Heat energy of `184kJ` is given to ice of mass `600g` at `-12^(@)C`.Specific heat of ice is `2222.3Jkg^(-1)"^@C^(-1)` and latent heat of ice in `336kJ/kg^(-1)`
A.Final temperature of system will be `0^(@)C`.
B.Final temperature of the system will be greater than `0^(@)C`.
C.The final system will have a mixture of ice and water in the ratio of `5:1`
D.the final system will have a mixture of ice and water in the ratio of `5:1`
E.The final system will have water only.
Choose the correct answer from the options given below:

Options

A. A` and `D` only `,
B. `B` and `D` only `
C. `A` and `E` only `
D. `A` and `C` only

Answer

To solve the problem, we need to analyze the heat transfer involved in warming the ice from -12°C to 0°C and then melting some of the ice into water. Let's break this down step by step.

### Step 1: Calculate the heat required to raise the temperature of ice from -12°C to 0°C

The formula to calculate the heat (Q) required to change the temperature of a substance is:
\[ Q = m \cdot c \cdot \Delta T \]
where:
- \( m \) = mass of the substance (in kg)
- \( c \) = specific heat capacity (in J/kg°C)
- \( \Delta T \) = change in temperature (in °C)

Given:
- Mass of ice, \( m = 600 \, \text{g} = 0.6 \, \text{kg} \)
- Specific heat of ice, \( c = 2222.3 \, \text{J/kg°C} \)
- Change in temperature, \( \Delta T = 0 - (-12) = 12 \, \text{°C} \)

Now, substituting the values:
\[ Q = 0.6 \, \text{kg} \cdot 2222.3 \, \text{J/kg°C} \cdot 12 \, \text{°C} \]
\[ Q = 0.6 \cdot 2222.3 \cdot 12 \]
\[ Q = 16000 \, \text{J} = 16 \, \text{kJ} \]

### Step 2: Calculate the remaining heat after warming the ice to 0°C

The total heat energy supplied is:
\[ Q_{\text{total}} = 184 \, \text{kJ} \]

After warming the ice to 0°C, the remaining heat energy is:
\[ Q_{\text{remaining}} = Q_{\text{total}} - Q_{\text{warmed}} \]
\[ Q_{\text{remaining}} = 184 \, \text{kJ} - 16 \, \text{kJ} = 168 \, \text{kJ} \]

### Step 3: Calculate the amount of ice that can be melted with the remaining heat

The latent heat of fusion of ice is given as:
\[ L = 336 \, \text{kJ/kg} \]

The amount of ice that can be melted is given by:
\[ m_{\text{melted}} = \frac{Q_{\text{remaining}}}{L} \]
Substituting the values:
\[ m_{\text{melted}} = \frac{168 \, \text{kJ}}{336 \, \text{kJ/kg}} = 0.5 \, \text{kg} \]

### Step 4: Determine the final amounts of ice and water

Initially, we had 600 g of ice. After melting 0.5 kg (or 500 g) of ice:
- Mass of ice remaining = \( 600 \, \text{g} - 500 \, \text{g} = 100 \, \text{g} \)
- Mass of water formed = 500 g

### Step 5: Calculate the ratio of ice to water

The ratio of ice to water is:
\[ \text{Ratio} = \frac{\text{mass of ice}}{\text{mass of water}} = \frac{100 \, \text{g}}{500 \, \text{g}} = \frac{1}{5} \]

### Conclusion

From the calculations, we find:
- The final temperature of the system is 0°C (as the ice has just melted).
- The final system has a mixture of ice and water in the ratio of 1:5.

### Final Answer

The correct options are:
- A. Final temperature of the system will be \( 0°C \).
- C. The final system will have a mixture of ice and water in the ratio of \( 1:5 \).