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In a calorimeter (water equivalent=40g) ...

In a calorimeter (water equivalent`=40g`) are 200g of water and 50 g of ice all at `0^@C`. 30 g of water at `90^@`C is poured into it. What will be the final condition of the system?

Text Solution

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Let us assume that all ice melts and temperature of water rises beyond `0^@`C. Thus we will assume that `Tgt0`
Heat lost by water added`=`heat gained by ice to melt
`+`Heat to warm water formed from ice and water added
`+`Heat gained by calorimeter can.
`implies30xx1xx(90-T)=50xx80+(50+200)xx1`
`xx(T-0)+40xx1xx(T-0)`
`implies2700-30T=4000+250T+40T`
`impliesT=-4.1^@C`
Hence our assumption that `Tgt0` is wrong, since hot water added is not able to melt all of the ice.
Therefore the final temperature will be `0^@`C.
Let `m=` mass of ice finally left in the can.
Heat lost by water`=`heat gained by melting ice
`implies30xx1xx(90-0)=(50-m)xx80impliesm=16.25g`
Finally there is `16.25g` of ice and `(200+30+33.75)=266.75g` of water at `0^@C`.
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