# Question #f4fd7

Oct 20, 2017

You need the heat of vaporization to answer the question.
About $1414$ kJ of heat is released.

#### Explanation:

First, you have to calculate the mass of water.
$29.82$ (mol)・$18.01 \left(\frac{g}{m o l}\right)$= $537.1$(g)

There are three steps to calculate the energy.

[Step1] Cool the steam 122.8℃ → 100℃.
Heat released in this step is $537.1$ (g) * $2.0$(J/g・K) * $\left(122.8 - 100\right)$ (K) = $2.449 \times {10}^{4}$ (J) = $24.49$ (kJ).

[Step2] Liquefy the steam to liquid water at 100℃.
According to a Japanese site (http://www.hakko.co.jp/qa/qakit/html/h01060.htm),
the heat of vaporization for ${H}_{2} O$ is $2257$ kJ/kg.

Thus, $0.5371$ (kg) * $2257$ (kJ/kg) = $1212$ (kJ) is released in Step2.

[Step3] Cool the liquid water 100℃→21.1℃.
Heat capacity of liquid water is $4.184$ J/g (since 1 cal=4.184 J).
Heat released in Step3 is $537.1$ (g) * $4.184$(J/g・K) * $\left(100 - 21.1\right)$ (K) =$1.773 \times {10}^{5}$ (J) =$177.3$ (kJ)

Sum up [Step1] to [Step3] and you find the answer.
$24.5 + 1212 + 177.3$= $1413.8$ (kJ)