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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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