Solution:

To calculate the heat required to convert 40 g of ice at –20°C into water at 20°C, we need to consider three steps:

1. Heating ice from -20°C to 0°C:
\[
Q_1 = m \times c_{\text{ice}} \times \Delta T = 40 \, \text{g} \times 0.5 \, \text{cal/g°C} \times (0 - (-20)) = 40 \times 0.5 \times 20 = 400 \, \text{cal}
\]

2. Melting ice at 0°C (latent heat of fusion):
\[
Q_2 = m \times L_f = 40 \, \text{g} \times 80 \, \text{cal/g} = 3200 \, \text{cal}
\]

3. Heating water from 0°C to 20°C:
\[
Q_3 = m \times c_{\text{water}} \times \Delta T = 40 \, \text{g} \times 1 \, \text{cal/g°C} \times (20 - 0) = 40 \times 1 \times 20 = 800 \, \text{cal}
\]

Total heat required:

\[
Q_{\text{total}} = Q_1 + Q_2 + Q_3 = 400 + 3200 + 800 = 4400 \, \text{cal}
\]

Thus, the total heat required is 4400 calories.