If 200gm ice at -50°c and 100gm water at 20°c mixed with 10gm steam at 110°c mixed with each other then what will be the temperature of mixture. Is ice remain left, if yes then find the amount of ice,vapour or water

 

Let us assume the final temperature to be 0 Celsius 

So for steam

Heat released to reach water at 100 celcius  = MCT = 10×0.48×10 = 48cal

Heat released by steam to condense water = 10×540 = 5400cal

Heat released by this water to reach 0 celcius  = 10× 1×100 = 1000cal

Net heat release by steam = +5400+1000+48 = 6488cal

For water

Heat release by water to reach 0 celcius from 20 celcius  = 100 ×1×20 = 2000cal

Net heat release by water and steam together = 6448+2000 = 8448cal

For ice 

Heat absorb by ice to reach from -50 to 0 celcius = 200×0.5×50 = 5000 cal

So heat released by water and steam will completely use in to change temperature of ice to 0 Celsius,  

Heat remain  = 8448-5000 = 3448 cal

So this much heat must be absorbed by ice to melt it

So 3448 = m × Lf

So m = 3448/80 = 43 gm 

So only 43 GM of ice will melt

The final temperature would be zero

Mass of ice left = 57 GM

Mass of water left = 10+100+43 = 153gm