The molar heat capacity of water at constant pressure, C, is `75 JK^(-1) mol^(-1)`. When 1.0 kJ of heat is supplied to 100 g water which is free to expand, the increase in temperature of water is :

Options

A. 4.8 K
B. 6.6 K
C. 1.2 K
D. 2.4 K

Explanation

To solve the problem, we need to determine the increase in temperature of 100 g of water when 1.0 kJ of heat is supplied, using the molar heat capacity at constant pressure.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Molar heat capacity of water, \( C_p = 75 \, \text{J K}^{-1} \text{mol}^{-1} \)
- Heat supplied, \( Q = 1.0 \, \text{kJ} = 1000 \, \text{J} \)
- Mass of water, \( m = 100 \, \text{g} \)

2. **Convert Mass of Water to Moles:**
- The molecular weight of water (H₂O) is approximately \( 18 \, \text{g mol}^{-1} \).
- Calculate the number of moles of water:
\[ n = \frac{m}{\text{Molar mass}} = \frac{100 \, \text{g}}{18 \, \text{g mol}^{-1}} \approx 5.56 \, \text{mol} \]

3. **Use the Formula for Heat Transfer:**
- The formula relating heat, molar heat capacity, and temperature change is:
\[ Q = n C_p \Delta T \]
- Rearranging for \( \Delta T \):
\[ \Delta T = \frac{Q}{n C_p} \]

4. **Substitute the Values:**
- Substitute \( Q \), \( n \), and \( C_p \) into the equation:
\[ \Delta T = \frac{1000 \, \text{J}}{5.56 \, \text{mol} \times 75 \, \text{J K}^{-1} \text{mol}^{-1}} \]

5. **Calculate \( \Delta T \):**
- Calculate the denominator:
\[ 5.56 \, \text{mol} \times 75 \, \text{J K}^{-1} \text{mol}^{-1} \approx 417 \, \text{J K}^{-1} \]
- Now calculate \( \Delta T \):
\[ \Delta T = \frac{1000 \, \text{J}}{417 \, \text{J K}^{-1}} \approx 2.40 \, \text{K} \]

### Final Answer:
The increase in temperature of the water is approximately \( \Delta T \approx 2.4 \, \text{K} \).

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