Calculations Data for iron (Fe) Boiling point: 2861C Melting point: 1538C heat of vaporization: 1.50 kcal/gram...

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Calculations Data for iron (Fe) Boiling point: 2861°C Melting point: 1538°C 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 25°C to 1538°C? Given: Heat - (grams) (specific heat)(AT) q=m.C.AT 10 9=10g = 0,105 m = 10 T = 1538°C .. 3) How many calories are needed to melt 10.0 grams of iron at its melting point (1538°C)? (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 25°C into liquid iron at 1538°C? q=m.c. AT 10g. 5) If 20.0 grams of molten, liquid iron was boiling hot (2861°C), how much more heat would be needed to vaporize the liquid iron? 6. How many grams of Fe can be warmed from 25°C to the melting point (1538°C) 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 22°C. Calculations Data for iron (Fe) Boiling point: 2861°C Melting point: 1538°C 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 25°C to 1538°C? Given: Heat - (grams) (specific heat)(AT) q=m.C.AT 10 9=10g = 0,105 m = 10 T = 1538°C .. 3) How many calories are needed to melt 10.0 grams of iron at its melting point (1538°C)? (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 25°C into liquid iron at 1538°C? q=m.c. AT 10g. 5) If 20.0 grams of molten, liquid iron was boiling hot (2861°C), how much more heat would be needed to vaporize the liquid iron? 6. How many grams of Fe can be warmed from 25°C to the melting point (1538°C) 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 22°C.

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1 a Up to its melting point of 1538C solid iron can exist at any temperature b Between the melting and boiling points of iron 1538C and 2861C liquid i... View the full answer