The molar heat capacity of water at constant pressure P is `75 J K ^(-1) mol ^(-1)`. When `1.0 kJ` of heat is supplied to 1000 g of water , which is free to expand, the increase in temperature of water is

Options

A. `1.2 K`
B. `2.4 K`
C. `4.8 K`
D. `0.24 K`

Answer

To solve the problem, we need to find the increase in temperature of water when a certain amount of heat is supplied. Here's a step-by-step solution:

### Step 1: Understand the Given Data
- Molar heat capacity of water at constant pressure (C) = 75 J K^(-1) mol^(-1)
- Heat supplied (Q) = 1.0 kJ = 1000 J
- Mass of water (m) = 1000 g

### Step 2: Calculate the Number of Moles of Water
To find the number of moles (n) of water, we use the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} \]

The molar mass of water (H₂O) is:
\[ 2 \times 1 + 16 = 18 \text{ g/mol} \]

Now substituting the values:
\[ n = \frac{1000 \text{ g}}{18 \text{ g/mol}} \approx 55.56 \text{ mol} \]

### Step 3: Use the Formula for Temperature Change
The formula relating heat (Q), number of moles (n), molar heat capacity (C), and change in temperature (ΔT) is:
\[ Q = n \cdot C \cdot \Delta T \]

Rearranging this formula to solve for ΔT gives:
\[ \Delta T = \frac{Q}{n \cdot C} \]

### Step 4: Substitute the Values into the Formula
Now we substitute the known values into the equation:
\[ \Delta T = \frac{1000 \text{ J}}{55.56 \text{ mol} \cdot 75 \text{ J K}^{-1} \text{ mol}^{-1}} \]

### Step 5: Calculate ΔT
Calculating the denominator:
\[ 55.56 \text{ mol} \cdot 75 \text{ J K}^{-1} \text{ mol}^{-1} = 4167 \text{ J K}^{-1} \]

Now substituting this back into the equation for ΔT:
\[ \Delta T = \frac{1000 \text{ J}}{4167 \text{ J K}^{-1}} \approx 0.24 \text{ K} \]

### Conclusion
The increase in temperature of the water is approximately **0.24 K**.

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