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| | | ### Statement |
| | | |
| | − | $5.9.3.$ How much will the entropy of $1 \text{ kg}$ of water at a temperature of $293 \text{ K}$ increase when it is converted to steam? |
| | + | $5.9.3.$ How much will the entropy of 1 kg of water at a temperature of 293 K increase when it is converted to steam? |
| | | |
| | | ### Solution |
| | | |
| | − | Resulting change of entropy can be represented as the sum of changes of the entropy in each stage: heating, then evaporation. |
| | + | The process of converting water into steam consists of two consecutive stages: heating the water to its boiling point, and the vaporization (boiling) process itself. The total change in the entropy of the system is the sum of the entropy changes in each stage: |
| | + | $$\Delta S = \Delta S_1 + \Delta S_2$$ |
| | | |
| | − | $$ |
| | − | \Delta S = \Delta S_{1} + \Delta S_{2} |
| | − | $$ |
| | + | <b>1. Heating the water</b> |
| | + | When heating from the initial temperature $T_1 = 293 \text{ K}$ to the boiling point $T_2 = 373 \text{ K}$, the temperature changes, so we find the entropy increment using the integral: |
| | + | $$\Delta S_1 = \int_{T_1}^{T_2} \frac{dQ}{T}$$ |
| | + | Given that the supplied heat is $dQ = c m dT$ (where $c \approx 4180 \text{ J/(kg}\cdot\text{K)}$ is the specific heat capacity of water): |
| | + | $$\Delta S_1 = \int_{T_1}^{T_2} \frac{c m dT}{T} = c m \ln\left(\frac{T_2}{T_1}\right)$$ |
| | + | $$\Delta S_1 = 4180 \cdot 1 \cdot \ln\left(\frac{373}{293}\right) \approx 4180 \cdot 0.2414 \approx 1009 \text{ J/K}$$ |
| | | |
| | − | I. Heating the water: As $\delta Q=cmdT$: |
| | + | <b>2. Vaporization</b> |
| | + | The boiling process occurs isothermally at a constant temperature $T_2 = 373 \text{ K}$. The entropy increment at this stage is: |
| | + | $$\Delta S_2 = \frac{Q_{\text{boil}}}{T_2} = \frac{L m}{T_2}$$ |
| | + | where $L \approx 2.26 \cdot 10^6 \text{ J/kg}$ is the specific heat of vaporization of water. |
| | + | $$\Delta S_2 = \frac{2260000 \cdot 1}{373} \approx 6059 \text{ J/K}$$ |
| | | |
| | − | $$ |
| | − | \Delta S_{1} =\int_{T_{1}}^{T_{2}}\frac{\delta Q}{T} = cm\int_{T_{1}}^{T_{2}}\frac{dT}{T} = cm\cdot \ln\frac{T_{2}}{T_{1}} |
| | − | $$ |
| | + | <b>3. Total entropy change</b> |
| | + | $$\Delta S = 1009 + 6059 = 7068 \text{ J/K} \approx 7 \text{ kJ/K}$$ |
| | | |
| | − | Where $T_{1} = 293~\text{K}$ and $T_{2} = 373~\text{K}$ and $c = 4.2~\mathrm{\frac{kJ}{kg\cdot \text{ K}}}$ is the specific heat of water. |
| | − | |
| | − | II. When water evaporates the temperature is constant. |
| | − | |
| | − | $$ |
| | − | \Delta S_{2} = \int_{1}^{2}\frac{\delta Q}{T} = \frac{1}{T_{2}}\int_{1}^{2}\delta Q = \frac{Q}{T_{2}} = \frac{mL}{T_{2}} |
| | − | $$ |
| | − | |
| | − | Here $L$ is the heat of vaporization of the water. |
| | − | |
| | − | The resulting change in entropy: |
| | − | |
| | − | $$ |
| | − | \boxed{\Delta S =cm\cdot \ln\frac{T_{2}}{T_{1}} + \frac{mL}{T_{2}}} |
| | − | $$ |
| | − | |
| | − | $$ |
| | − | \Delta S = 7189 ~\mathrm{\frac{J}{K}} \approx 7.2 ~\mathrm{\frac{kJ}{K}} |
| | − | $$ |
| | − | |
| | | #### Answer |
| | − | |
| | − | $$ |
| | − | \Delta S = 7.2 ~\mathrm{\frac{kJ}{K}} |
| | − | $$ |
| | − | |
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| | − | |
| | + | $\Delta S \approx 7 \text{ kJ/K}$ |
