An electric heater which produces 900W of power is used to vaporize water. How much water at 100c can be changed to steam in 3 mins by the heater? (Heat of vaporization=2.26×106j/kg, specific heat capacity of water=4.2×103 j/kg.k).

H=Pt

P=900w
t=3min
3*60=180s
H=900*180=162,000J

H=ml
162,000 = m*2.26*10^6
m=162,000/2.26*10^6
=0.072kg

if it is already at 100 temp will not change, just boils

2.26*10^6 m = 900 (3 *60)

m = (900*3*60 / 2.26) 10^-6

To calculate the amount of water that can be changed to steam in 3 minutes by the electric heater, we can use the formula:

Heat energy = Mass of water * Specific heat capacity of water * Temperature change + Mass of water * Heat of vaporization

First, let's calculate the heat energy produced by the electric heater in 3 minutes:

Energy (E) = Power (P) * Time (t)
E = 900W * 3 minutes

Since 1 minute is 60 seconds, we need to convert the time to seconds:
E = 900W * (3 * 60 seconds)
E = 900W * 180 seconds

Next, let's calculate the amount of energy required to raise the temperature of 1kg of water from 100°C to its boiling point (assuming atmospheric pressure), which is 100°C:
Energy (E1) = Mass of water (m) * Specific heat capacity of water (c) * Temperature change (ΔT)
E1 = 1kg * 4.2×10^3 J/kg·K * (100°C - 0°C)
E1 = 1kg * 4.2×10^3 J/kg·K * 100°C

Now, let's calculate the amount of energy required to vaporize the water:
Energy (E2) = Mass of water (m) * Heat of vaporization (L)
E2 = 1kg * 2.26×10^6 J/kg

The total energy required to change 1kg of water into steam is the sum of E1 and E2:
Total Energy = E1 + E2

Since the heater produces the total energy in 3 minutes, we can set up the equation:
Total Energy = Energy (E) from the heater

Now, we can solve for the mass of water (m):
Total Energy = Energy (E) from the heater
E1 + E2 = 900W * 180 seconds

m * 4.2×10^3 J/kg·K * 100°C + m * 2.26×10^6 J/kg = 900W * 180 seconds

Simplifying the equation, we have:
m * 4.2×10^5 J + m * 2.26×10^6 J = 900W * 180 seconds

m * (4.2×10^5 J + 2.26×10^6 J) = 900W * 180 seconds

m * 2.66×10^6 J = 900W * 180 seconds

Now, we can substitute the values:
m = (900W * 180 seconds) / (2.66×10^6 J)

Evaluating the expression using these values, we get:
m = (900W * 180 seconds) / 2.66×10^6

m ≈ 0.061 kg

Therefore, approximately 0.061 kg (or 61 grams) of water can be changed to steam in 3 minutes by the electric heater.

To find out how much water can be changed to steam in 3 minutes by the electric heater, we need to calculate the amount of energy required to convert water at 100°C to steam and compare it with the power of the electric heater.

Here's how we can proceed:

Step 1: Calculate the energy needed to vaporize the water.
The heat of vaporization, denoted by "L," is the amount of energy required to convert one kilogram of a substance from its liquid phase to its gaseous phase without a change in temperature. In this case, the heat of vaporization (L) for water is given as 2.26×10^6 J/kg.

Step 2: Calculate the energy input by the electric heater.
The power of the electric heater is given as 900W (watts). To calculate the energy input, we multiply the power by the time. In this case, the time is 3 minutes, which is equal to 180 seconds.

Energy input = Power × Time
Energy input = 900W × 180s

Step 3: Calculate the mass of water that can be vaporized.
We know that the specific heat capacity (c) of water is 4.2×10^3 J/kg.K. This means it takes 4.2×10^3 J of energy to raise the temperature of 1 kilogram of water by 1 degree Celsius.

To convert 100°C water to steam, we need to raise its temperature by T (the temperature change), which is equal to 100°C.

Therefore, the energy required to raise the temperature of the water to its boiling point is:
Energy required = Mass × Specific heat capacity × Temperature change
Energy required = Mass × 4.2×10^3J/kg.K × 100K

Step 4: Compare the energy needed to vaporize the water with the energy input by the electric heater.
Since the energy required to convert water to steam is equal to the energy input from the electric heater, we can set the two expressions equal to each other:

Mass × 2.26×10^6J/kg = 900W × 180s + Mass × 4.2×10^3J/kg.K × 100K

Step 5: Solve for the mass of water.
Rearrange the equation to solve for the mass of water (Mass):
Mass × (2.26×10^6J/kg - 4.2×10^3J/kg.K × 100K) = 900W × 180s
Mass = (900W × 180s) / (2.26×10^6J/kg - 4.2×10^3J/kg.K × 100K)

Now you can substitute the given values into the equation and calculate the mass of water that can be changed to steam in 3 minutes by the electric heater.