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How much ice melts at equilibrium

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3 kg of ice whose temperature is -10 C is dropped into 10 kg of water whose temperature is 10 C. How much of the ice melts when the mixture is brought to equilibrium?

 

I wanted to find the equilibrium temperature first so I used q = cm T

And c (water) = 1 kcal/kg C, c (ice) = 0.5 kcal/kg C

 

So

(0.5)(3)(x - (-10)) = (1)(10)(10 - x)

That turns into 1.5x + 15 = 100 - 10x

x = 7.4 C

 

So I then determined how much energy outputted when lowering the temperature of the water is

q = cm T = (1)(10)(10 - 7.4) = 26 kcal

 

Then, to bring the ice to 0 C requires

q = cm T = (0.5)(3)(0 - (-10)) = 15 kcal

 

Then, to cause the phase change, I found the latent heat.

Heat of fusion = 80 kcal/kg

q = mL = (3 kg)(80 kcal/kg) = 240 kcal

 

But 240 is so much higher than 26, so how is this possible. What am I doing wrong?

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(0.5)(3)(x - (-10)) = (1)(10)(10 - x)

That turns into 1.5x + 15 = 100 - 10x

x = 7.4 C

 

 

Think again about this.

 

Can you say why it's wrong?

Edited by studiot

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Ok so the equation I picked out is a heat balance

 

Heat gained = Heat lost.

 

But

 

Heat gained by what? = Heat lost by what?

 

Heat gained by raising 3kg ice from -10oC to 0oC, plus heat gained by converting 3kg ice to water at 0oC, plus heat gained raising 3kg of water fom 0oC to +xoC degrees centigrade.

 

Heat lost by lowering 10kg water from +10oC to +xoC

 

Can you now put the correct figures into this ?

Edited by studiot

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So if the net heat is 0

q = heat gained by raising 3kg ice from -10oC to 0oC + heat gained by converting 3kg ice to water at 0oC + heat gained raising 3kg of water fom 0oC to +xoC degrees + Heat lost by lowering 10kg water from +10oC to +xoC = 0

q = (1/2)(3)(0 - (-10)) + (3)(80) + (1)(3)(x-0) + (1)(10)(x-10) = 0

 

But after I solved for x now, I get 11.9 degrees C, which doesn't make sense either since it's greater than 10.

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But after I solved for x now, I get 11.9 degrees C, which doesn't make sense either since it's greater than 10.

 

 

Which should tell you that there is no solution to your equatiion.

That is, it does not happen!

 

So let us look more closely at the question, still thinking about heat lost = heat gained.

 

The maximum heat the 10kg water can loose, and still remain as water is 10 * (10-0) * 1 kcals = 100kc.

 

It takes 3 * (0-(-10)) * 0.5 = 15kc to raise 3kg of of ice to 0oC

 

To melt all this ice takes 3 * 80 = 240 kc.

 

But there is only (100 - 15) = 85kc available from cooling the water so this can only melt just over 1kg of ice, leaving 2kg unmelted.

 

Any mixture of ice and water is at what temperature in equilibrium?

 

You should be able to complete this now.

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