Suppose 12.67 g water at 40.0 °C is flash-evaporated by allowing it to fall into an iron crucible maintained at 129.7 °C.
You are given that
ΔHfus = 6007.000 J mol-1;
ΔHvap = 40.680 kJ mol-1;
cP(ice) = 38.0 J K-1 mol-1;
cP(steam) = 36.0 J K-1 mol-1; and
cP(water) = 75.4 J K-1 mol-1.
Calculate these quantities for the process.
ΔSsys = 
 J K-1( ##Enter 87.9 as the answer here##)

ΔSsurr = 
 J K-1
(Use the previous answer and the temperature of the iron crucible in calculation #2.
ΔSuniv = 
 J K-1

To calculate the entropy changes, we need to follow these steps:

### Step 1: Calculation of ΔSsys

To calculate the entropy change of the system (\(\Delta S_{\text{sys}}\)) for water as it undergoes phase transitions and temperature changes, we break it into three parts:

1. **Heating of water from 40.0 °C to 100.0 °C (boiling point):**
   \[
   \Delta S_1 = n \cdot c_P(\text{water}) \cdot \ln\left(\frac{T_{\text{final}}}{T_{\text{initial}}}\right)
   \]
   where \(n\) is the number of moles of water, \(c_P(\text{water})\) is the specific heat capacity of liquid water, \(T_{\text{final}}\) is 373.15 K (boiling point of water), and \(T_{\text{initial}}\) is 313.15 K (40.0 °C).

2. **Phase change from liquid water to steam (evaporation):**
   \[
   \Delta S_2 = \frac{\Delta H_{\text{vap}}}{T_{\text{boiling}}}
   \]
   where \(\Delta H_{\text{vap}}\) is the heat of vaporization and \(T_{\text{boiling}}\) is 373.15 K.

3. **Heating of steam from 100.0 °C to the temperature of the iron crucible (129.7 °C):**
   \[
   \Delta S_3 = n \cdot c_P(\text{steam}) \cdot \ln\left(\frac{T_{\text{final}}}{T_{\text{initial}}}\right)
   \]
   where \(T_{\text{final}}\) is 402.85 K (129.7 °C) and \(T_{\text{initial}}\) is 373.15 K.

The total entropy change of the system is:
\[
\Delta S_{\text{sys}} = \Delta S_1 + \Delta S_2 + \Delta S_3
\]
This will give us the system's entropy change.

You already provided \(\Delta S_{\text{sys}} = 87.9 \, \text{J K}^{-1}\), so we will use this in the next steps.

### Step 2: Calculation of ΔSsurr

The entropy change of the surroundings (\(\Delta S_{\text{surr}}\)) can be calculated using the heat exchanged with the surroundings and the temperature of the surroundings, which is the temperature of the iron crucible (129.7 °C = 402.85 K). 

\[
\Delta S_{\text{surr}} = -\frac{\Delta H_{\text{total}}}{T_{\text{surr}}}
\]
where \(\Delta H_{\text{total}}\) is the total enthalpy change during the process and \(T_{\text{surr}} = 402.85 \, \text{K}\).

To find \(\Delta H_{\text{total}}\), we need to calculate the total heat involved in the entire process. This includes:

1. Heating water from 40.0 °C to 100.0 °C.
2. Vaporizing the water.
3. Heating steam from 100.0 °C to 129.7 °C.

We can break down the heat calculation similarly to the steps used for \(\Delta S_{\text{sys}}\), but for now, I’ll summarize the process.

### Step 3: Calculation of ΔSuniv

The entropy change of the universe is the sum of the entropy changes of the system and the surroundings:
\[
\Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}}
\]

Let me calculate \(\Delta S_{\text{surr}}\) and \(\Delta S_{\text{univ}}\).The results of the entropy calculations are:

- \(\Delta S_{\text{surr}} = -80.78 \, \text{J K}^{-1}\)
- \(\Delta S_{\text{univ}} = 7.12 \, \text{J K}^{-1}\)

The positive \(\Delta S_{\text{univ}}\) indicates that the process is overall spontaneous, as the total entropy of the universe increases.

Would you like any further details or explanations on these results?

Here are some related questions to expand your understanding:

1. What would happen if the temperature of the iron crucible was lower than 100°C?
2. How does the heat capacity of water affect the total heat required in this process?
3. Can you explain why \(\Delta S_{\text{surr}}\) is negative in this case?
4. How does the entropy change of vaporization differ from that of fusion?
5. What would happen to the entropy of the system if more water was added?

**Tip:** Entropy is a measure of disorder, and phase changes like vaporization usually lead to significant increases in entropy.

Suppose 12.67 g water at 40.0 °C is flash-evaporated by allowing it to fall into an iron crucible maintained at 129.7 °C. You are given that ΔHfus = 6007.000 J mol-1; ΔHvap = 40.680 kJ mol-1; cP(ice) = 38.0 J K-1 mol-1; cP(steam) = 36.0 J K-1 mol-1; and cP(water) = 75.4 J K-1 mol-1. Calculate these quantities for the process. ΔSsys = 87.9 J K-1; ΔSsurr = ? J K-1; ΔSuniv = ? J K-1.

Calculate the entropy changes in a system where 12.67 g of water at 40°C is flash-evaporated into a crucible at 129.7°C. Explore key thermodynamic concepts like phase changes, entropy, and heat capacity with detailed formulas and step-by-step calculations. Ideal for undergraduate students studying chemistry or physics.

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