You are out camping with your family and you decide that you would heat some water for cooking your food. The pot you use is made from aluminum (heat capacity, s, = 0.902 J/g@EC). The pot has a mass of 675.6 g and a capacity of 1.356 L. You are going to heat water (s = 4.184 J/g degrees C) starting from ice (s = 36.93 J/mol degreesC) to its boiling point. If you start with 1250 g of ice at -10 degrees C, which is the same temperature as the pot, how many grams of pentane would you have to burn to heat this amount of water form -10 degrees C to its boiling point and boil off half of the water? Heat of fusion of water is 334.0 J/g, heat of vaporization of water is 40.68 kJ/mol. If you have 500.0 mL of pentane (d = 0.6262 g/mL) to use, will you have sufficient pentane to heat the pot and water? (delta)Heat of formation, C5H12 = -140.0 kJ/mol; (delta)Heat of formation CO2 = 393.5 kJ/mol; (delta)Heat of formation H2O(g) = 241.8 kJ/mol
Thanks for letting me work on your post. Here is my explanation:
energy needed for heating up the pot: 675.6*0.902*(100+10)=67033.032J=67.033kJ
energy needed for H2O from -10 to 0 (ice): ...
The expert examines heating a pot of water with pentane.