3.2 kg of ice at -10 degree Celsius just melts with a mass m of steam. the value of m is
a.900
b.425
c.800
d.650
PSN03:
do you know the value of latent heat for ice?
cause that is required to solve this question
for fusion or vaporization?
both
fusion - 80 cal/g vaporization- 540 cal/g
Answers
Answered by
12
mass of ice=3200gm
temp=-10
first the ice will get converted to ice at 0 degree celsius and then to 0 degree water
fusion=80
vapourisation=540
therefore
total heat given by steam=heat required to convert ice to 0 degree celsius +heat required to convert into zero degree water
=mS1T1+mL
=3200(10*1+80)
=3200(90)
=288000
this heat should be equal to heat given by steam
m*L=288000
m*540=288000
m=533.33
then u are my senior
fine
what is the other way to solve
we have to then take even the heat required to convert water to 100 degree celcius
did u ask your teacher
or friends
we tried...but not with teacher
I will give the correct answer tomorrow
hey
did you get the correct answer
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