Question

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.​

Answer 1

Answer:

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

Explanation:

1)

change in entropy is given by $\Delta s\:=\frac{Q}{T}$

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

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

2) $\Delta S=\frac{-Q}{T}\:$

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

$\Delta S=\frac{Q}{T}\:=\frac{0}{T}=0$