Question

Calculate the change in entropy when 1.00 kg of water at 100 ∘c is vaporized and converted to steam at 100 ∘c. Assume that the heat of vaporization of water is 2256×103j/kg.

Answer 1

Answer:6.04 kJ/K

Step-by-step explanation:

Given

mass of water
`m=1 kg`

Latent heat of vaporization is
`L=2256* 10^(3) J/kg`

entropy at constant temperature is given by

`\Delta S=(Q)/(T_0)`

`Q=m* L`

`Q=1* 2256* 10^3`

`\Delta s=(2256* 10^3)/(100+273)`

`\Delta s=6.04kJ/K`