1.

`A` `0.50` kg ice cube at `-10^(@)C` is placed in `3.0` kg of coffce at `20^(@)C` . What will be the final temperature of mixture? Assume that specifec heat of tea is same as that of water.A. `5.1^(@)C`B. `8^(@)C`C. `10^(@)C`D. `6^(@)C`

Answer» Correct Answer - A
In this situation there are three possibilities regarding the final state of the mixture:
1. All ice `2 .A` mixture of ice water at `0_(0)C`
3. All waer Energy release required to brong the `3.0` kg of water at `20^(@)` down to `0^(@)` ,
`Q_(1)=m_(w)C_(w)(20^(@)C-0^(@)C)`
`=(3)(4186)(20)=251kJ`
Energy required to change the ice from `-10^(@)C to 0^(@)C` ,
`Q_(2)=m_(ice)C_(ice)[0^(@)C-(-10^(@)C)]`
`=(0.50)(2100)(10)=10.5kJ`
Energy required to change the ice to water at `0^(@)C`
`Q_(3)=m_(ice)L_(f)=(0.50)(333)=167kJ`
`A` total of `10.5kJ+167kJ=177kJ`
energy is required to bring ice at `-10^(@)C` to water at `20^(@)C` down to `0^(@)C` is `250kJ` . Thus the mixture must end up all water somewhere between `0^(@)C and 20^(@)C` . Let the final temperature be `T` Now applying energy conservation,
`(("Heat to raise"),(0.50 "kg of ice"),("from" -10^(0) C "to"),(0^(0)C))+(("Heat to change"),(0.50 kg),("of ice"),("to water"))+(("Heat to raise"),(0.50 kg),("of water"),("from" 0^(0) C "to" T))=(("Heat lost by"),(3.0 "kg of"),("water colling"),("from" 20^(0) C "to" T))`
Hence, `10.5+167+0.50(4184)T=(3)(4186)(20^(@)C-T)`
Which on solving yilds `T=5.1^(@)C`


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