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

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Sorry i dont understand the question can you rephase it?