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72.0 grams of ice has changed from −10.0 °C to 120.0 °C. The energy calculation for this change required five steps
The following table summarizes the five steps and their results. Each step number is a link back to the explanation of the calculation.
Step q 72.0 g of H2O 1 1483.2 J Δt = 10 (solid) 2 24.08 kJ melting 3 30124.8 J Δt = 100 (liquid) 4 162.8 kJ boiling 5 2908.8 J Δt = 20 (gas)
Converting to kJ gives us this:
1.4832 kJ 24.08 kJ 30.1248 kJ 162.8 kJ 2.9088 kJ
Summing up gives 221.3968 kJ and proper significant digits gives us 221.4 kJ for the answer.
Notice how all units were converted to kJ before continuing on. Joules is a perfectly fine unit, it's just that 221,396.8 J is an awkward number to work with. Usually Joules is used for values under 1000, otherwise kJ is used.
By the way, on other sites you may see kj used for kilojoules. I've also seen Kj used. Both of these are wrong symbols. kJ is the only correct symbol. Of course, if you use a wrong symbol, most people will know what you are saying, it will just look slightly silly.