At 1 atm, how much energy is required to heat 79.0 g of H2O(s) at –20.0 °C to H2O(g) at 131.0 °C?

Question

At 1 atm, how much energy is required to heat 79.0 g of H2O(s) at –20.0 °C to H2O(g) at 131.0 °C?

Answer 1

1. heat to raise temp of 79.0 g H2O from -20.0ºC to 0ºC: q = mC∆T = (79.0g)(2.09 J/g-deg)(20deg) = 3302 J
2. heat to melt 79.0 g of ice at 0ºC: q = m∆Hfus = (79.0g)(334 J/g) = 26,386 J
3. heat to raise 70.0 g of liquid water from 0ºC to 100ºC: q = mC∆T = (79.0g)(4.184 J/g-deg)(100deg) = 33,054 J
4. heat to convert 79.0 g liquid to gas at 100ºC: q = m∆Hvap = (79.0g)(2260 J/g) = 178,540 J
5. heat to raise temp of steam from 100ºC to 131ºC: q = mC∆T = (79.0g)(1.89 J/g-deg)(31 deg) = 4629 J

Add up all the joules (J) to get total amount of heat needed. If desired units are kJ, then divide the answer by 1000.