contestada

if 0.45kg is changed by a reversible process to steam at 100°C, determine the entropy of: 1. the water 2. the surrounding. 3. the universe (water+surroundings) as a whole.​

Respuesta :

Answer:

1) 2726.5 J/K      2) decrease                     3) 0

Explanation:

1)

change in entropy is given by [tex]\Delta s\:=\frac{Q}{T}[/tex]

here Q=mL where L is the latent heat of vaporization. therefore

[tex]Q=\frac{mL}{T}[/tex][tex]\Delta s\:=\frac{0.45X22.5X10^5}{100+273}=2726.5\:\:\frac{J}{K}[/tex]

2) [tex]\Delta S=\frac{-Q}{T}\:[/tex]

since the heat is taken from surroundings, the change in entropy is negative therefore entropy of surrounding will decrease.

3) since there is no heat exchange with universe  therefore , Q=0 in this case

[tex]\Delta S=\frac{Q}{T}\:=\frac{0}{T}=0[/tex]