What is the change in entropy when `2.5` mole of water is heated from `27^(@)C` to `87^(@)C`?
Assume that the heat capacity is constant `(C_(p))_(m)(H_(2)O)=4.2J//g=k,ln(1.2)=0.18)`

Options

A. (A)16.4 `JK^(-1)`
B. (B) 34.02 `JK^(-1)`
C. (C) 2.89 `JK^(-1)`
D. (D) 18.2 `JK^(-1)`

Answer

To calculate the change in entropy when 2.5 moles of water is heated from 27°C to 87°C, we can follow these steps:

### Step 1: Convert temperatures from Celsius to Kelvin
To use the formula for entropy change, we need to convert the temperatures from Celsius to Kelvin.
- \( T_1 = 27°C + 273.15 = 300.15 K \)
- \( T_2 = 87°C + 273.15 = 360.15 K \)

### Step 2: Identify the molar heat capacity
The molar heat capacity \( C_{p,m} \) of water is given as \( 4.2 \, \text{J/g·K} \). The molar mass of water (H₂O) is approximately \( 18 \, \text{g/mol} \).

### Step 3: Calculate the heat capacity at constant pressure
To find the heat capacity for 1 mole of water:
- \( C_p = C_{p,m} \times \text{molar mass} \)
- \( C_p = 4.2 \, \text{J/g·K} \times 18 \, \text{g/mol} = 75.6 \, \text{J/mol·K} \)

### Step 4: Use the entropy change formula
The change in entropy \( \Delta S \) can be calculated using the formula:
\[ \Delta S = n \cdot C_p \cdot \ln\left(\frac{T_2}{T_1}\right) \]
Where:
- \( n = 2.5 \, \text{moles} \)
- \( C_p = 75.6 \, \text{J/mol·K} \)
- \( T_1 = 300.15 \, K \)
- \( T_2 = 360.15 \, K \)

### Step 5: Calculate the natural logarithm
Calculate \( \frac{T_2}{T_1} \):
\[ \frac{T_2}{T_1} = \frac{360.15}{300.15} \approx 1.200 \]
Now, calculate \( \ln(1.200) \):
- Given \( \ln(1.2) \approx 0.18 \)

### Step 6: Substitute values into the entropy formula
Now substitute the values into the entropy change formula:
\[ \Delta S = 2.5 \, \text{moles} \times 75.6 \, \text{J/mol·K} \times 0.18 \]

### Step 7: Perform the calculation
\[ \Delta S = 2.5 \times 75.6 \times 0.18 \approx 34.02 \, \text{J/K} \]

### Final Answer
The change in entropy when 2.5 moles of water is heated from 27°C to 87°C is approximately:
\[ \Delta S \approx 34.02 \, \text{J/K} \]

---