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

 

Asked by Aryan.suthar | 20th Apr, 2019, 09:34: PM

Expert Answer:

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

Answered by Ankit K | 20th Apr, 2019, 11:47: PM