Chemistry Help

[f1br0pt1kl]

Ok, we are doing energy, such as calories and stuff...maybe you can figure this out for me.

If 3.2 Cal of heat is added to 1kg of ice at 0 (Degrees Celcius), how much water at 0 (Degrees Celcius), is producted and how much ice is left.

What I do know that 3.2 Cal is 3,200 calories (or "cal&quot.

Any takers? Answers should be in grams.

I'm lost, and need some guidance.

 

[Handle]

Find out how many calories are in a kJ.

Find out the heat of fusion of water (or I guess if you have a constant for calories, then you can skip the first step).

Then use factor label and you should have your answer.

 

[Cyberian]

I believe the definition of 'calorie' is - the amount of energy required to raise the temperature of 1 gram of water 1 degree Celsius.

You have 1,000 grams of ice and 3,200 calories.
Sounds to me like you'd end up with a big puddle.
But I must be missing something, that's too easy!

 

[Handle]

Since the question doesn't deal with temperature changes, (I think) he has to work with the heat of fusion as opposed to the specific heat capacity. So the "1 calorie is the amount of energy required to change the temperature of 1 gram of water by 1 degree Celsius" won't work here.

 

[pudcox]

hated chemistry then and now lol

 

[maziar]

If you have a mixture of ice and water, before the net temperature can go up, the ice first has to be melted. The heat of fusion for ice is 6.01 kJ/mol.

You are putting in 3.2 Cal which is equivalent to 13.4 kJ since (1 Cal=4.184 kJ). So now we see how many mols of ice are melted by: 13.4 kJ/(6.01 kJ/mol)=2.2 mol. Since the mol. weight of ice is 18.0 g/mol: 2.2 mol (18.0 g/mol)=39.6 g are melted. 960.4 g ice remain.

 

[shenea]

I vote for maziar's answer!

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nimemaliza Chemistry mtihani yangu leo mchana... sasa nimemaliza shule yangu kabisa!

 

[Shantanu]

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[FrontlineWarrior]

general chemistry is suck.