19 g of water at `30^@C` and 5 g of ice at `-20^@C` are mixed together in a calorimeter. What is the final temperature of the mixture? Given specific heat of ice `=0.5calg^(-1) (.^(@)C)^(-1)` and latent heat of fusion of ice `=80calg^(-1)`

Options

A. `0^(@)C`
B. `-5^(@)C`
C. `5^(@)C`
D. `10^(@)C`

Explanation

To find the final temperature of the mixture when 19 g of water at 30°C is mixed with 5 g of ice at -20°C, we will follow these steps:

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

The formula to calculate the heat (q) is:
\[ q = m \cdot c \cdot \Delta T \]
Where:
- \( m \) = mass of ice = 5 g
- \( c \) = specific heat of ice = 0.5 cal/g°C
- \( \Delta T \) = change in temperature = \( 0 - (-20) = 20°C \)

Calculating:
\[ q_1 = 5 \, \text{g} \cdot 0.5 \, \text{cal/g°C} \cdot 20 \, \text{°C} = 50 \, \text{cal} \]

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

The formula for the heat required to melt ice is:
\[ q = m \cdot L_f \]
Where:
- \( m \) = mass of ice = 5 g
- \( L_f \) = latent heat of fusion of ice = 80 cal/g

Calculating:
\[ q_2 = 5 \, \text{g} \cdot 80 \, \text{cal/g} = 400 \, \text{cal} \]

### Step 3: Calculate the heat lost by the water as it cools from 30°C to 0°C.

Using the same formula:
\[ q = m \cdot c \cdot \Delta T \]
Where:
- \( m \) = mass of water = 19 g
- \( c \) = specific heat of water = 1 cal/g°C
- \( \Delta T \) = change in temperature = \( 30 - 0 = 30°C \)

Calculating:
\[ q_3 = 19 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 30 \, \text{°C} = 570 \, \text{cal} \]

### Step 4: Determine the total heat required to convert ice at -20°C to water at 0°C.

Total heat required for the ice to become water at 0°C:
\[ q_{\text{total}} = q_1 + q_2 = 50 \, \text{cal} + 400 \, \text{cal} = 450 \, \text{cal} \]

### Step 5: Determine the heat available from the water.

The water can lose a maximum of 570 cal as it cools to 0°C. Since the heat required to convert the ice to water (450 cal) is less than the heat available from the water (570 cal), the ice will completely melt, and there will still be some heat left.

### Step 6: Calculate the remaining heat after the ice has melted.

Remaining heat after melting the ice:
\[ \text{Remaining heat} = 570 \, \text{cal} - 450 \, \text{cal} = 120 \, \text{cal} \]

### Step 7: Calculate the final temperature of the mixture.

Now, this remaining heat will be used to raise the temperature of the resulting water (from melted ice) from 0°C. The total mass of the water after melting the ice is:
\[ m_{\text{total}} = 19 \, \text{g} + 5 \, \text{g} = 24 \, \text{g} \]

Using the formula:
\[ q = m \cdot c \cdot \Delta T \]
Where:
- \( q \) = remaining heat = 120 cal
- \( m \) = total mass of water = 24 g
- \( c \) = specific heat of water = 1 cal/g°C

Let \( T \) be the final temperature:
\[ 120 \, \text{cal} = 24 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot (T - 0) \]

Solving for \( T \):
\[ T = \frac{120 \, \text{cal}}{24 \, \text{g}} = 5 \, \text{°C} \]

### Final Answer:
The final temperature of the mixture is **5°C**.

---