How much heat is added to 10.0 grams of ice at -20.0º Celsius to convert it to steam at 120.0º Celsius?

Question

I'm having trouble starting this problem.
Please help me.

Answer 1

You may be missing the extra heat during the transition from ice to water, and then from water to steam.

You break the problem down into several steps.
1. from -20 C to 0 C as ICE, H1 = mass * specific heat of ICE* (0 - (-20))
2. from 0 C ice to 0 C water, H 2= mass * latent heat of ice
3. from 0 C water to 100 C water, H3= mass * specific heat of water * (100-0)
4, from 100 C water to 100C steam , H4 = mass * latent heat of steam
5, from 100C steam to 120C steam, H5= mass * specific heat of steam * (120-100)

Add them up.

Answer 2

Specific heat capacity is: c=Q/(m*(Tf-Ti) where c is that value, m is the mass and Tf-Ti is the temperature increment. The solving depends of your units, you should get Q ``solved´´.

The water case is c= 1 calorie / (1gr * 1ºC)

In the changes of state you need to know how many calories takes (or gives) a gram of substance to change it's state: the Latent Heat.
Q = Lh * m (Lh because I'm spanish and I don't know how do you write it shortly, i've had to look fot ``latent heat´´).

The water case (call me your oddparent) is from solid to liquid, 80 cal/g and from liquid to steam 540 cal/g.

Adding the 3 Qs (you can get all the temp. raise at one time, and 2 state changes) you get total Q (heat, in english) used.

I've answered 2 problems before this and i'm sorry but i'm not going to do the calculus again, i hope you can do it with this.
Well, and maybe be one more cause the fact that I cant get help with my quantum field theory problems. (That was a joke).