What could be the final temperature of a mixture of 100 g of water at `90^@C` and 600 g of water at `20^@C`
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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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