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. `16.6J//K`
B. `9 J//K`
C. `34.02J//K`
D. `1.89J//K`

Explanation

To find the change in entropy (ΔS) when 2.5 moles of water is heated from 27°C to 87°C, we will use the formula for entropy change at constant pressure:

\[ \Delta S = N C_P \ln\left(\frac{T_2}{T_1}\right) \]

Where:
- \(N\) = number of moles
- \(C_P\) = molar heat capacity at constant pressure
- \(T_1\) = initial temperature in Kelvin
- \(T_2\) = final temperature in Kelvin

### Step 1: Convert 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: Calculate the number of grams of water
The molar mass of water (H₂O) is approximately 18 g/mol. Therefore, for 2.5 moles:

\[ \text{Mass of water} = 2.5 \, \text{moles} \times 18 \, \text{g/gmol} = 45 \, \text{g} \]

### Step 3: Use the given heat capacity
The heat capacity at constant pressure for water is given as \(C_P = 4.2 \, \text{J/g·K}\).

### Step 4: Calculate the change in entropy
Now we can substitute the values into the entropy change formula:

\[ \Delta S = N C_P \ln\left(\frac{T_2}{T_1}\right) \]

Substituting in our values:

\[ \Delta S = 45 \, \text{g} \times 4.2 \, \text{J/g·K} \times \ln\left(\frac{360.15}{300.15}\right) \]

### Step 5: Calculate the natural logarithm
Using the hint provided in the question, we know:

\[ \ln\left(\frac{360.15}{300.15}\right) \approx \ln(1.2) = 0.18 \]

### Step 6: Final calculation
Now substituting this value back into the equation:

\[ \Delta S = 45 \times 4.2 \times 0.18 \]

Calculating this gives:

\[ \Delta S = 34.02 \, \text{J/K} \]

### Final Result
The change in entropy when 2.5 moles of water is heated from 27°C to 87°C is approximately:

\[ \Delta S \approx 34.02 \, \text{J/K} \]