
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#### Chestermiller

Mentor
Are you sure about that -5C for the water?

#### iAlexN

Are you sure about that -5C for the water?
Thank you. It is supposed to be +5 C for the water.

#### Chestermiller

Mentor
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

#### iAlexN

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!

"Thermodynamics: Mixing ice and water"

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