12 kg of water in an insulated container is at 30 C. Initially, only 18 percent of the water is in the liquid phase.
An electric heater, connected to a 120-V source and 9 A, is used to heat the system. Determine the time required to vaporize the entire water. Express your answer in minutes.
Hmmm. water at 30C, but not in the liquid state. How does that happen?
i think so
To determine the time required to vaporize the entire water, we need to calculate the amount of heat needed to vaporize 12 kg of water. This can be achieved by using the formula:
Q = mL
Where Q is the heat required, m is the mass of the water, and L is the latent heat of vaporization.
Given that the latent heat of vaporization of water is approximately 2.26 x 10^6 J/kg, we can calculate the heat required as follows:
Q = (12 kg)(2.26 x 10^6 J/kg) = 27.12 x 10^6 J
Next, we need to calculate the electrical power consumed by the heater using the formula:
P = VI
Where P is the power consumed, V is the voltage, and I is the current.
Given that the voltage is 120 V and the current is 9 A, we can calculate the power consumed as follows:
P = (120 V)(9 A) = 1080 W
Now we can determine the time required to provide the heat needed to vaporize the entire water by dividing the heat required by the power consumed:
t = Q / P = (27.12 x 10^6 J) / (1080 W) ≈ 25,111 seconds
To convert this time from seconds to minutes, we can divide it by 60:
t ≈ 25,111 seconds / 60 ≈ 418.52 minutes
Therefore, the time required to vaporize the entire water is approximately 418.52 minutes.