Answer : The change in entropy is [tex]6.05\times 10^3J/K[/tex]
Explanation :
Formula used :
[tex]\Delta S=\frac{m\times L_v}{T}[/tex]
where,
[tex]\Delta S[/tex] = change in entropy = ?
m = mass of water = 1.00 kg
[tex]L_v[/tex] = heat of vaporization of water = [tex]2256\times 10^3J/kg[/tex]
T = temperature = [tex]100^oC=273+100=373K[/tex]
Now put all the given values in the above formula, we get:
[tex]\Delta S=\frac{(1.00kg)\times (2256\times 10^3J/kg)}{373K}[/tex]
[tex]\Delta S=6048.25J/K=6.05\times 10^3J/K[/tex]
Therefore, the change in entropy is [tex]6.05\times 10^3J/K[/tex]
