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The document provides steps to calculate the number of moles of liquid water needed to freeze and remove 100 kJ of heat. It first converts the heat given in kJ to joules. Using the heat of fusion value of 334 J/g, it calculates the grams of ice formed, and then the number of moles of water needed using the molar mass of water. The calculation determines that 16.6 moles of liquid water are needed to freeze and remove 100 kJ of heat.
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Lets solve a problem
- 1. Lets Solve a Problem
- 2. • How many moles of liquid water must freeze to remove 100 kJ of heat? (Hf = 334 J/g)
- 3. Know Values • Q= • Hf=334j/g • Molar mass of water=18g/mole • Do we need to convert? • Equations – how do we solve?
- 4. Convert? • How many moles of liquid water must freeze to remove 100 kJ of heat? (Hf = 334 J/g) • YES, Q is given in kJ and Hf is in joules • So convert 100kJ to joules • There are a 1000 joules per kilojoule so • 100kJ x 1000j/kJ 100,000 j
- 5. Solving us Hf • First find out how many grams by using Hf=joules of heat/grams of ice • Rearrange first • Grams of ice=Joules of heat/Hf • Grams of ice =100,000j/334j/g • Divide (joules cancel) • =299.4 grams
- 6. How many moles? • 18 g/mole or 1mole/18 g • We have 299.4 grams • 299.4gx 1mole/18g (grams cancel) • =16.6 moles of water