Question #f4fd7

1 Answer
yumean1119
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(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)

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