High quality coal (anthracite) is almost pure carbon. The combustion of carbon to carbon dioxide releases 393kJ per mol of carbon burnt. If 1kg of anthracite is burnt:

a) How much heat is released?
b) How much ice (in kg) at 273K could be melted to give water at 273K?
c) How much ice at 273K could be melted to give water at 100degrees celcius?
d) How much ice at 273K could be melted to give steam at 373K?

I found a) to be 32723kJ.
But I don't know how to do the rest.
I know it has something to do with enthalpy of fusion.

Please help.
Thanks

b.

q = mass x heat fusion
q about 32,723,000 J
heat fusion about 334 J/g
Solve for mass ice.
You should confirm the 334 value.

c.
q = [mass ice x heat fusion ice] + [mass water x specific heat water x delta T]
Let mass ice = mass water = x and solve for x.

d. Same procedure but include heat vaporization at 100.
Note the correct spelling of celsius.

Your answer to (a) looks correct.

For (b), divided the heat release by the heat of fusion of water, 333 Kj per kg.
For (c), divide the heat realse by (333 +418 kJ) = 551 kJ/kg
For (c) add the heat of vaporization to the amount of heat required per kg, and divde that into the heat release.

Your answer to (a) looks correct.

For (b), divided the heat release by the heat of fusion of water, 333 Kj per kg.
For (c), divide the heat realse by (333 +418 kJ) = 551 kJ/kg
Note:I believe drwls make a typo here. I think he meant to type 751 and not 551. :-)
For (c) add the heat of vaporization to the amount of heat required per kg, and divde that into the heat release.

Thank you,

I still can't seem to get the last one.
We are given our data in kJ/mol

So would
q = n (heat fusion ice + heat sublimation ice + (heat capacity water x delta T)) ?

We also don't have the heat vaporisation value. Only for sublimation - is it the same? They only give vaporisation values for specific elements, such as Hydrogen, Oxygen etc, but not compounds.

To solve the remaining parts of the question, you are correct in identifying that the enthalpy of fusion is involved. Enthalpy of fusion is the amount of heat required to convert a substance from a solid to a liquid at its melting point. In this case, we need to calculate the amount of ice that can be melted by the heat released from burning 1 kg of anthracite.

b) To determine how much ice at 273K can be melted to give water at 273K, we need to find the enthalpy of fusion of ice. The enthalpy of fusion for ice is 334 J/g. Since we are given the mass of anthracite burned (1 kg), we need to convert it to grams: 1 kg = 1000 g.

We can calculate the amount of ice melted using the following formula:
Amount of ice melted = Heat released from burning anthracite / Enthalpy of fusion of ice

First, we know that the heat released from burning anthracite is 32723 kJ. To convert it to J, multiply by 1000:
Heat released from burning anthracite = 32723 kJ × 1000 = 32,723,000 J

Then, to calculate the amount of ice melted, divide the heat released by the enthalpy of fusion of ice:
Amount of ice melted = 32,723,000 J / 334 J/g

To obtain the answer in kg, divide the result by 1000 (since there are 1000 grams in a kilogram):
Amount of ice melted = (32,723,000 J / 334 J/g) / 1000 kg

Performing the calculations gives us:
Amount of ice melted = 97.99 kg

Therefore, approximately 98 kg of ice at 273K could be melted to give water at 273K.

c) To determine the amount of ice at 273K that could be melted to give water at 100 degrees Celsius, we need to consider both the enthalpy of fusion and the enthalpy of heating. The enthalpy of heating, also known as the specific heat capacity, is the amount of heat required to raise the temperature of a substance by 1 degree Celsius.

In this case, we need to calculate the heat required to raise the temperature of the ice from 273K to 0 degrees Celsius, melt it into water at 0 degrees Celsius, and then raise the temperature of the water from 0 degrees Celsius to 100 degrees Celsius.

First, let's calculate the heat required to raise the temperature of the ice from 273K to 0 degrees Celsius:
Heat = Mass × Specific heat capacity × Temperature change

The specific heat capacity of ice is 2.09 J/g·°C. Again, we need to convert the mass of ice to grams:
Mass of ice = Amount of ice melted × 1000 g/kg

Therefore, the heat required to raise the temperature of the ice from 273K to 0 degrees Celsius is:
Heat = (Amount of ice melted × 1000 g/kg) × 2.09 J/g·°C × (0 - 273) °C

Next, let's calculate the heat required to melt the ice into water at 0 degrees Celsius. This amount of heat is given by the enthalpy of fusion of ice, which is 334 J/g. We already know the mass of ice to be melted, which is the same as in part b).

The heat required to melt the ice is:
Heat = Amount of ice melted × 334 J/g

Finally, let's calculate the heat required to raise the temperature of the water from 0 degrees Celsius to 100 degrees Celsius. The specific heat capacity of water is 4.18 J/g·°C.

The heat required to raise the temperature of the water is:
Heat = Amount of water produced × Specific heat capacity × Temperature change

Amount of water produced = Mass of ice melted

Heat = Amount of water produced × 4.18 J/g·°C × (100 - 0) °C

To find the total heat required for the entire process (raising the temperature, melting the ice, and raising the temperature of the resulting water), sum up all the individual heat values calculated above.

Total Heat = Heat (raising the temperature of the ice) + Heat (melting the ice) + Heat (raising the temperature of the water)

d) Similar to part c), we need to consider both the enthalpy of fusion and the enthalpy of vaporization in this case. The enthalpy of vaporization is the amount of heat required to convert a substance from a liquid to a gas at its boiling point.

To solve for how much ice at 273K could be melted to give steam at 373K, we need to calculate the heat required to raise the temperature of the ice from 273K to 0 degrees Celsius, melt it into water at 0 degrees Celsius, raise the temperature of the water from 0 degrees Celsius to 100 degrees Celsius, and then convert the water into steam at 100 degrees Celsius.

Using the same formulas and principles as in part c), you can calculate the total heat required for this process accordingly.

Remember to convert the mass of ice melted from kg to grams and divide the result by 1000 to obtain it in kg.

I hope this explanation helps you solve the remaining parts of the question. Let me know if you have any further questions!