Question

How many kJ are required to heat 45.0 g of H2O at 25.0°C and then boil it all away?

Answer 1

The answer is `"116 kJ"`.

You need to go through two stages in order to find the total heat required to completely boil `"45.0 g"` of water. First, you must provide enough heat to get the water to `"100"^@"C"`. This is calculated by using

`q_1 = m * c * DeltaT`, where

`c` - water's specific heat - `"4.184 J/g" * ^@"C"`.

You get

`q_1 = "45.0 g" * 4.184 J/(g*^@C) * (100.0 -25.0)^@C = 14121` `"J"`

Then, you must provide enough heat to get all the water from liquid to steam. This is calculated using water's enthalpy of vaporization, `DeltaH_("vap")` - `"40.68 kJ/mol"`.

We can go to moles of water using its molar mass, `"18.0 g/mol"`

`n_("water") = m/("molar mass") = ("45.0 g")/("18.0" ("g")/("mol")) = 2.50` `"moles"`

`q_2 = n_("water") * DeltaH_("vap") = "2.50 moles" * "40.68" ("kJ")/("mol") = 101.7` `"kJ"`

Therefore, the total energy in the form of heat that is needed is

`q_("TOTAL") = q_1 + q_2 = 14121*10^(-3)` `"kJ" + 101.7` `"kJ"`

`q_("TOTAL") = 116` `"kJ"`