Practice Energy Calculation Quiz

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/mol heat of vaporization = 40.7 kJ/mol specific heat capacity of ice = 2.1 J/g∙  C specific heat capacity of steam = 1.8 J/g∙  C

Label the graphs with the correct temperatures!

Step 1: Convert the mass in grams to moles. mole H 2 O 1 g H 2 O 18.0 g H 2 O = 40.1 mol H 2 O 722

Step 2 Heat 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 10 5 J

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

Step 4 Heat 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

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

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

Step 7: Add the heats! q total = q 2 + q 3 + q 4 + q 5 + q 6 q total = 3,240,000 J or 3.24 x 10 6 J q total = 3,240 kJ or 3.24 x 10 3 kJ q total = 3.2 x 10 5 J + 241, 000 J + 302,000 J + 1, 630,000 J + 747,000 J