1989,2M

JEE Advanced 2015 Guidance › Mechanics questions › 1989,2M

300g of water at 25C is aded to 100g ice at OC.Final temp of mix is ??

#Physics #Mechanics

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2 Answers

29

govind ·2010-03-15 05:16:32

heat energy required for change of temperature of water from 25Â°C to 0Â°C = 25 * 300 *1 cal = 7500
heat energy liberated on conversion of ice to water at 0Â°C = 100 *80 =8000
Since 7500 < 8000..so not all ice will be converted into water...
And the temperature of the mixture will be 0Â°C....
Final situation
Amount of ice = 6.25 g
Amount of water = 393.75 g

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24

eureka123 ·2010-03-15 10:45:59

kk..thx

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Basic Integral Calculus Operators Symbols Matrix Cases

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