When m gram of steam at 100C is mixed with 200 gm of ice at 0C. it results in the water at 40C. Find the value of m in gram. (given : Latent heat of fusion (Lf) = 80 cal/gm, Latent heat of vaporisation (Lv) = 540 cal/gm., specific heat of water (Cw)= 1 cal/gm/C)
Answer (1)
22nd Mar, 2020
Hi,
As the system is thermally insulated ,Heat given out by steam is heat absorbed by the system
Heat given out by steam during condensation=M540
Heat given out by water at100C to go to40C=MCpwater(10040)
Heat absorbed by ice during melting=20080
Heat absorbed by water at0Cto go to40C=200Cpwater(400)
Heat balance givens
M540+MCpwater60=20080+200Cpwater40
Cpwater=1cal/gmC.
M540+M60=20080+20040
M600=16000+8000
=24000
M=60024000
M=40gm
As the system is thermally insulated ,Heat given out by steam is heat absorbed by the system
Heat given out by steam during condensation=M540
Heat given out by water at100C to go to40C=MCpwater(10040)
Heat absorbed by ice during melting=20080
Heat absorbed by water at0Cto go to40C=200Cpwater(400)
Heat balance givens
M540+MCpwater60=20080+200Cpwater40
Cpwater=1cal/gmC.
M540+M60=20080+20040
M600=16000+8000
=24000
M=60024000
M=40gm
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