Question

The latent heat of vaporization for water at room temperature is 2430 J/g.

Answer 1

Answer:

1)   $kinectic energy=7.26*10^-^2^0J$

2)  $V= 2.0m/s$

3)  $T=3.5*10^3K$

4)  The Molecules do not burn because of the presences of hydrogen bond in place

Explanation:

From the question we are told that

latent heat of vaporization for water at room temperature is 2430 J/g.

1)Generally in determining the molar mass of water evaporated we have that

-One mole (6.02 x 10. 23 molecules)

-Molar mass of water is 18.02 g/mol

Mathematically the mass of water is give as

   

  $M=\frac{18.02}{6.02*10^-^2^6}$

  $M=3*10^-^2^3g$

Therefore

  $kinectic energy=2430J/g*3*10^-^2^3g$

 $kinectic energy=7.26*10^-^2^0J$

b)Generally the evaporation speed V is given as$V= \sqrt{\frac{K.E*2}{m} }$

Mathematically derived from the equation

$\frac{1}{2} mv^2 =K.E$

To Give

$V= \sqrt{\frac{K.E*2}{m} }$

$V= \sqrt{\frac{7.26*10^-^2^0J*2}{3*10^-^2^3g} }$

$V= 2.0m/s$

c)Generally the equation for velocity   $Vrms=\sqrt{\frac{3RT}{M} }$

Therefore

Effective temperature T is given by

      $T=\frac{\sqrt{v}*m}{R}$

where

     $T=\frac{\sqrt{2.0m/s}*6.02*10^-^2^6}{0.082057 L atm mol-1K-1}$

     $T=3.5*10^3K$

4) The Molecules do not burn because of the presences of hydrogen bond in place