10 g of ice at `0^@C` is mixed with 100 g of water at `50^@C`. What is the resultant temperature of mixture

Options

A. `31.2^@C`
B. `32.8^@C`
C. `36.8^@C`
D. `38.2^@C`

Explanation

To find the resultant temperature when 10 g of ice at 0°C is mixed with 100 g of water at 50°C, we can use the principle of conservation of energy, which states that the heat lost by the warmer water will be equal to the heat gained by the ice as it melts and then warms up.

### Step-by-Step Solution:

1. **Identify the known values:**
- Mass of ice, \( m_{ice} = 10 \, g \)
- Mass of water, \( m_{water} = 100 \, g \)
- Initial temperature of ice, \( T_{ice} = 0 \, °C \)
- Initial temperature of water, \( T_{water} = 50 \, °C \)
- Latent heat of fusion of ice, \( L_f = 80 \, cal/g \)
- Specific heat of water, \( c_w = 1 \, cal/g°C \)

2. **Set up the heat balance equation:**
- Heat lost by water = Heat gained by ice
- The heat lost by the water as it cools down from 50°C to the final temperature \( \theta \) is given by:
\[ Q_{lost} = m_{water} \cdot c_w \cdot (T_{water} - \theta) = 100 \cdot 1 \cdot (50 - \theta) \]
- The heat gained by the ice includes two parts: the heat required to melt the ice and the heat required to raise the temperature of the melted ice (water) from 0°C to \( \theta \):
\[ Q_{gained} = m_{ice} \cdot L_f + m_{ice} \cdot c_w \cdot (\theta - T_{ice}) = 10 \cdot 80 + 10 \cdot 1 \cdot (\theta - 0) \]
\[ Q_{gained} = 800 + 10\theta \]

3. **Equate the heat lost and gained:**
\[ 100 \cdot (50 - \theta) = 800 + 10\theta \]
\[ 5000 - 100\theta = 800 + 10\theta \]

4. **Rearranging the equation:**
\[ 5000 - 800 = 100\theta + 10\theta \]
\[ 4200 = 110\theta \]

5. **Solve for \( \theta \):**
\[ \theta = \frac{4200}{110} \approx 38.18 \, °C \]

6. **Final Result:**
The resultant temperature of the mixture is approximately \( 38.2 \, °C \).