# Convert grams of water to steam

I'm working through problems in my textbook. How much heat is required to convert 2.55 g of $\ce{H2O}$ at 28 °C to steam at 100 °C?

In simpler problems I've used the fact that $\Delta H_\text{fus} = 6.02\ \mathrm{kJ/mol}$ for $\ce{H2O}$ at a melting point of 0 °C to do these kinds of calculations. But here the temperatures are different. How can I approach this problem? Is there some other conversion factor I should be aware of?

You need the vaporization $\Delta H_\text{vap}$ and you need to use this equation: $$Q = m~c~\Delta T$$

Where $Q$ equals the energy to heat $1\ \mathrm{mol}$ of water up to $\mathrm{100~^\circ C}$. $m$ is the mass of water, $c$ is the specific heat capacity and $\Delta T$ is the difference between the initial and final temperature.

In this stage you have the energy that is used to heat the liquid water up to $\mathrm{100~^\circ C}$ but the water is still liquid and you need steam water.

So you need to calculate the energy for phase transition from liquid to steam and the equation for this is: $$\Delta H_\text{vap} \cdot n = Q$$

$Q$ is the necessary energy for the phase transition, $n$ is the amount of substance and $\Delta H_\text{vap}$ is the energy used for phase change of $1\ \mathrm{mol}$ of the substance.

You also need to convert the mass to amount of substance (or use latent steam heat instead) and to calculate the $Q$ in the two stages.

This is the energy used for heat and change the phase.