Calculations Data for iron (Fe) Boiling point: 2861C Melting point: 1538C heat of vaporization: 1.50 kcal/gram heat of fusion: 59 cal/gram specific heat of Fe = 0.105 cal/g C 1. a) Solid iron can exist at any temperature up to melting pointoc. b) Molten, liquid iron can exist between the temperatures of 153 8P) C and 2861 bP C. 2) How many calories are needed to warm 10.0 grams of iron from 25C to 1538C? Given: Heat - (grams) (specific heat)(AT) q=m.C.AT 10 9=10g = 0,105 m = 10 T = 1538C .. 3) How many calories are needed to melt 10.0 grams of iron at its melting point (1538C)? (Hint - there is no AT. The iron is already at the melting point.) 15 13%-1588.65 cal 4) How many calories are needed to convert 10.0 grams of iron at 25C into liquid iron at 1538C? q=m.c. AT 10g. 5) If 20.0 grams of molten, liquid iron was boiling hot (2861C), how much more heat would be needed to vaporize the liquid iron? 6. How many grams of Fe can be warmed from 25C to the melting point (1538C) with 2000 dietary Calories (equal to 2.00 x 106 scientific calories) of heat energy? 7. What temperature change (AT) would be expected if a 15.0 gram iron object (at room temperature) absorbed one dietary Calorie (equal to 1000 scientific calories) of heat energy? Q = m. c. AT? AT = Q Accal m.c 1 dietary 0,105 15 cal = 1000 cal = 1000 * 4.184 ) = 4184 ] (since I cal = 4.124))~ 1 kcal 8. What would be the final temperature if 455 scientific calories of heat was absorbed by a 50.0 gram block of aluminum at 22C.