Final answer:
To melt 8.86 kJ of heat, 110.75 grams of propane would be required.
Step-by-step explanation:
To calculate the number of grams of propane that would melt with 8.86 kJ of heat, we can use the heat of fusion of propane. The heat of fusion is the amount of heat required to melt one gram of a substance. In this case, the heat of fusion of propane is given as 80.0 J/g.
To convert the heat of fusion from joules to kilojoules and to find the number of grams that would melt with 8.86 kJ of heat, we can use the following calculations:
80.0 J/g * (1 kJ/1000 J) = 0.08 kJ/g
8.86 kJ / 0.08 kJ/g = 110.75 g
Therefore, 110.75 grams of propane would melt with 8.86 kJ of heat.