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In an insulated vessel, 250 g of ice at ...

In an insulated vessel, 250 g of ice at `0^(@)C` is added to 600 g of water at `18^(@)C`
(a) What is the final temperature of the system?
(b) How much ice remains when the system reaches equilibrium?
Useful data:
Specific heat capacity of water: 4190 K/K.kg
Speicific heat capacity of ice: 2100J/K.kg
Latent heat of fusio of ice: `3.34xx10^(5)J//kg`

Text Solution

Verified by Experts

`Q_("required")=mL`
`=(250xx10^(-3))xx3.34xx10^(5)`
`=8.35xx10^(4)J`.
Heat available from water is
`Q_("available")=MCDeltaT=600xx10^(-2)xx4190xx1187`
`=4.525xx10^(4)J`.
(a) As heast available is not sufficient to melt ice, final temperature is zero. (b) the amount of ice that will melt to bring down the temperature of water from `18^(@)C` to `0^(@)C` is calculated as shown.
`Q_("available")=m_(ice)xxL_(ice)`
`4.525xx10^(4)=mxx3.34xx10^(5)`
`impliesm_(ice)=135xx10^(-1)kg~~135g`. ltBrgt So, ice remaining=250-135
=115g
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