The average molar heat capacities of ice and water are respectively `37.8 (J)/(mol)` and `75.6 (J)/(mol)` and the enthalpy of fusion of ice is `6.012 (kJ)/(mol)`.The amount of heat required to change 10 g of ice at `-10^(@)C` to water at `10^(@)C` would be

Options

A. 2376 J
B. 4752 J
C. 3970 J
D. 1128 J

Explanation

To find the total amount of heat required to change 10 g of ice at -10°C to water at 10°C, we will break the process down into three steps:

1. Heating the ice from -10°C to 0°C.
2. Melting the ice at 0°C to water.
3. Heating the water from 0°C to 10°C.

### Step 1: Heating the Ice from -10°C to 0°C

**Formula:**
\[ q_1 = n \cdot C_{p, \text{ice}} \cdot \Delta T \]

Where:
- \( n \) = number of moles of ice
- \( C_{p, \text{ice}} = 37.8 \, \text{J/(mol·°C)} \)
- \( \Delta T = 0 - (-10) = 10 \, \text{°C} \)

**Calculating the number of moles:**
The molar mass of water (H₂O) is 18 g/mol. Thus, for 10 g of ice:
\[ n = \frac{10 \, \text{g}}{18 \, \text{g/mol}} = \frac{10}{18} \, \text{mol} \approx 0.5556 \, \text{mol} \]

**Calculating \( q_1 \):**
\[ q_1 = 0.5556 \, \text{mol} \cdot 37.8 \, \text{J/(mol·°C)} \cdot 10 \, \text{°C} \]
\[ q_1 \approx 210 \, \text{J} \]

### Step 2: Melting the Ice at 0°C

**Formula:**
\[ q_2 = n \cdot \Delta H_{fusion} \]

Where:
- \( \Delta H_{fusion} = 6.012 \, \text{kJ/mol} = 6012 \, \text{J/mol} \)

**Calculating \( q_2 \):**
\[ q_2 = 0.5556 \, \text{mol} \cdot 6012 \, \text{J/mol} \]
\[ q_2 \approx 3340 \, \text{J} \]

### Step 3: Heating the Water from 0°C to 10°C

**Formula:**
\[ q_3 = n \cdot C_{p, \text{water}} \cdot \Delta T \]

Where:
- \( C_{p, \text{water}} = 75.6 \, \text{J/(mol·°C)} \)
- \( \Delta T = 10 - 0 = 10 \, \text{°C} \)

**Calculating \( q_3 \):**
\[ q_3 = 0.5556 \, \text{mol} \cdot 75.6 \, \text{J/(mol·°C)} \cdot 10 \, \text{°C} \]
\[ q_3 \approx 420 \, \text{J} \]

### Total Heat Required

Now, we sum all the heat quantities calculated:
\[ q_{total} = q_1 + q_2 + q_3 \]
\[ q_{total} = 210 \, \text{J} + 3340 \, \text{J} + 420 \, \text{J} \]
\[ q_{total} = 3970 \, \text{J} \]

### Final Answer
The total amount of heat required to change 10 g of ice at -10°C to water at 10°C is approximately **3970 J**.

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