When $1.0 \mathrm {~kg}$ of steam at $100 ^ { \circ } \mathrm { C }$ condenses to water at $100 ^ { \circ } \mathrm { C }$, what is the change in entropy of the steam? The latent heat of vaporization of water is $22.6 \times 10 ^ { 5 } \mathrm {~J} / \mathrm { kg }$.
A) $22.6 \times 10 ^ { 5 } \mathrm {~J} / \mathrm { K }$
B) $- 6.1 \times 10 ^ { 3 } \mathrm {~J} / \mathrm { K }$
C) zero
D) $6.1 \times 10 ^ { 3 } \mathrm {~J} / \mathrm { K }$
E) $- 22.6 \times 10 ^ { 5 } \mathrm {~J} / \mathrm { K }$

Question

Quizplus · Accepted Answer

The change in entropy ($\Delta S$) during the condensation of steam can be calculated using the formula $\Delta S = \frac{Q}{T}$, where $Q$ is the heat exchanged and $T$ is the temperature in Kelvin. Since $Q = -22.6 \times 10^5 \, \text{J}$ (it's negative because heat is released) and $T = 100^\circ \text{C} = 373 \, \text{K}$, $\Delta S = \frac{-22.6 \times 10^5}{373} \approx -6.1 \times 10^3 \, \text{J/K}$.

When 1.0 kg1.0 \mathrm {~kg}1.0 kg Of Steam At 100∘C100 ^ { \circ } \mathrm { C }100∘C

When $1.0 \mathrm {~kg}$ Of Steam At $100 ^ { \circ } \mathrm { C }$