If water heat of fusion is 80 cal/g, how many heat is required to covert 30g of ice at 0 degree C to liquid water at 30 degree C?

A 800 cal

B 600 cal

C 2200 cal

D 1000 cal

E 3300 cal

To find the heat, you need to find the heat of ice at 0 and when it melts into liquid

heat ice = heat of fusion x mass

heat of liquid = specific heat x mass x ΔT

specific heat of water = 1.00 cal/g C

mass = 30 grams

ΔT = 30 - 0 = 30C

heat of ice = 80 x 30 = 2400 calories

heat of liquid = 1.00 x 30 x 30C = 900 calories

total heat = 2400 + 900 = 330 calories

answer is E

The answer this question is 3300 cal