# What is the minimum amount of energy released in kilojoules when 450.0 grams of water vapor condenses to a liquid at 100°C?

Apr 13, 2016

Approx. ${10}^{3}$ $k J$ of energy are released

#### Explanation:

${H}_{2} O \left(g\right) \rightarrow {H}_{2} O \left(l\right) + \text{energy}$

Now we need investigate only the phase change, because both ${H}_{2} O \left(g\right)$ and ${H}_{2} O \left(l\right)$ ARE both at $100$ ""^@C.

So, we have been given the heat of vaporization as $2300$ $J \cdot {g}^{-} 1$. And, $\text{energy}$ $=$ $450.0 \cdot g \times 2300 \cdot J \cdot {g}^{-} 1$ $=$ ??

Because the energy is RELEASED, the calculated energy change is NEGATIVE.