`10` grams of stream at `100^@ C` is mixed with `50 gm` of ice at `0^@ C` then final temperature is

Options

A. `20^@ C`
B. `50^@ C`
C. `40^@ C`
D. `100^@ C`

Explanation

To solve the problem of mixing 10 grams of steam at 100°C with 50 grams of ice at 0°C, we will apply the principle of conservation of energy. The heat lost by the steam will be equal to the heat gained by the ice until thermal equilibrium is reached.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of steam, \( m_s = 10 \, \text{g} \)
- Initial temperature of steam, \( T_s = 100^\circ C \)
- Mass of ice, \( m_i = 50 \, \text{g} \)
- Initial temperature of ice, \( T_i = 0^\circ C \)
- Latent heat of vaporization of steam, \( L_v = 540 \, \text{cal/g} \)
- Latent heat of fusion of ice, \( L_f = 80 \, \text{cal/g} \)
- Specific heat of water, \( c_w = 1 \, \text{cal/g°C} \)

2. **Calculate the Heat Lost by Steam:**
The steam will first condense into water and then cool down to the final temperature \( T_f \).
- Heat lost during condensation:
\[ Q_{\text{condensation}} = m_s \times L_v = 10 \, \text{g} \times 540 \, \text{cal/g} = 5400 \, \text{cal} \]
- Heat lost while cooling from \( 100^\circ C \) to \( T_f \):
\[ Q_{\text{cooling}} = m_s \times c_w \times (T_s - T_f) = 10 \, \text{g} \times 1 \, \text{cal/g°C} \times (100 - T_f) = 1000 - 10T_f \, \text{cal} \]
- Total heat lost by steam:
\[ Q_{\text{steam}} = Q_{\text{condensation}} + Q_{\text{cooling}} = 5400 + (1000 - 10T_f) = 6400 - 10T_f \, \text{cal} \]

3. **Calculate the Heat Gained by Ice:**
The ice will first melt into water and then warm up to the final temperature \( T_f \).
- Heat gained during melting:
\[ Q_{\text{melting}} = m_i \times L_f = 50 \, \text{g} \times 80 \, \text{cal/g} = 4000 \, \text{cal} \]
- Heat gained while warming from \( 0^\circ C \) to \( T_f \):
\[ Q_{\text{warming}} = m_i \times c_w \times (T_f - T_i) = 50 \, \text{g} \times 1 \, \text{cal/g°C} \times (T_f - 0) = 50T_f \, \text{cal} \]
- Total heat gained by ice:
\[ Q_{\text{ice}} = Q_{\text{melting}} + Q_{\text{warming}} = 4000 + 50T_f \, \text{cal} \]

4. **Set Up the Equation for Heat Transfer:**
According to the principle of conservation of energy:
\[ Q_{\text{lost by steam}} = Q_{\text{gained by ice}} \]
Therefore,
\[ 6400 - 10T_f = 4000 + 50T_f \]

5. **Solve for \( T_f \):**
Rearranging the equation:
\[ 6400 - 4000 = 50T_f + 10T_f \]
\[ 2400 = 60T_f \]
\[ T_f = \frac{2400}{60} = 40^\circ C \]

### Final Answer:
The final temperature of the mixture is \( 40^\circ C \).