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

Jan 18, 2015

The answer is $\text{116 kJ}$.

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

${q}_{1} = m \cdot c \cdot \Delta T$, where

$c$ - water's specific heat - $\text{4.184 J/g" * ^@"C}$.

You get

${q}_{1} = \text{45.0 g} \cdot 4.184 \frac{J}{g {\cdot}^{\circ} C} \cdot {\left(100.0 - 25.0\right)}^{\circ} C = 14121$ $\text{J}$

Then, you must provide enough heat to get all the water from liquid to steam. This is calculated using water's enthalpy of vaporization, $\Delta {H}_{\text{vap}}$ - $\text{40.68 kJ/mol}$.

We can go to moles of water using its molar mass, $\text{18.0 g/mol}$

n_("water") = m/("molar mass") = ("45.0 g")/("18.0" ("g")/("mol")) = 2.50 $\text{moles}$

${q}_{2} = {n}_{\text{water") * DeltaH_("vap") = "2.50 moles" * "40.68" ("kJ")/("mol}} = 101.7$ $\text{kJ}$

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

${q}_{\text{TOTAL}} = {q}_{1} + {q}_{2} = 14121 \cdot {10}^{- 3}$ $\text{kJ} + 101.7$ $\text{kJ}$

${q}_{\text{TOTAL}} = 116$ $\text{kJ}$