Question

Question #f4fd7

Answer 1

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(g/(mol))`= `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) * `(122.8-100)` (K) = `2.449xx10^4` (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_2O` 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) * `(100-21.1)` (K) =`1.773xx10^5` (J) =`177.3` (kJ)

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