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1 kg ice at -10^(@)C is mixed with 0.1 k...

`1 kg` ice at `-10^(@)C` is mixed with `0.1 kg` of steam at `200^(@)C`. If final temperature of mixture at equilibrium is `T_(eq)=(58x)/11`, then fill the vallue of `x`
Latent heat of fusion of ice `=80` cal/gram, latent heat of vaporization of water `=540` cal/gram, specific heat capacity of ice `~=` specific heat of water `=0.5` cal/gram-K`

Text Solution

Verified by Experts

The correct Answer is:
5
`1xx 1/2 xx 10+1xx80+1xx1xxT_("eq")=0.1xx1/2xx100+0.1xx540+0.1xx(100-T_("eq"))`
`T_("eq")=290/11 K=58/11xx5`
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Knowledge Check

  • In a container of negligible heat capacity, 200 gm ice at 0^(@)C and 100gm steam at 100^(@)C are added to 200 gm of water that has temperature 55^(@)C . Assume no heat is lost to the surrpundings and the pressure in the container is constant 1.0 atm . (Latent heat of fusion of ice =80 cal//gm , Latent heat of vaporization of water = 540 cal//gm , Specific heat capacity of ice = 0.5 cal//gm-K , Specific heat capacity of water =1 cal//gm-K) What is the final temperature of the system ?

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