Posted by ridhi on Tuesday, May 29, 2012 at 1:03am.
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
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