A 50.0 g sample of liquid water at 0.0°c ends up as ice at -20.0°c. How much energy is involved in this change.

-4200J

Explanation:

Since q = mc∆T

where:

m = mass of the substance

c = specific heat capacity of the substance

∆T = change in temperature in Kelvin

but;

the specific heat capacity of water = 4.2J/K/g

Therefore q = 50*4.2*(253-273)

q = -4200J

Note: the negative is only used to indicate heat loss

Hope this helps