Heat required = 220 cal
Further explanation
Given
2 g ice at -20°C into water at 20°C
c ice = 0.5 cal g-1 °C-1
Lf = 80 cal/g
Required
Heat
Solution
The heat needed to raise the temperature
Q = m . c . Δt
1. heat to raise temperature from -20 °C to 0 °C
Q = 2 x 0.5 x 20
Q = 20 cal
2. phase change(ice to water)
Q = m.Lf
Q = 2 . 80 cal/g
Q = 160 cal
3. heat to raise temperature from 0 °C to 20 °C ( c water = 1 cal/g.C)
Q = 2 . 1 . 20
Q = 40 cal
Total heat = 220 cal