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120 g of ice at 0^(@)C is mixed with 100...

120 g of ice at `0^(@)C` is mixed with 100 g of water at `80^(@)C`. Latent heat of fusion is 80 cal/g and specific heat of water is 1 cal/`g-.^(@)C`. The final temperature of the mixture is

A

`8g 100^(@) C`

B

`100 g , 90^@C`

C

`92 g, 100^(@)C`

D

`80g, 100^@C`

Text Solution

Verified by Experts

The correct Answer is:
C
`underset((0^@C))(Ice) overset(Lr) rarr underset((0^@C))(Water ) overset(Qs)underset(Heat)rarr underset((100^@C))"Water" overset(Lv)rarrunderset((100^@C))"Steam"`
`Q = L_i + Q_s + L_v = m.L_(f) + mSDelta T + m xx 540`
`rArr 22320 = 100 xx 80 + 100 xx 1(100-0) + mxx 540`
`rArr m. = 8g`
Mass of vapour = 8g
Amount of water = 100-8 = 92 g.
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