How long would it take a 1.00 x 10^3 W heating element to melt 2.00kg of ice at -20 deg C, assuming all energy is absorbed by the ice?

To determine how long it would take a 1.00 x 10^3 W heating element to melt 2.00 kg of ice at -20°C, we can use the formula for the heat required to melt a substance:

Q = m * L

Where:
Q - heat energy required (in Joules)
m - mass of the substance (in kilograms)
L - latent heat of fusion (in J/kg)

First, let's calculate the heat energy required to melt the ice. The latent heat of fusion for water is 334,000 J/kg.

Q = (2.00 kg) * (334,000 J/kg)
Q = 668,000 J

Now, we know the power of the heating element is 1.00 x 10^3 W (Watts). We can use the formula for power to find the time required:

P = Q / t

Where:
P - power (in Watts)
t - time (in seconds)

Since we want to find the time it takes to melt the ice, we can rearrange the formula to solve for t:

t = Q / P

Substituting the values we have:

t = (668,000 J) / (1.00 x 10^3 W)
t = 668 seconds

Therefore, it would take approximately 668 seconds (or 11 minutes and 8 seconds) for the 1.00 x 10^3 W heating element to melt 2.00 kg of ice at -20°C, assuming all the energy is absorbed by the ice.