To solve the problem, we need to find the increase in temperature (ΔT) of 1 kg of water when 10 kJ of heat is supplied, using the molar heat capacity at constant pressure (C_p).
### Step-by-Step Solution:
1. **Identify Given Values:**
- Molar heat capacity of water, \( C_p = 75 \, \text{J K}^{-1} \text{mol}^{-1} \)
- Heat supplied, \( Q = 10 \, \text{kJ} = 10,000 \, \text{J} \)
- Mass of water, \( m = 1 \, \text{kg} \)
2. **Calculate the Number of Moles of Water:**
- The molecular mass of water (H₂O) is approximately \( 18 \, \text{g/mol} \).
- Convert the mass of water from kg to g: \( 1 \, \text{kg} = 1000 \, \text{g} \).
- Calculate the number of moles (\( n \)):
\[
n = \frac{\text{mass}}{\text{molecular mass}} = \frac{1000 \, \text{g}}{18 \, \text{g/mol}} \approx 55.56 \, \text{mol}
\]
3. **Use the Heat Transfer Formula:**
- The formula relating heat, molar heat capacity, number of moles, and temperature change is:
\[
Q = n \cdot C_p \cdot \Delta T
\]
- Rearranging for \( \Delta T \):
\[
\Delta T = \frac{Q}{n \cdot C_p}
\]
4. **Substitute the Known Values:**
- Substitute \( Q = 10,000 \, \text{J} \), \( n \approx 55.56 \, \text{mol} \), and \( C_p = 75 \, \text{J K}^{-1} \text{mol}^{-1} \):
\[
\Delta T = \frac{10,000 \, \text{J}}{55.56 \, \text{mol} \cdot 75 \, \text{J K}^{-1} \text{mol}^{-1}}
\]
5. **Calculate \( \Delta T \):**
- Calculate the denominator:
\[
55.56 \cdot 75 \approx 4167 \, \text{J K}^{-1}
\]
- Now calculate \( \Delta T \):
\[
\Delta T = \frac{10,000}{4167} \approx 2.4 \, \text{K}
\]
### Final Answer:
The increase in temperature of the water is approximately \( \Delta T \approx 2.4 \, \text{K} \).
---
