What is the entropy change of 0.6kg of steam at 1atm pressure and 100oC, when it condenses to 0.6kg of water at 100oC?

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1 answer

ΔS = ΔQ/T. ΔQ = -mL, m = mass of water, L= latent heat of vaporization = xxxxxx J/kg.
ΔS = -(0.6 kg)(xxxxx J /kg)/273 K = xxxxx J/K.
No change of temparature. Put in the values.