How Much Heat is Needed to Convert Ice at -10C to Steam at 110C?

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SUMMARY

To convert 20.0g of H2O(s) at -10°C to steam at 110°C, a total of 12,765.4 joules of heat must be added. This process involves five distinct steps: heating the ice from -10°C to 0°C, melting the ice to liquid water, heating the liquid water to 100°C, vaporizing the water to steam, and finally heating the steam to 110°C. The specific heats used are 2.09 J/g°C for ice, 4.184 J/g°C for liquid water, and 1.84 J/g°C for steam, along with the heat of fusion (334.7 J/g) and heat of vaporization (2259.4 J/g).

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How much heat (in joules) must be added to change 20.0g of H2O(s) at -10C to steam at 110C? The specific heat of H2O(s) is 2.09 J/gC; of H2O(L) is 4.184 J/gC; and of H2O(g) is 1.84 J/gC. The heat fusion of water is 334.7 J/g and the heat vaporization of water is 2259.4J/g.I feel like we have to find the temp for s ---> l and then l ---> g. But I'm completely lost; I don't understand Thermochemistry.
 
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Split it into separate parts - you have to warm up ice till it can be melted, you have to melt it, you have to warm the the water till boiling point, you have to boil it, you finally have to warm up steam. Then add everything.

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