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 the time it takes for a heating element to melt ice, we need to consider the amount of energy required and the rate at which the heating element supplies that energy.
First, let's calculate the amount of energy needed to melt the ice. The energy required to change the state of a substance from solid to liquid is given by the equation:
Q = m * L
where Q is the energy, m is the mass, and L is the latent heat of fusion. For ice, the latent heat of fusion is 334,000 J/kg.
Given that the mass of the ice is 2.00 kg, we can calculate the energy required:
Q = (2.00 kg) * (334,000 J/kg) = 668,000 J
Now, let's determine the time required to supply this amount of energy using the power of the heating element. Power is defined as the rate at which energy is transferred, given by:
P = W / t
where P is power, W is work (energy), and t is time.
We are given that the power of the heating element is 1.00 x 10^3 W. By rearranging the formula, we can solve for time:
t = W / P
Substituting the values, we have:
t = (668,000 J) / (1.00 x 10^3 W)
t = 668 seconds
Since the time is given in seconds, we can convert it to minutes by dividing by 60:
t = 668 seconds / 60 seconds/minute ≈ 11.1 minutes
Therefore, it would take approximately 11.1 minutes 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.