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.
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?
PLS HELP ME!!!
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