# Can you help me with these questions?

How many calories are required to melt 75 grams of ice at 0 degrees Celsius? & How many calories are required to evaporate 75 grams of water at 100 degrees Celsius?

### 2 Answers

- billrussell42Lv 75 months agoFavorite Answer
334 J/g x 75 g x 1 cal/4.184 J = 5990 cal

2256 J/g x 75 g x 1 cal/4.184 J = 40400 cal

you do know that the calorie is an obsolete unit?

specific heat of water is 4.186 kJ/kgC = 4.186 J/gC

= 1 calorie/gC = 5.375 J/mol·K = 1 Btu/lb-F

specific heat of ice is 2.06 kJ/kgC = 2.06 J/gC

specific heat of steam is 2.1 kJ/kgK = 2.1 J/gK

heat of fusion of ice is 334 kJ/kg = 334 J/g

heat of vaporization of water is 2256 kJ/kg = 2256 J/g

- Roger the MoleLv 75 months ago
(333.6 J/g) x (75 g) / (4.184 J/cal) = 5980 cal to melt

(2257 J/g) x (75 g) / (4.184 J/cal) = 40458 cal to evaporate

[Adjust the number of significant digits as you see fit.]