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The amount of heat required to convert 1 gnm of ice at- 10^(@)C to steam at 100^(@)C is: |
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Answer» 725 CAL `Q=mC_(1)DeltaT_(1)+mL_(1)+mC_(2)DeltaT_(2)+mL_(2).` `=1xx0 cdot 5xx10+1xx80+ bot xx1xx100+1xx540.` `=5+80+100+540=725 cal`. HENCE, correct choice is (a). |
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