An electric heater that produces 860 W (1 W = 1 J/s) of power is used to vaporize water. How much water at 100 0C can be changed to steam in 5 min by the heater? Use Lv = 2.25x106 J/kg for water at 100 0C and express your final answer in grams.
masswater/5min=860*5*60/2.25E6 (1000) grams
=115 grams. check all that.
answer please because i really don't know that all thank you
To find out how much water can be changed to steam, we need to calculate the amount of heat required to vaporize the water.
The formula to calculate the heat required is:
Q = m * Lv
Where:
Q = Heat required (in Joules)
m = Mass of water (in grams)
Lv = Latent heat of vaporization (in J/kg)
We can rearrange the formula to solve for the mass of water:
m = Q / Lv
First, let's convert the power of the electric heater to Joules:
Power = 860 W
Time = 5 min = 5 * 60 s = 300 s
Energy = Power * Time
Energy = 860 J/s * 300 s = 258,000 J
Next, substitute the values into the formula to find the mass of water:
m = 258,000 J / 2.25 x 10^6 J/kg
m = 0.1147 kg
Lastly, convert the mass from kilograms to grams:
Mass in grams = 0.1147 kg * 1000 g/kg
Mass in grams = 114.7 g
Therefore, the electric heater can change 114.7 grams of water at 100°C to steam in 5 minutes.