What is the final temperature of a 100g ice at -15°C which absorbs 2000 calories of heat energy?

Question

Specific heat of ice: ci=0.50cal/g°C
Specific heat of steam: cs=0.50cal/g°C
Specific heat of water: cw=1.00cal/g°C
Latent heat of fusion of ice: Lf=80Cal/g
Latent heat of Vaporization of Water: Lv=540Cal/g

Answer 1

No way, please check for typos
Here is what happens for 1 gram
-15 to 0 ==== 0.50*15 = 7.5
melt ======= 80, sum = 87.5
0 to 100===== 100 * 1 = 100 sum = 187.5
boil ======== 540 * 1 = 540 sum = 727.5
heat steam ======== .5 ( T-100)
so for just 1 gram you are up to 727.5 + 0.5(T-100)
for 100 grams you needed 72,750 just to get to steam at 100 C
sure it is not ten grams or 20,000 calories ?

Answer 2

nope. It's really 100g and 2000 calories.

Answer 3

since there is no mass of water...

q=mcΔT
2000J=100(0.50)(Tf-Ti)
2000J=100(0.50)(Tf-(-15))
Tf=25°C / 298K