Answer:
-594.87666 J/kg
Explanation:
m = Mass of Helium = 0.12 kg
T = Temperature = 4.216 K
[tex]H_v[/tex] = Heat of vaporization = [tex]-2.09\times 10^{4}\ J/kg[/tex]
Change in entropy is given by
[tex]\Delta S=\dfrac{mH_v}{T}\\\Rightarrow \Delta S=\dfrac{0.12\times -2.09\times 10^{4}}{4.216}\\\Rightarrow \Delta S=-594.87666\ J/kg[/tex]
The change in entropy is -594.87666 J/kg