Solution:

To calculate the heat required to convert 5 g of ice at 0°C to steam at 100°C, we need to consider the following steps:

1. Heat to melt ice (latent heat of fusion):
\[
Q_1 = m \times L_f = 5 \, \text{g} \times 80 \, \text{cal/g} = 400 \, \text{cal}
\]

2. Heat to raise temperature of water from 0°C to 100°C:
\[
Q_2 = m \times c \times \Delta T = 5 \, \text{g} \times 1 \, \text{cal/g°C} \times (100 - 0) = 5 \times 100 = 500 \, \text{cal}
\]

3. Heat to convert water at 100°C to steam (latent heat of vaporization):
\[
Q_3 = m \times L_v = 5 \, \text{g} \times 540 \, \text{cal/g} = 2700 \, \text{cal}
\]

Total heat required:
\[
Q_{\text{total}} = Q_1 + Q_2 + Q_3 = 400 + 500 + 2700 = 3600 \, \text{cal}
\]

Thus, the total heat required is 3600 calories.