1.

What could be the finaltemperature ofa mixture of 100 g of water at 90^@C and 600 g of water at 20^@C

Answer»

Solution :To FIND final temperature :`DeltaQ=me`
100 g of WATER originally at `90^@C` will loose an amount of heat ,
`DeltaQ=me DeltaT`
`DeltaT=100 xx c xx (90-T)`
The same amount of heat will be absorbed by 600 g of water originally at `20^@C` to raise its temperature to T .
`DeltaQ`= 600 x c x (T-30)
`600 C (T-20^@)=100C (90^@-T)`
`6T-120^@=90^@-T`
`6T+T=120^@+90^@`
`7T= 210^@ rArr` T=210/7
`T=30^@C`


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