Education Boiling water and melting iron

So, while my Dad and I were waiting for the aluminum to melt in the furnace, we got into a discussion about the idea that it takes more energy to boil a pound of water than it does to melt a pound of iron. I heard this someplace and believe it to be true, but Dad thinks it's incorrect. A couple of tries on Google didn't yield a definitive answer, so I figured I'd throw it out there for you guys.

 

I think that there may be a little bit of Chinese whispers with this one.

The equation to calculated heat is Q = m * c * dT
Where Q = Heat Energy, m = Mass in Kg, c = Specific Heat Capacity (J per Kg per °K), dT = Temperature change

For ease of calculation we’ll assume that your 1lb is 1 Kg.

The Specific Heat Capacity of water is 4190 J kg-1 K-1
The Specific Heat Capacity of Iron is 450 J kg-1 K-1
For the water we’ll do the temperature change at 80°K from 20°C to 100°C
The melting point of iron is 1538°C so the change will be 1518°C

For water: 1kg x 4190 J kg-1 K-1 x 80°C = 335,200 Joules or 335.2 KJ

For iron: 1kg x 450 J kg-1 K-1 x 1518°C = 683,100 Joules or 683.1 KJ

So, you need about twice as much energy to melt a kilo of iron over boiling a kilo of water.
Where I think your dad may be coming from is heating kilo of iron to 100°C from room temperature would only take 36KJ which is only a 10th of the energy required to heat a Kg of water the same.

 

^^^ Science!

 

How much energy would it take to boil that kilo of water completely away I.E. convert it all to steam. That is how I always understood the statement, though it may have been worded wrong.

Thanks for the Science, BTW.

 

You need to include the enthalpy of state change (or whatever it's called) which I believe is highly significant for boiling water (about 2kJ/g apparently).

 

Bah, you just have to make thing awkward!

Ok, so now to factor in a change of state, you’ll have to include the Latent Heat of Vaporisation (Lw) which for water is 2.26x10^6 J kg-1, note no °K as that during this phase there is no temperature change all the energy is absorbed while the liquid is in its 100°C boiling state.

The Equation here will be Q = m x Lw
1kg x 2.26x10^6 J kg-1 = 2,260,000 J or 2,260 KJ

Now we’ll have to have to include the last little bit where the liquid does finally turn in to Steam. We can use the original equation here, but the specific heat capacity of steam is more than half that of water at 2010 J kg-1 K-1. For the purpose of this equation we’ll also have the temperature finish at 120°C so the temperature of change will be 20°K.
1kg x 2010 J kg-1 K-1 x 20°K = 40,200 J or 40.2KJ

To finish off we’ll combine all the results of all the equations together to give the ultimate required energy.
335,200 + 2,260,000 + 40,200 = 2,975,400 J or 2,975 KJ

So yes, to boil a pan dry of a Kg of water you would use over 4 times the energy required to melt Iron.

 

I don't want to play any more

 

Witchcraft!

Burns the lots of you I says.

 

And how much energy would that take ?

 

Depends If residual body fat is part of the equation!

 

Whoa, slow down and define your boundary conditions properly!

What partial pressure are we at? Does the closed system have finite volume?
I'm sure this can be made needlessly more difficult.

 

You also forgot to ask if he's currently invested in a game of Mornington Cresent.

Because, as we all know, the Parsons Green paradox would dramatically change results.

 

Game over, Mankz wins.

 

HOLY BALLS.

I've been a member 'ere for a decade.

 

Pffft. Part timer

 

Says him with over 10 times less posts

 

Quality not quantity, see above.

 

As a 2008'er, I feel young.

 

You are well thanked