1.

The amount of heat required to convert 1 gnm of ice at- 10^(@)C to steam at 100^(@)C is:

Answer»

725 CAL
1000 cal
800 cal
80 cal

Solution :Heat required `Q=Q_(1)+Q_(2)+Q_(3)+Q_(4)`.
`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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