10 gm of ice at `– 20^(@)C` is added to 10 gm of water at `50^(@)C`. Specific heat of water `= 1 cal//g–.^(@)C`, specific heat of ice = `0.5 cal//gm-.^(@)C`. Latent heat of ice = 80 cal/gm. Then resulting temperature is -

Options

A. `-20^(@)C`
B. `15^(@)C`
C. `0^(@)C`
D. `50^(@)C`

Answer

To solve the problem, we need to analyze the heat exchanges between the ice and the water. Let's break it down step by step.

### Step 1: Calculate the heat required to raise the temperature of ice from -20°C to 0°C.

**Formula:**
\[ Q_1 = m \cdot c \cdot \Delta T \]
Where:
- \( m = 10 \, \text{g} \) (mass of ice)
- \( c = 0.5 \, \text{cal/g°C} \) (specific heat of ice)
- \( \Delta T = 0 - (-20) = 20 \, \text{°C} \) (temperature change)

**Calculation:**
\[ Q_1 = 10 \, \text{g} \cdot 0.5 \, \text{cal/g°C} \cdot 20 \, \text{°C} = 100 \, \text{cal} \]

### Step 2: Calculate the heat required to melt the ice at 0°C.

**Formula:**
\[ Q_2 = m \cdot L \]
Where:
- \( L = 80 \, \text{cal/g} \) (latent heat of fusion of ice)

**Calculation:**
\[ Q_2 = 10 \, \text{g} \cdot 80 \, \text{cal/g} = 800 \, \text{cal} \]

### Step 3: Calculate the total heat required for the ice to become water at 0°C.

**Total Heat:**
\[ Q_{\text{total}} = Q_1 + Q_2 = 100 \, \text{cal} + 800 \, \text{cal} = 900 \, \text{cal} \]

### Step 4: Calculate the heat released by the water when it cools from 50°C to 0°C.

**Formula:**
\[ Q_3 = m \cdot c \cdot \Delta T \]
Where:
- \( m = 10 \, \text{g} \) (mass of water)
- \( c = 1 \, \text{cal/g°C} \) (specific heat of water)
- \( \Delta T = 50 - 0 = 50 \, \text{°C} \)

**Calculation:**
\[ Q_3 = 10 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 50 \, \text{°C} = 500 \, \text{cal} \]

### Step 5: Compare the heat required and the heat released.

- Heat required to convert ice to water at 0°C: \( 900 \, \text{cal} \)
- Heat released by water: \( 500 \, \text{cal} \)

Since the heat released by the water (500 cal) is less than the heat required to convert all the ice (900 cal), not all the ice will melt.

### Step 6: Calculate the remaining heat after the water cools down.

**Remaining Heat:**
\[ Q_{\text{remaining}} = Q_3 - Q_{\text{total}} = 500 \, \text{cal} - 900 \, \text{cal} = -400 \, \text{cal} \]

This indicates that 400 cal is still needed to melt the remaining ice.

### Step 7: Calculate how much ice can be melted with the available heat.

**Heat required to melt ice:**
\[ Q = m \cdot L \]
Where \( Q = 400 \, \text{cal} \)

**Calculation:**
\[ m = \frac{Q}{L} = \frac{400 \, \text{cal}}{80 \, \text{cal/g}} = 5 \, \text{g} \]

### Step 8: Determine the final state of the system.

- Initial ice: 10 g
- Ice melted: 5 g
- Remaining ice: \( 10 \, \text{g} - 5 \, \text{g} = 5 \, \text{g} \)

### Final Result:
The final state of the system consists of:
- 5 g of ice at 0°C
- 5 g of water at 0°C

Thus, the resulting temperature of the system is **0°C**.

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