Yes, it is just the ratio of heat of vaporization to heat of fusion.
I get
2260 J/gram / 334 J/gram = 6.77 g
how many gram of ice at 0 c can be melted by the heat released when one gram of steam condense to water at 100 c
check my answer please and you deman i am sorry by accident i post it the previous question
solution
Qcold=-Qhot
MLf(ice)=-MlV(water)
Mice = -(-0.001*2256000)/(334000)
Mice melt=6.75 g
3 answers
Heat out of steam= 20 {2260+4.2(100-T)}
Heat into water = 80 (T-20)
so
4(T-20) = 2260 + 4.2(100-T)
4 T - 80 = 2260 + 4200 - 4.2 T
8.2 T = 6540
T = 797 which is silly of course
What it means is that 80 grams of water at 20 degrees simply can not condense 20 grams of steam. The final temp is 100 deg C and you still have some stem left :)
Heat into water = 80 (T-20)
so
4(T-20) = 2260 + 4.2(100-T)
4 T - 80 = 2260 + 4200 - 4.2 T
8.2 T = 6540
T = 797 which is silly of course
What it means is that 80 grams of water at 20 degrees simply can not condense 20 grams of steam. The final temp is 100 deg C and you still have some stem left :)
There was another question in here that seems to have disappeared but my answer remained ?