Solution:

To find the final temperature when 5 g of ice at 0°C is mixed with 5 g of steam at 100°C, we need to consider the energy exchange between the ice and the steam.

We proceed step by step:

1. Heat required to melt the ice into water at 0°C:

\[
Q_1 = m_{\text{ice}} \times L_f = 5 \, \text{g} \times 80 \, \text{cal/g} = 400 \, \text{cal}
\]
This is the heat required to convert 5 g of ice at 0°C into 5 g of water at 0°C.

2. Heat released by steam as it condenses into water at 100°C:

\[
Q_2 = m_{\text{steam}} \times L_v = 5 \, \text{g} \times 540 \, \text{cal/g} = 2700 \, \text{cal}
\]
This is the heat released when 5 g of steam condenses into water at 100°C.

3. Heat required to raise the temperature of 5 g of water from 0°C to 100°C:

\[
Q_3 = m_{\text{water}} \times c \times \Delta T = 5 \, \text{g} \times 1 \, \text{cal/g°C} \times (100 - 0) = 5 \times 100 = 500 \, \text{cal}
\]

 Total heat available from the steam:

- The steam releases 2700 cal by condensing.
- The ice requires 400 cal to melt, and then 500 cal to be heated from 0°C to 100°C, totaling 900 cal.

Since the heat available from the steam (2700 cal) is more than the 900 cal required to melt the ice and raise its temperature to 100°C, the final temperature will be 100°C.

In conclusion, all the ice melts and the final temperature of the mixture is 100°C.