Solution:

Here’s a short solution:

Given

- Mass of water = 54 gm
- Initial temperature of water = 30°C, Final temperature = 90°C
- Latent heat of steam = 530 cal/gm
- Specific heat of water = 1 cal/gm°C

Step 1: Heat gained by water to reach 90°C:

\[
Q_{\text{gained}} = m_{\text{water}} \times c_{\text{water}} \times \Delta T = 54 \times 1 \times (90 - 30) = 3240 \, \text{cal}
\]

Step 2: Heat lost by steam:

\[
Q_{\text{lost}} = m_s \times (530 + 10) = 540 m_s
\]

Step 3: Equating heat gained and heat lost:

\[
3240 = 540 m_s \quad \Rightarrow \quad m_s = \frac{3240}{540} = 6 \, \text{gm}
\]

Step 4: Total mass of the mixture:

\[
m_{\text{mixture}} = 54 \, \text{gm} + 6 \, \text{gm} = 60 \, \text{gm}
\]

Final Answer:
The mass of the mixture is 60 grams.