# What is the energy released by 20 grams of 100°C steam that condenses and then cools to 0°C?

Jul 11, 2018

$53. 6 \textcolor{w h i t e}{l} \text{kJ}$

#### Explanation:

The energy released came from two discrete processes:

1. the steam condenses to release some latent heat of condensation at $100 {\textcolor{w h i t e}{l}}^{\text{o""C}}$
2. the water cools down to $0 {\textcolor{w h i t e}{l}}^{\text{o""C}}$ to $100 {\textcolor{w h i t e}{l}}^{\text{o""C}}$ without solidfying.

The amount of energy released in the first process is dependent on the $\text{L"_"v}$ latent heat of vaporization for water and the mass of the sample:

${\text{E"("phase change") = m * "L"_"v" = 20 color(white)(l) "g" xx 2, 260 color(white)(l) "J" * "g}}^{- 1}$
color(white)("E"("phase change"))= 45, 200 color(white)(l) "J"

On the other hand, the amount of energy released in the second process depends on both the specific heat of water, the mass of the sample, and the magnitude of the change in temperature.

"E"("cooling") = m * c * Delta T
$\textcolor{w h i t e}{\text{E"("cooling") ) = 20 color(white)(l) "g" * 4.2 color(white)(l) "J" * "g" ^(-1) * "K" ^(-1) * (100 color(white)(l) "K" - 0 color(white)(l) "K}}$
color(white)("E"("cooling") ) = 8, 400 color(white)(l) "J"
Taking the sum of the energy change of the two processes gives the total amount of energy released:

$\text{E"("released") = "E"("phase change") + "E"("cooling") = 53, 600 color(white)(l) "J" = 53.6 color(white)(l) "kJ}$