- #1
northh
- 2
- 0
1. Problem, data and attempt at a solution
How much energy is required to boil off 3 litres of water from 20*C?
This is not really homework, but I still thought this would be the best place to post:
To get something to compare to in terms of energy (joule/watt), I would like to know approximately how much energy it takes to boil all the water in a standard kettle (3 litres).
I tried doing this with data from log p H (mollier) diagrams for water:
Enthalpy of water at 20 *C and athmospheric pressure from mollier: 86,6 kJ/kg
Enthalpy of saturated steam (100*C): 2676,1 kJ/kg
DeltaH= 2676,1 - 86,6 = 2589,5 kJ/kg
3 litres of water = 3kg so: 2589,5 kJ/kg * 3 kg = 7768,5 kJ required to boil 3 litres of water.
However when searching for this on google i only found another solution to the problem that did not match my answer using constant specific heat capacity and heat of vaporization:
http://wiki.answers.com/Q/How_much_energy_is_needed_to_boil_water#ixzz18El72mGe
However according to wikipedia: "The specific heat capacities of substances comprising molecules (as distinct from monatomic gases) are not fixed constants and vary somewhat depending on temperature."
So using a constant heat capacity value from 20 *C to 100 *C would be wrong, no?
Are any of these methods viable? Which is better? They don't match up when i use his equation for my case (although I know it's partly because we have different values for heat of vaporization and it seems like he has mixed up some units in his description).
How much energy is required to boil off 3 litres of water from 20*C?
This is not really homework, but I still thought this would be the best place to post:
To get something to compare to in terms of energy (joule/watt), I would like to know approximately how much energy it takes to boil all the water in a standard kettle (3 litres).
I tried doing this with data from log p H (mollier) diagrams for water:
Enthalpy of water at 20 *C and athmospheric pressure from mollier: 86,6 kJ/kg
Enthalpy of saturated steam (100*C): 2676,1 kJ/kg
DeltaH= 2676,1 - 86,6 = 2589,5 kJ/kg
3 litres of water = 3kg so: 2589,5 kJ/kg * 3 kg = 7768,5 kJ required to boil 3 litres of water.
However when searching for this on google i only found another solution to the problem that did not match my answer using constant specific heat capacity and heat of vaporization:
http://wiki.answers.com/Q/How_much_energy_is_needed_to_boil_water#ixzz18El72mGe
However according to wikipedia: "The specific heat capacities of substances comprising molecules (as distinct from monatomic gases) are not fixed constants and vary somewhat depending on temperature."
So using a constant heat capacity value from 20 *C to 100 *C would be wrong, no?
Are any of these methods viable? Which is better? They don't match up when i use his equation for my case (although I know it's partly because we have different values for heat of vaporization and it seems like he has mixed up some units in his description).
Last edited: