In a container of negligible mass 30g of steam at `100^@C` is added to 200g of water that has a temperature of `40^@C` If no heat is lost to the surroundings, what is the final temperature of the system? Also find masses of water and steam in equilibrium. Take `L_v = 539 cal//g and c_(water) = 1 cal//g-^@C.`

Explanation

To solve the problem step by step, we will follow the principles of calorimetry, where the heat lost by the steam will equal the heat gained by the water.

### Step 1: Calculate the heat required to raise the temperature of water from 40°C to 100°C.

The formula to calculate the heat (Q) required to change the temperature of a substance is given by:

\[ Q = m \cdot c \cdot \Delta T \]

Where:
- \( m \) = mass of the water (200 g)
- \( c \) = specific heat capacity of water (1 cal/g°C)
- \( \Delta T \) = change in temperature (100°C - 40°C = 60°C)

Substituting the values:

\[ Q = 200 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 60 \, \text{°C} = 12000 \, \text{cal} \]

### Step 2: Calculate the mass of steam that condenses to provide this heat.

The heat released by the condensation of steam can be calculated using the formula:

\[ Q = m_0 \cdot L_v \]

Where:
- \( m_0 \) = mass of steam condensed
- \( L_v \) = latent heat of vaporization (539 cal/g)

Rearranging for \( m_0 \):

\[ m_0 = \frac{Q}{L_v} = \frac{12000 \, \text{cal}}{539 \, \text{cal/g}} \approx 22.24 \, \text{g} \]

### Step 3: Determine if all the steam has condensed.

Initially, we have 30 g of steam. Since the mass of steam condensed (22.24 g) is less than the initial mass of steam (30 g), it indicates that not all steam has condensed.

### Step 4: Determine the final temperature of the system.

Since some steam has condensed to heat the water to 100°C, and the remaining steam is still at 100°C, the final temperature of the system will be:

\[ T_f = 100 \, \text{°C} \]

### Step 5: Calculate the new mass of water and the remaining mass of steam.

The new mass of water after condensation is:

\[ M_w' = M_w + m_0 = 200 \, \text{g} + 22.24 \, \text{g} = 222.24 \, \text{g} \]

The remaining mass of steam is:

\[ m_s = 30 \, \text{g} - m_0 = 30 \, \text{g} - 22.24 \, \text{g} = 7.76 \, \text{g} \]

### Final Answers:
- Final Temperature \( T_f = 100 \, \text{°C} \)
- New mass of water \( M_w' = 222.24 \, \text{g} \)
- Remaining mass of steam \( m_s = 7.76 \, \text{g} \)

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