Homework Help: Calculate the amount of heat

1. Apr 13, 2012

domyy

1. The problem statement, all variables and given/known data

Calculate the amount of heat to be added to warm 150g of ice from -10C to produce steam at 110C.

2. Relevant equations

Ice = cice = 0.5cal/gC, L = 80 cal/g.
Water = Cw = 1.0 cal/gC;
Steam = csteam = 0.5 cal/gC; Lv = 540 cal/g

3. The attempt at a solution

I know the answer should be 109.5kcal.

I have found 12,000 cal. BUT I HAVE ABSOLUTELY NO IDEA HOW TO GET TO 109kcal. I JUST NEED SOME DIRECTION.

2. Apr 13, 2012

Staff: Mentor

Re: 109.5kcal?

Show how you got 12 kcal.

3. Apr 13, 2012

domyy

Re: 109.5kcal?

Q = mLf = 150g x 80 Cal/g = 12,000

4. Apr 13, 2012

domyy

Re: 109.5kcal?

I know this problem is more complex than that.

Another equation that is possible to use is q=mcdeltat

I remember my professor drawing a graph, with equations for solid, liquid and gas...
I think I am supposed to have that in mind.

5. Apr 13, 2012

domyy

Re: 109.5kcal?

The graph shows increase, then constant, then increase and constant, then increase.

6. Apr 13, 2012

Staff: Mentor

Re: 109.5kcal?

That's just a melting part.

And that's a good hint at equations that should be used. Have you copied that graph?

What must happen to the ice so that water becomes a steam?

7. Apr 13, 2012

Pengwuino

Re: 109.5kcal?

There are 5 processes at work here. First, you must heat the water from -10C to 0C. Then more energy must be provided to turn the 0C ice to 0C water. Then you must heat the water from 0C to 100C. Then you must turn the 100C water to 100C steam. Finally, you heat the steam to the final 110C. Phase changes require energy as well as temperature changes. The equations you have can be used to determine exactly how these processes occur.

8. Apr 17, 2012

domyy

Re: 109.5kcal?

One of the solutions would be q=mlf
=150g x 540=81,000?

9. Apr 17, 2012

domyy

Re: 109.5kcal?

I will also have q=150g x 80cal/g = 12,000

10. Apr 17, 2012

domyy

Re: 109.5kcal?

Am i doing it right? I will have to sum the values later?

11. Apr 17, 2012

Staff: Mentor

Re: 109.5kcal?

You can sum them all as the last step.

12. Apr 17, 2012

domyy

Re: 109.5kcal?

13. Apr 17, 2012

domyy

Re: 109.5kcal?

Can I ask you something? I read that as bubbles rise in deep column of water the diameter increases. Could explain this more clear to me?

14. Apr 17, 2012

Staff: Mentor

Re: 109.5kcal?

PV=nRT, P=ρh