Edit to “Solution”
en/5.9.2.md
+1 −1
| ### Statement | |||
| $5.9.2.$ Find the entropy increment of $1\text{ kg}$ of ice as it melts. | |||
| ### Solution | |||
| The ice melts at the constant temperature. Using the expression for the entropy: | |||
| $$ | |||
| @@ -10,7 +10,7 @@Solution | |||
| \Delta S = \int_{1}^{2}\frac{\delta Q}{T} | |||
| $$ | |||
| − | Considering the | ||
| + | Considering the melting temperature $T$ of ice constant: | ||
| $$ | |||
| \Delta S = \frac{1}{T}\int_{1}^{2}\delta Q = \frac{Q}{T} | |||
| $$ | |||
| For melting, $Q = \lambda m$, where $\lambda$ is a heat of fusion of ice, $Q$ is the heat the ice absorbs. This yields: | |||
| $$ | |||
| \boxed{\Delta S = \frac{\lambda m}{T}} | |||
| $$ | |||
| Substituting $m = 1 \text{ kg}$, $T = 273 \text{ K}$, $\lambda = 3.33\cdot 10^5~\mathrm{\frac{J}{kg}}$ | |||
| $$ | |||
| \Delta S = 1219~\mathrm{\frac{J}{K}}\approx 1.2~\mathrm{\frac{kJ}{K}} | |||
| $$ | |||
| #### Answer | |||
| $$ | |||
| \Delta S = 1.2~\mathrm{\frac{kJ}{K}} | |||
| $$ | |||
| unchanged lines 25 | |||
| ### Statement | ### Statement | ||
| $5.9.2.$ Find the entropy increment of $1\text{ kg}$ of ice as it melts. | $5.9.2.$ Find the entropy increment of $1\text{ kg}$ of ice as it melts. | ||
| ### Solution | ### Solution | ||
| The ice melts at the constant temperature. Using the expression for the entropy: | The ice melts at the constant temperature. Using the expression for the entropy: | ||
| $$ | $$ | ||
| @@ -10,7 +10,7 @@Solution | |||
| \Delta S = \int_{1}^{2}\frac{\delta Q}{T} | \Delta S = \int_{1}^{2}\frac{\delta Q}{T} | ||
| $$ | $$ | ||
| Considering the |
Considering the melting temperature $T$ of ice constant: | ||
| $$ | $$ | ||
| \Delta S = \frac{1}{T}\int_{1}^{2}\delta Q = \frac{Q}{T} | \Delta S = \frac{1}{T}\int_{1}^{2}\delta Q = \frac{Q}{T} | ||
| $$ | $$ | ||
| For melting, $Q = \lambda m$, where $\lambda$ is a heat of fusion of ice, $Q$ is the heat the ice absorbs. This yields: | For melting, $Q = \lambda m$, where $\lambda$ is a heat of fusion of ice, $Q$ is the heat the ice absorbs. This yields: | ||
| $$ | $$ | ||
| \boxed{\Delta S = \frac{\lambda m}{T}} | \boxed{\Delta S = \frac{\lambda m}{T}} | ||
| $$ | $$ | ||
| Substituting $m = 1 \text{ kg}$, $T = 273 \text{ K}$, $\lambda = 3.33\cdot 10^5~\mathrm{\frac{J}{kg}}$ | Substituting $m = 1 \text{ kg}$, $T = 273 \text{ K}$, $\lambda = 3.33\cdot 10^5~\mathrm{\frac{J}{kg}}$ | ||
| $$ | $$ | ||
| \Delta S = 1219~\mathrm{\frac{J}{K}}\approx 1.2~\mathrm{\frac{kJ}{K}} | \Delta S = 1219~\mathrm{\frac{J}{K}}\approx 1.2~\mathrm{\frac{kJ}{K}} | ||
| $$ | $$ | ||
| #### Answer | #### Answer | ||
| $$ | $$ | ||
| \Delta S = 1.2~\mathrm{\frac{kJ}{K}} | \Delta S = 1.2~\mathrm{\frac{kJ}{K}} | ||
| $$ | $$ | ||
| unchanged lines 25 | |||