174520 = (massboiled) (2260)

.1716 = massboiled

so .1716?

Suppose a burner transfers 325 kJ of energy to 450 kg of liquid water at 20 C degrees. What mass of the water would be boiled away?

The first energy going in heats the water:

energy=mass*c*(100-20) Then subtract that from the intial 325kJ.

Then, you can figure the mass of water boiled away:

heatleft=massboiled*Lv Where Lv is the heat of vaporization.

To solve this question, we first need to find the energy required to heat the water from 20°C to its boiling point. We can use the formula:

energy = mass * c * (final temperature - initial temperature)

Where:
- "mass" is the mass of the water (in kg)
- "c" is the specific heat capacity of water (approximately 4.18 kJ/kg°C)
- "final temperature" is the boiling point of water (100°C)
- "initial temperature" is the initial temperature of the water (20°C)

Let's plug in the values:

energy = mass * 4.18 * (100 - 20)

Now, subtract this energy from the initial energy of 325 kJ to find out how much energy is left for boiling:

heat_left = 325 kJ - energy

Next, we can calculate the mass of water that will be boiled away using the equation:

heat_left = mass_boiled * Lv

Where:
- "mass_boiled" is the mass of the water boiled away (in kg)
- "Lv" is the heat of vaporization of water (approximately 2260 kJ/kg)

Let's rearrange the equation to solve for "mass_boiled":

mass_boiled = heat_left / Lv

Now, we can substitute the value of heat_left into the equation:

mass_boiled = (325 kJ - energy) / Lv

Finally, to find the mass_boiled, you can substitute energy with the value we calculated earlier:

mass_boiled = (325 kJ - mass * 4.18 * (100 - 20)) / Lv

You can then simplify this equation to find the mass_boiled.