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 : The change in entropy is
`6.05* 10^3J/K`

Explanation :

Formula used :

`\Delta S=(m* L_v)/(T)`

where,

`\Delta S` = change in entropy = ?

m = mass of water = 1.00 kg

`L_v` = heat of vaporization of water =
`2256* 10^3J/kg`

T = temperature =
`100^oC=273+100=373K`

Now put all the given values in the above formula, we get:

`\Delta S=((1.00kg)* (2256* 10^3J/kg))/(373K)`

`\Delta S=6048.25J/K=6.05* 10^3J/K`

Therefore, the change in entropy is
`6.05* 10^3J/K`