`20 gm` ice at `-10^(@)C` is mixed with `m gm` steam at `100^(@)C`. The minimum value of `m` so that finally all ice and steam converts into water is: `("Use " s_("ice") = 0.5 "cal gm"^(@)C, S_("water") = 1 cal//gm^(@)C, L`) (melting) `= 80 cal// gm` and `L ("vaporization") = 540 cal//gm`)

Options

A. `(85)/(32)gm`
B. `(85)/(64)gm`
C. `(32)/(85)gm`
D. `(64)/(85)gm`

Explanation

To solve the problem, we need to calculate the heat absorbed by the ice and the heat released by the steam, and then set them equal to find the minimum mass of steam (m) required for all the ice to melt and the steam to condense into water.

### Step-by-Step Solution:

1. **Calculate the heat absorbed by the ice**:
- The ice is initially at `-10°C` and needs to be heated to `0°C` before it melts.
- The specific heat of ice (s_ice) = `0.5 cal/g°C`.
- The mass of ice (m_ice) = `20 g`.
- The temperature change (ΔT) = `0 - (-10) = 10°C`.

The heat absorbed by the ice to reach `0°C`:
\[ Q_1 = m_{\text{ice}} \cdot s_{\text{ice}} \cdot \Delta T = 20 \, \text{g} \cdot 0.5 \, \text{cal/g°C} \cdot 10 \, \text{°C} = 100 \, \text{cal} \]

- Next, the ice at `0°C` melts into water at `0°C`.
- The latent heat of fusion (L_f) = `80 cal/g`.

The heat absorbed during melting:
\[ Q_2 = m_{\text{ice}} \cdot L_f = 20 \, \text{g} \cdot 80 \, \text{cal/g} = 1600 \, \text{cal} \]

- Total heat absorbed by the ice:
\[ Q_{\text{ice}} = Q_1 + Q_2 = 100 \, \text{cal} + 1600 \, \text{cal} = 1700 \, \text{cal} \]

2. **Calculate the heat released by the steam**:
- The steam is initially at `100°C` and needs to condense to water at `100°C`.
- The latent heat of vaporization (L_v) = `540 cal/g`.

The heat released during condensation:
\[ Q_3 = m \cdot L_v = m \cdot 540 \, \text{cal/g} \]

- After condensation, the water at `100°C` cools down to `0°C`.
- The specific heat of water (s_water) = `1 cal/g°C`.
- The temperature change (ΔT) = `100°C - 0°C = 100°C`.

The heat released during cooling:
\[ Q_4 = m \cdot s_{\text{water}} \cdot \Delta T = m \cdot 1 \, \text{cal/g°C} \cdot 100 \, \text{°C} = 100m \, \text{cal} \]

- Total heat released by the steam:
\[ Q_{\text{steam}} = Q_3 + Q_4 = m \cdot 540 \, \text{cal/g} + 100m \, \text{cal} = 640m \, \text{cal} \]

3. **Set the heat absorbed equal to the heat released**:
\[ Q_{\text{ice}} = Q_{\text{steam}} \]
\[ 1700 \, \text{cal} = 640m \, \text{cal} \]

Solving for m:
\[ m = \frac{1700}{640} = \frac{85}{32} \, \text{g} \]

### Final Answer:
The minimum value of \( m \) so that finally all ice and steam converts into water is:
\[ m = \frac{85}{32} \, \text{g} \]