A 5 g piece of ice at `-20^@C` is put into 10 g of water at `30^@C`. Assuming that heat is exchanged only between the ice and water. The final temperature of the mixture.

Options

A. `10^@C`
B. `20^@C`
C. `0^@C`
D. `15^@C`

Explanation

To solve the problem of finding the final temperature of a mixture of ice and water, we will follow these steps:

### Step 1: Calculate the heat required to raise the temperature of the ice from -20°C to 0°C
We use the formula:
\[ Q_1 = m \cdot s \cdot \Delta T \]
Where:
- \( m \) = mass of ice = 5 g = 0.005 kg (since we need to convert to kg)
- \( s \) = specific heat of ice = 2.09 J/g°C (or 0.5 cal/g°C)
- \( \Delta T \) = change in temperature = 0°C - (-20°C) = 20°C

Calculating \( Q_1 \):
\[ Q_1 = 5 \, \text{g} \times 2.09 \, \text{J/g°C} \times 20 \, \text{°C} = 209 \, \text{J} \]

### Step 2: Calculate the heat required to melt the ice at 0°C
We use the formula:
\[ Q_2 = m \cdot L_f \]
Where:
- \( L_f \) = latent heat of fusion for ice = 334 J/g

Calculating \( Q_2 \):
\[ Q_2 = 5 \, \text{g} \times 334 \, \text{J/g} = 1670 \, \text{J} \]

### Step 3: Calculate the total heat required to convert the ice at -20°C to water at 0°C
\[ Q_{\text{total}} = Q_1 + Q_2 = 209 \, \text{J} + 1670 \, \text{J} = 1879 \, \text{J} \]

### Step 4: Calculate the heat released by the water as it cools from 30°C to 0°C
We use the formula:
\[ Q_3 = m \cdot s \cdot \Delta T \]
Where:
- \( m \) = mass of water = 10 g = 0.01 kg
- \( s \) = specific heat of water = 4.18 J/g°C
- \( \Delta T \) = change in temperature = 30°C - 0°C = 30°C

Calculating \( Q_3 \):
\[ Q_3 = 10 \, \text{g} \times 4.18 \, \text{J/g°C} \times 30 \, \text{°C} = 1254 \, \text{J} \]

### Step 5: Compare the heat required and the heat released
- Heat required to convert ice to water: \( Q_{\text{total}} = 1879 \, \text{J} \)
- Heat released by water: \( Q_3 = 1254 \, \text{J} \)

Since \( Q_{\text{total}} > Q_3 \), not all the ice will melt.

### Step 6: Determine the final state
The water can only provide 1254 J of heat. This will raise the temperature of the ice to 0°C and partially melt it. The remaining heat after raising the ice to 0°C and melting some of it will be:
\[ Q_{\text{remaining}} = Q_3 - Q_1 = 1254 \, \text{J} - 209 \, \text{J} = 1045 \, \text{J} \]

This remaining heat will be used to melt the ice. The amount of ice that can be melted with this remaining heat is:
\[ \text{mass melted} = \frac{Q_{\text{remaining}}}{L_f} = \frac{1045 \, \text{J}}{334 \, \text{J/g}} \approx 3.13 \, \text{g} \]

### Conclusion
After the heat exchange, some ice remains (5 g - 3.13 g = 1.87 g). The final temperature of the mixture will be 0°C, as the system reaches thermal equilibrium at the melting point of ice.

### Final Answer
The final temperature of the mixture is **0°C**.

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