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# At 1 atm, how much energy is required to heat 83.0 g of H2O(s) at –18.0 °C to H2O(g) at 169.0 °C?

At 1 atm, how much energy is required to heat 83.0 g of H2O(s) at –18.0 °C to H2O(g) at 169.0 °C?

Where do I begin?

There are three phases to consider here. The solid water is heated to 0 C. This makes use of the specific heat of ice. At 0 C, the ice melts, which requires a bunch more heat than just warming the ice. For the s to l phase change, use the heat of fusion. At 1 atm, the water will be liquid from 0 C up to 100 C, so calculate the heat to raise the waters temperature to 100 C with the specific heat of liquid water. At 100 C, the water boils (l to g transition). Use the heat of vaporization for this phase change. Finally, calculate the energy to heat the gas (steam) to 169 C using the specific heat of water as a gas. There are 2 specific heats of water (gas) to choose from. Use the one for constant pressure since this all occurs at 1 atm.

To summarize:

water(-18 to 169) = water (-18 to 0) + heat of fusion + water (0 to 100) +heat of vaporization + water (100 to 169)

Do this for 1 g and multiply by 83 g or do this for 1 mol and multiply by the number of moles you have (83/18). It all depends on the units you have for the three Cp's and the two latent heats (fusion, vaporization). The Cp of water varies greatly for three different phases.

This should get you started.

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