Question

Calculate the change in entropy when 1.00 kgkg of water at 100∘C∘C is vaporized and converted to steam at 100∘C∘C. Assume that the heat of vaporization of water is 2256×103J/kg2256×103J/kg. Express your answer in joules per kelvin.

Answer 1

`\Delta S=6045.8(J)/(K)`

Step-by-step explanation:

Hello,

In this case, we can compute the change in the entropy for vaporization processes in term of the enthalpy of vaporization as shown below:

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

Whereas the temperature is in Kelvins. In such a way, the entropy results:

`\Delta S=(1.00kg*2256x10^3(J)/(kg) )/((100+273.15)K)\\\\\Delta S=6045.8(J)/(K)`

Best regards.