a mixture of 250g of water and 200g of ice at 0deg C is kept in a calorimeter which has a water equivalent of 50g. if 200g of steam at 100deg C is passed through this mixture, calculate the final temperature and the mass of contents of the calorimeter. ans 572.22g

Question

please explain me the steps.
thanks in advance :)

Answer 1

Let the final temperature of the mixture = 100ºC.
Heat gained by ice and iced water Q1 is
Q1 = m•λ + m•c•ΔT = 200•80 + 200•1•100 = 3600 cal.
Q2 = (m1•c +W) •(100º -0º) =(250•1 + 50) •100 = 30000 cal.
Q =Q1+Q2 = 36000 +30000 = 66000 cal.
If entire steam condensed the total heat given by it is
Q3 = m•L = 200•540= 108000 cal.
Now the total heat available is 108000 cal, but only 66000 cal are required. Therefore? all the steam will not get condensed.
Final temperature of the mixture is 100ºC.
Mass of steam condensed is 66000/540=122.2 g.
Mass of the contents is 250 +200 + 122.2 = 572.2 g

Answer 2

Why did u assume that final temperature is 100 why not 70 or 80

Answer 3

Why do you take 66000/540 (at last).
You should take 108000-66000=42000
So,
42000/540???