Heat Requirements for Melting Ice

If you leave it at room temperature, not very much!!

Seriously, I think you mean how much energy is required to melt the ice...

I'm afraid I can't remember what the specific heat of water is... or maybe its the latent heat... hmmm

John.

As the previous post said, I understand "heat" as energy required to convert ice to water at given temps. There are 3 components to the calculation:

1. How much energy is required to raise the temp of 215g H2O by 2 degC. Use standard heat balance eq.

2. How much energy is required to transform 215g of ice to water. i.e. you need to know the heat of fusion of H2O.

3. What are the losses in your system.

The sum of these 3 components should give you the heat required. Obviously you can discard the losses if you are looking for a purely theoretical answer to your question and are happy to ignore these.

Hi

hope you would get the required answer.

ALi Jamali

The melting of the ice, and 2. the heating of the water. The heat taken up in the complete process is the sum of the heat taken up in each stage.

1. Heat taken up for converting ice
at -0ºC to water at 0ºC mass of water = 215 g = 0.215 Kg the specific latent heat of fusion of ice = 334 Kj/kg we can calculate it by mass of water x latent heat of fusion
= 0.215 (kg) x 334 (kJ.kg^-1)
= 71.81 kJ 2. Heat taken up heating the water
from 0 ºC to the boiling point, 1 ºC
mass of water = 215 g = 0.215 Kg specific heat of vaporization of water = 2260 kJ.kg^-1. we can calculate it by mass of water x specific heat capacity x temperature change
= 0.215 (kg) x 4.18 (kJ.kg^-1. K^-1)x 1 (º C)
= 0.8987 kJ so your desired value is 0.8987 x 71.81 = 64.535kJ APX

Ali_jamali,

I disagree with your calculation somewhat as the latent heat of vaporization doesn't come into account in this situation as we aren't vaporizing the water. Therefore

Energy_Total = Energy_Heat + Energy_Melting

Energy_Melting = LatentHeat_Fusion x Mass = 335.5j/g x 215g = 72,132.5j

Energy_Heat = Specific_Heat x Mass x Temperature_Rise = 4.1868j/g/ºC x 215g x 2ºC = 1,800.324j

Energy_Total = 1,800j + 72,135j = 73,995.824j ≈ 74Kj

So the answer is that it will take just under 74Kj to melt the 215g of water and heat it up from -1ºC to +1ºC.

If you wish to read up on the theory behind this follow this link.

http://en.wikipedia.org/wiki/Latent_heat

Masu, I'm sure you should be using kg not grammes in those equations....

Afterall the definition of a calorie is the amount of heat to raise 1 kg of water 1 degree Celcius...

74000 Joules is an awful lot of energy!!!

What is the conversion factor from Joules to Calories? I think its only about 4.8 isn't it??

74 Joules sounds a little more reasonable??

John.

I just went and checked and here are the figures

1 Calorie = 4.1868 joules

1 Calorie = energy to heat 1 gram of water from 14.5ºC to 15.5º at a pressure of 1013.2 Hpa

H₂O_FusionHeat = 335.5 joules per gram = 79.72 calories per gram

You may have been thinking of food calories which are

1 Calorie_Food = Energy required to raise 1Kg of water 1ºC ≈ 1,000 Calories_Energy

So if you do the calculation using calories you get

Energy_Melting = 79.72Cal x 215g = 17,138.8

Energy_Heating = 1Cal x 2ºC x 215g = 430Cal

Total = Energy_Melting + Energy_Heating = 17,138.8 +430 = 17,458Cal = 17,569.8Cal = 73,561 joules

Which is pretty close to the original figures so yes it's correct. It's surprising how much energy it takes to actually melt the ice, its about the same amount that it takes to boil water from room temperature..

Interestingly though I have looked at about 10 sites to get the latent heat of fusion for water and I get figures that range from 79 to 80 calories per gram where as if you convert the 335.5 you get 80.133. I suspect the 335.5 joules per gram is the correct figure.

Hi,

your result is not absolutely exact because ice and liquid water heat specific are differents because :

ice : Specific_{ice heat} =2.04 j/(g°C)

therefore

Energy_Total = Energy_{ice heat} + Energy_{Melting +}Energy_{liquid water heat}

Energy_Melting = LatentHeat_Fusion x Mass = 335.5j/g x 215g = 72,132.5j

Energy_{ice heat} = Specific_{ice heat} x Mass x Temperature_Rise = 2.040j/g/ºC x 215g x 1ºC = 438.6 j

Energy_{liquid water heat} = Specific_{liquid water heat} x Mass x Temperature_Rise = 4.1868j/g/ºC x 215g x 1ºC = 900.16 j

Energy_Total = 438j+72,135j +900j ≈ 74Kj

the numeric answer is quite the same because melting energy is more important than the others.

kind regard from France

Tugdual

tugdual2@aol.com

Actually Tugdual is correct, well spotted. I had forgotten that ice and steam have different specific heats. All the values for H₂O are as follows

Specific Heat H₂0_Solod = 2.09jg^-1K^-1

Specific Heat H₂0_Liquid = 4.18jg^-1K^-1

Specific Heat H₂0_Vapour = 1.84jg^-1K^-1

Latent Heat_Fusion = 6.01Kjmol^-1 ≈ 335.5jg^-1

Latent Heat_Vaporization = 44.0Kjmol^-1 ≈ 2.603Kjg^-1

You're right Masu.... I was reading this morning in Laboratory News about 'the mystery of the Calorie'...

http://www.labnews.co.uk

and it was very interesting reading, unfortunately it also described the problems of calories and Calories one being for nutritian and the other for engineering, 'they' apparently have changed from capital C to kilocalories for the food one now, which makes more sense...

Sorry, I was confusing the Calorie with my calories!!!

Confused or what? - John.

Now you have the technically correct answer, here's a different way to melt the ice: - if the ambient temperature is as high as -1degC, you could just add some salt and let the environment provide the energy.