1. Given a liter of water just at boiling, how much energy does it take to boil it all?
2. If that liter of water was originally at 10°C, how much extra energy would it take to boil
the water?
Would it be:
1. Q=mL=(1kg)(2.26e6)=2.26e6
2. Q=mC(Tf-Ti)=(1kg)(4186J)(100C-10C)=376740J
did not check arithmetic but that is the right idea
Goodness. Your use of units leaves much to be desired. Let units work for you.
a. energy=Hv*mass=2257kJ/kg*1liter*1kg/liter= 2257kJ
b. mcDeltaTemp=1kg*4.184kJ/kgC*90C=
= 90*4.184kJ
To answer your questions:
1. The amount of energy required to boil a given mass of water at its boiling point is calculated using the formula Q = m * L, where Q is the amount of energy, m is the mass of the water, and L is the specific latent heat of vaporization for water. The specific latent heat of vaporization for water is approximately 2.26 million joules per kilogram (J/kg) at standard atmospheric pressure.
To find the energy required to boil a liter of water, you need to convert the volume to mass. Since 1 liter of water has a mass of 1 kilogram (kg), you can apply the formula:
Q = m * L = 1 kg * 2.26 million J/kg = 2.26 million joules (J)
So, it would take approximately 2.26 million joules of energy to boil a liter of water at its boiling point.
2. If the initial temperature of the water is not at its boiling point, you also need to consider the amount of energy required to raise the temperature of the water from the initial temperature (Ti) to the boiling point (Tf). This can be calculated using the formula Q = m * C * (Tf - Ti), where Q is the amount of energy, m is the mass of the water, C is the specific heat capacity of water, and (Tf - Ti) is the temperature change.
The specific heat capacity of water is approximately 4186 J/kg°C.
To find the extra energy required to boil a liter of water initially at 10°C, you can apply the formula:
Q = m * C * (Tf - Ti) = 1 kg * 4186 J/kg°C * (100°C - 10°C) = 376,740 joules (J)
So, it would take an additional 376,740 joules of energy to bring the water from 10°C to its boiling point and then boil it.
In summary:
1. To boil a liter of water at its boiling point, it would take approximately 2.26 million joules of energy.
2. To boil a liter of water initially at 10°C, it would take approximately 2.26 million joules for boiling plus 376,740 joules for heating from 10°C to the boiling point, resulting in a total of around 2.64 million joules of energy.