Verified answer

This answer:

"51g H2O x 4.184 J/gram deg celsius x 124 deg celsius = 26,459.616 J or 26.46 kJ"

is incorrect.

There are five separate calculations you need to carry out, then you'll add the results together for the final answer.

The five are these:

1) heat ice from -14 to 0

2) melt ice at 0

3) heat liquid from 0 to 100

4) boil liquid at 100

5) heat steam from 100 to 110

Here is a link to an explanation of the five steps:

http://www.chemteam.info/Thermochem/Time-Temperatu...

My example uses 72 g and goes from -10 to 120. Replacing the appropriate values with yours will be simple. Remember to note that answers 2 and 4 will be in kJ and 1, 3 and 5 will be in J. Convert the J to kJ, then add them.

Step 1: Heat ice to 0° = 51g x 14° x 2.11J/g° =

Step2: Melt ice. = 51g x 334J/g =

Step 3: Heat water from 0 to 100 = 51g x 100° x 4.18J/g° =

Step 4: boil water = 51g x 2260J/g =

Step 5: heat steam = 51g x 10° x 2.00J/g° =

Step6: add 5 steps together.

Constants from

Engineeringtoolbox.com/water-thermal-properties-d_162.html

51g H2O x 4.184 J/gram deg celsius x 124 deg celsius = 26,459.616 J or 26.46 kJ

4.184= specific heat of h2o