The heat of fusion of water is 6.01 kJ/mol. The heat capacity of liquid water is 75.3 J/mol ⋅ K. The conversion of 50.0 g of ice at 0.00 °C to liquid water at 15.0°C requires ________ kJ of heat.

Question

The heat of fusion of water is 6.01 kJ/mol. The heat capacity of liquid water is 75.3 J/mol ⋅ K. The conversion of 50.0 g of ice at 0.00 °C to liquid water at 15.0°C requires ________ kJ of heat.

Answer 1

Using a bit of dimensional analysis and resolving and combining units can help you solve this problem.

You are given in the problem that there was a temperature change of 15 degrees C. Now whether this stays in Celsius or is converter to Kelvin, it will still be a change in 15 degrees because the difference from Celsius to Kelvin is 273.

So with this information you will use it to cancel out units and get it to where all you have is kJ.

Take the 75.3 J/mol *K and multiply by 15 K. (express it as Kelvin, this way you will get your units to cancel out.)

75.3 J/mol*K. x 15 K. Kelvin temperature units cancel each other out and you will have a product of, with the following units:

1129.5 J/mol

Now you can take these to kJ by multiplying by the right equivalency:

1 kJ = 1000 J

1129.5 J/mol. x 1 kJ/1000 J. Joules cancel top to bottom and you have:.

1.1295 kJ/mol

Now combine this with the heat of fusion because we are tallying up all the energy associated or connected with this change, you have:

1.1295 kJ/mol + 6.01 kJ/mol. = 7.1395 kJ/mol

Now you must account for the 50.0 g of ice, which is water give in the problem. To do this you must get grams to moles by using H2O atomic weight.

Atomic weight H2O; 18.02g/mol

50 g * 1mol/18.02 g. grams cancel and you will have moles at:

2.77469 moles

Now take this to the 7.1395 kJ/mol so that you can finally be left with kJ:

7.1395 kJ/mol x 2.77469 mol. moles cancel top to bottom and you have:

19.8 kJ for 3 sig fig purposes.