The molar heat capacity of water at constant pressure, `C_(p)` is `"75 J K"^(-1)"mol"^(-1)`. When 10 kJ of heat is supplied to 1 kg water which is free to expand, the increase in temperature of water is

Options

A. (a) 2.4 K
B. (b) 4.8 K
C. (c) 3.2 K
D. (d) 10 K

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Molar heat capacity at constant pressure, \( C_p = 75 \, \text{J K}^{-1} \text{mol}^{-1} \)
- Heat supplied, \( Q = 10 \, \text{kJ} = 10 \times 10^3 \, \text{J} = 10000 \, \text{J} \)
- Mass of water, \( m = 1 \, \text{kg} \)

### Step 2: Calculate the number of moles of water
To find the number of moles, we use the molar mass of water, which is approximately \( 18 \, \text{g/mol} \) or \( 0.018 \, \text{kg/mol} \).

\[ \text{Number of moles} (n) = \frac{\text{mass}}{\text{molar mass}} = \frac{1 \, \text{kg}}{0.018 \, \text{kg/mol}} = \frac{1000 \, \text{g}}{18 \, \text{g/mol}} \approx 55.56 \, \text{mol} \]

### Step 3: Use the formula for molar heat capacity
The formula relating heat, molar heat capacity, number of moles, and change in temperature is given by:

\[ Q = n \cdot C_p \cdot \Delta T \]

Rearranging the formula to find the change in temperature \( \Delta T \):

\[ \Delta T = \frac{Q}{n \cdot C_p} \]

### Step 4: Substitute the known values into the equation
Now we can substitute the known values into the equation:

\[ \Delta T = \frac{10000 \, \text{J}}{55.56 \, \text{mol} \cdot 75 \, \text{J K}^{-1} \text{mol}^{-1}} \]

Calculating the denominator:

\[ n \cdot C_p = 55.56 \, \text{mol} \cdot 75 \, \text{J K}^{-1} \text{mol}^{-1} = 4167 \, \text{J K}^{-1} \]

Now substituting back into the equation for \( \Delta T \):

\[ \Delta T = \frac{10000 \, \text{J}}{4167 \, \text{J K}^{-1}} \approx 2.4 \, \text{K} \]

### Step 5: Conclusion
The increase in temperature of the water is approximately \( 2.4 \, \text{K} \).

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