how much energy is needed to raise the temperature of 38 grams of ice at -27°C to water vapor at 134°C

Question

how much energy is needed to raise the temperature of 38 grams of ice at -27°C to water vapor at 134°C?

Answer 1

This problem requires several steps:

Step 1:  raise 38 g ice from -27ºC to 0ºC

   q = mC∆T = (38 g)(2.09 J/g/deg)(27 deg) = 2144 J

 

Step 2:  melt 38 g of ice at 0ºC

   q = m∆Hf = (38g)(334 J/g) = 12,692 J

 

Step 3:  heat 38 g of water from 0º to 100º

   q = mC∆T = (38g)(4.184 J/g/deg)(100 deg) = 15,899 J

 

Step 4:  convert 38 g at 100º to gas at 100º

   q = m∆Hap = (38 g)(2260 J/g) = 85,880 J

 

Step 5:  raise temp of steam from 100º to 134º

   q = mC∆T = (38 g)(1.89 J/g/deg)(34 deg) = 2442 J

 

Add up all the joules to get 119,057 J = 119 kJ