Changes of State ice to steam problem

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To convert 0.500 kg of ice at -20°C to steam at 250°C, the energy calculations must include the heat of fusion and heat of vaporization. The specific heat capacities for ice, liquid water, and steam are utilized to calculate energy for each phase change. The user initially calculated energy for temperature changes but overlooked the heat of fusion (Hfus) and heat of vaporization (Hvap) in the process. Incorporating these values is essential for an accurate total energy calculation. The discussion highlights the importance of considering all phase changes and associated energies in thermodynamic problems.
jKotha
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Ok, I am having some trouble with this problem:

How much energy does it take to convert .500 kg ice at -20C to steam at 250 C? Specific heat capacities: ice: 2.1 j/g C, liquid: 4.2 j/g C, steam: 2.0 j/g C. Hvap = 40.7 kj/mol, and Hfus = 6.02 Kj/mol.


What I have done so far is use q=mst for all 3 phases to obtain the energy in 3 phases.

q= (.500)(2.1)(20) = 21000 J, this is for the ice from -20C to 0C
q= (.500)(4.2)(100) = 210000 J, this is from 0 C to 100 C because of liquid water to boiling pt (steam).
q= (.500)(2.0)(150) = 150000 J, this is from 100C to 250C

Am I on the right track? Or am I doing this all wrong? I don't have a clue what to do next. However I think it must use the Clausius-Clapeyron equation in some context. I'm not sure what to do with the Hfus either. Please HELP!

Thanks,
Jkotha
 
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One also has to account for the heat of fusion (ice absorbs energy (heat) at constant temperature while changing to liquid) and heat of vaporization (steam condenses (by releasing heat) to liquid at constant temperature).
 
I Got It!

Hey, thanks. YOU ARE THE MAN. I can't believe I forgot about those.
 
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