`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. `(185)/(27) gm`
B. `(135)/(17) gm`
C. `(85)/(32) gm`
D. `(113)/(17) gm`

Explanation

To solve the problem, we need to calculate the minimum mass of steam (m) required to convert 20 g of ice at -10°C into water at 100°C. We will use the principle of conservation of energy, where the heat lost by the steam will equal the heat gained by the ice.

### Step 1: Calculate the heat required to raise the temperature of ice from -10°C to 0°C.
The formula for heat (Q) is given by:
\[ Q = m \cdot s \cdot \Delta T \]
where:
- \( m \) = mass of ice = 20 g
- \( s \) = specific heat capacity of ice = 0.5 cal/g°C
- \( \Delta T \) = change in temperature = 0 - (-10) = 10°C

Calculating the heat required:
\[ Q_1 = 20 \, \text{g} \cdot 0.5 \, \text{cal/g°C} \cdot 10 \, \text{°C} = 100 \, \text{cal} \]

### Step 2: Calculate the heat required to melt the ice at 0°C.
The heat required to melt the ice is given by:
\[ Q = m \cdot L \]
where:
- \( L \) = latent heat of fusion = 80 cal/g

Calculating the heat required for melting:
\[ Q_2 = 20 \, \text{g} \cdot 80 \, \text{cal/g} = 1600 \, \text{cal} \]

### Step 3: Calculate the total heat required to convert ice at -10°C to water at 0°C.
The total heat required (Q_total) is the sum of the heat required to raise the temperature of the ice and the heat required to melt it:
\[ Q_{\text{total}} = Q_1 + Q_2 = 100 \, \text{cal} + 1600 \, \text{cal} = 1700 \, \text{cal} \]

### Step 4: Calculate the heat released by steam when it condenses to water at 100°C.
The heat released by the steam when it condenses is given by:
\[ Q = m \cdot L_v \]
where:
- \( L_v \) = latent heat of vaporization = 540 cal/g

Thus, the heat released by m grams of steam is:
\[ Q_{\text{steam}} = m \cdot 540 \, \text{cal/g} \]

### Step 5: Set the heat gained by ice equal to the heat lost by steam.
Using the conservation of energy:
\[ Q_{\text{steam}} = Q_{\text{total}} \]
\[ m \cdot 540 = 1700 \]

### Step 6: Solve for m.
Rearranging the equation gives:
\[ m = \frac{1700}{540} \]
\[ m = \frac{170}{54} \]
\[ m = \frac{85}{27} \approx 62.96 \, \text{g} \]

### Step 7: Conclusion
The minimum value of m so that finally all ice and steam converts into water is approximately 63 g.