Practice Energy Calculation Quiz

1. Practice Energy Calculation Quiz

2. How much energy does it take to convert 722 grams of ice at 211C to steam at 675C?(Be sure to draw and label the appropriate heating or cooling curve.)Provided information:heat of fusion = 6.0 kJ/molheat of vaporization = 40.7 kJ/molspecific heat capacity of ice = 2.1 J/g∙Cspecific heat capacity of steam = 1.8 J/g∙C

3. Label the graphs with the correct temperatures!

4. Step 1: Convert the mass in grams to moles. 722 mole H2O 1 g H2O = 40.1 molH2O 18.0 g H2O

5. Step 2Heat the ice from 211C to its melting point of 0C. q = mcΔT q = (722 g)(2.1 J/g∙C)(0  (211C)) q = 3.20 x 105 J

6. Step 3Convert ice to liquid water - (melt the ice!!) q = ΔHfusion ∙moles q = (6.0 kJ/mol)(40.1 mol) q = 241 kJ = 241,000 J

7. Step 4Heat the ice from 0C to its boiling point of 100C. q = mcΔT q = (722 g)(4.18 J/g∙C)(100  0C) q = 302,000 J

8. Step 5Convert water to steam - (boil the water!!) q = ΔHvaporization ∙moles q = (40.7 kJ/mol)(40.1 mol) q = 1630 kJ = 1,630,000 J

9. Step 6Heat the steam from 100C to 675C. q = mcΔT q = (722 g)(1.8 J/g∙C)(675  100C) q = 747,000J

10. qtotal = q2+ q3 +q4 +q5+ q6 Step 7: Add the heats! qtotal= 3.2 x 105 J + 241, 000 J + 302,000 J + 1, 630,000 J + 747,000 J qtotal= 3,240,000 J or 3.24 x 106 J qtotal = 3,240 kJ or 3.24 x 103 kJ