The temperature of 100g of water is to be raised from `24^@C` to `90^@C` by adding steam to it. Calculate the mass of the steam required for this purpose.

Explanation

To solve the problem of calculating the mass of steam required to raise the temperature of 100g of water from 24°C to 90°C, we can use the principle of calorimetry, which states that the heat lost by the steam will be equal to the heat gained by the water.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of water (m_w) = 100 g
- Initial temperature of water (T_initial) = 24°C
- Final temperature of water (T_final) = 90°C
- Latent heat of vaporization of steam (L) = 540 cal/g
- Specific heat capacity of water (s_w) = 1 cal/g°C

2. **Calculate the Change in Temperature of Water:**
\[ \Delta T' = T_{final} - T_{initial} = 90°C - 24°C = 66°C \]

3. **Calculate the Change in Temperature of the Steam:**
- The steam will condense at 100°C and then cool down to 90°C.
\[ \Delta T = 100°C - 90°C = 10°C \]

4. **Set Up the Heat Transfer Equation:**
The heat lost by the steam (Q_steam) is equal to the heat gained by the water (Q_water):
\[ Q_{steam} = Q_{water} \]
\[ m_s \cdot L + m_s \cdot s_w \cdot \Delta T = m_w \cdot s_w \cdot \Delta T' \]
where \( m_s \) is the mass of the steam.

5. **Substitute the Known Values:**
\[ m_s \cdot 540 + m_s \cdot 1 \cdot 10 = 100 \cdot 1 \cdot 66 \]
Simplifying the equation:
\[ m_s \cdot 540 + m_s \cdot 10 = 6600 \]
\[ m_s (540 + 10) = 6600 \]
\[ m_s \cdot 550 = 6600 \]

6. **Solve for the Mass of the Steam (m_s):**
\[ m_s = \frac{6600}{550} = 12 \text{ g} \]

### Final Answer:
The mass of the steam required to raise the temperature of 100g of water from 24°C to 90°C is **12 grams**.