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)

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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)

Rakesh 22nd Mar, 2020

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Answer (1)

lubhan cherwoo 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

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