When m gram of steam at `100^@ C` is mixed with 200 gm of ice at `0^@`C. it results in water at `40^@ C`. Find the value of m in gram . (given : Latent heat of fusion (`L_f`) = 80 cal/gm, Latent heat of vaporisation (`L_v`) = 540 cal/gm., specific heat of water (`C_w`)= 1 cal/gm/C)

Answer

To solve the problem of mixing steam and ice to find the mass of steam (m) that results in water at 40°C, we will use the principles of heat transfer and the conservation of energy. Here's a step-by-step solution:

### Step 1: Identify the heat released by steam
When steam at 100°C condenses into water at 100°C, it releases heat. The heat released (Q1) can be calculated using the formula:
\[ Q_1 = m \times L_v \]
where \( L_v \) is the latent heat of vaporization (540 cal/g).

### Step 2: Calculate the heat released by the steam when it cools to 40°C
After the steam condenses into water at 100°C, it further cools down to 40°C. The heat released during this cooling (Q2) can be calculated using the formula:
\[ Q_2 = m \times C_w \times \Delta T \]
where:
- \( C_w \) is the specific heat of water (1 cal/g°C),
- \( \Delta T \) is the change in temperature (100°C - 40°C = 60°C).

Thus,
\[ Q_2 = m \times 1 \times 60 = 60m \]

### Step 3: Identify the heat absorbed by ice
The ice at 0°C absorbs heat to melt into water at 0°C. The heat absorbed (Q1') can be calculated using the formula:
\[ Q_1' = m_{ice} \times L_f \]
where \( m_{ice} \) is the mass of ice (200 g) and \( L_f \) is the latent heat of fusion (80 cal/g).
So,
\[ Q_1' = 200 \times 80 = 16000 \text{ cal} \]

### Step 4: Calculate the heat absorbed by the melted ice to reach 40°C
After melting, the water at 0°C absorbs heat to reach 40°C. The heat absorbed (Q2') can be calculated as:
\[ Q_2' = m_{water} \times C_w \times \Delta T \]
where:
- \( m_{water} = 200 \) g,
- \( \Delta T = 40°C - 0°C = 40°C \).

Thus,
\[ Q_2' = 200 \times 1 \times 40 = 8000 \text{ cal} \]

### Step 5: Set up the energy balance equation
According to the conservation of energy, the heat lost by the steam must equal the heat gained by the ice:
\[ Q_1 + Q_2 = Q_1' + Q_2' \]
Substituting the values we calculated:
\[ m \times 540 + 60m = 16000 + 8000 \]
\[ 600m = 24000 \]

### Step 6: Solve for m
Now, we can solve for m:
\[ m = \frac{24000}{600} = 40 \text{ g} \]

Thus, the mass of steam (m) is **40 grams**.