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Grade 12th passMarks and Rank in IIT JEE

A calorimeter of specific heat 0.42j/qxm℃
Weighing 40qxm contains 50qxm of water mixed with 50qxm of ice. Steam at 100℃ is passed into the mixture until the temp rises to 20℃. Find mass of steam condensed.take Of=80cal/qxm
Lv=540cal/qxm and J =4.2j/cal .

Profile image of Aditya Pathak
7 Years agoGrade 12th pass
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1 Answer

Profile image of Aditya Pathak
7 Years ago
Q1= mlv +ms∆t 
=m(540)+m(80)=m(620)
Mixture = 40qxmcal+50qxmwater+50qxm of ice )
= mc.sc(20)+me.sw(20)+milf +misw(20)
Where lf is latent heat of fussion 
Heat required for 0℃to 20℃ rise
= 40×0.42/j×20+50×20+50×80+50×20
80+1000+4000+1000=6080=m(620)
M=9.8 qxm 
I hope your help
Heat required for 0

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