How much heat is required to vaporize 50.0 g of water if the initial temperature of the water is 25.0 C and the water is heated to its boiling point where its converted to steam? The specific heat of water is 4.18 J and the standard enthalpy of vaporization of water at its boiling point is 40.7 KJ.
1: 169 KJ
2: 129 KJ
3: 23.5 KJ
4: 64.2 KJ
5: 40.7 KJ
1: 169 KJ
2: 129 KJ
3: 23.5 KJ
4: 64.2 KJ
5: 40.7 KJ
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This is really just a heat problem, where you need to calculate heat for both warming of liquid water and the phase change from liquid to gas. So you will need both Q = m*c*delta T and Q = m*Hvap. One thing to be keen on is units. Make sure they are the same. One more thing- you are given the standard enthalpy of vaporization... that means per mole of water, so you will have to consider that when forming the equation.
Q = (50 g)(4.18J/gC)(100C-25C) this is the warming of the liquid part, and your answer for this part will be in joules.
plus the phase change heat...
Q = m*Hvap = (50 g H2O)(40.7 kJ/mol) / (18 g/mol) std Hvap is per mole, so you must multiply by moles, and we accomplish this by dividing mass by molar mass of water.
so, part 1 Q = 15675 J = 15.675 kJ
part 2 Q = 113.06 kJ
add the 2 together, get 128.7 kJ, which matches choice 2.
hope that helps
Q = (50 g)(4.18J/gC)(100C-25C) this is the warming of the liquid part, and your answer for this part will be in joules.
plus the phase change heat...
Q = m*Hvap = (50 g H2O)(40.7 kJ/mol) / (18 g/mol) std Hvap is per mole, so you must multiply by moles, and we accomplish this by dividing mass by molar mass of water.
so, part 1 Q = 15675 J = 15.675 kJ
part 2 Q = 113.06 kJ
add the 2 together, get 128.7 kJ, which matches choice 2.
hope that helps