Assume 12,500 J of energy is added to 2.0 moles (36 grams) of H2O as an ice sample at O°C. The molar heat of fusion is 6.02 kJ/mol. The specific heat ofliquid water is 4.18 J/g°C. The molar heat of vaporization is 40.6 kJ/mol. The resulting sample contains which of the following?A. only iceB. ice and waterC. only waterD. water and water vaporE. only water vapor
The first thing that happens is the ice melting. It takes a lot of energy to melt all the ice. 6.02 kJ/mol * 2. 0 mol = 12. 04 kJ = 12,040 J We do have that amount of energy and more: 12,500 J - 12,040 J = 460 J. Just by intuition, we can answer the question because 460 J isn't nearly enough energy to warm the water to the boiling point, so we can say C. Even though the bp hasn't been reached, some of the ice will melt and some of the water will evaporate, so D would be a more realistic answer. Water Vapor due to these processes are ignored in these types of questions.
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