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M grams of steam at 100^(@)C is mixed wi...

M grams of steam at `100^(@)C` is mixed with 200g of ice at its melting point in a thermally insulated container. If it produces liquid water at `40^(@)`C [heat of vaporization of water is 540cal/g and heat of fusion of ice is 80 cal/gl, the value of M is

A

30

B

40

C

50

D

60

Text Solution

Verified by Experts

The correct Answer is:
B
Suppose mass of ice M= 200gm,
mass of steam= m
temperature of ice `T= 0^(@)C`
temperature of steam `T= 100^(@)C`
`ML_(f) + M(T. -T) C_(W) = mL_(v) + m(T.. - T.) C_(w)`
`:. 200 xx 80 + 200 (40-0)1 = m xx 540 + m(100-40)(1)`
`16000+ 8000= 540 m + 60m`
`:. 24000= 600m`
`:.m= (24000)/(600)= 40gm`
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