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`4 gm` of steam at `100^(@)C` is added to `20 gm` of water at `46^(@)C` in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of water in container at thermal equilibrium is. Latent heat of vaporisation `= 540 cal//gm`. Specific heat of water `=1 cal//gm-^(@)C`.A. 18 gmB. 20 gmC. 22 gmD. 24 gm |
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Answer» Correct Answer - C Heat released by steam inconversion to water at `100^(@)C` is `Q_(1)=mL=4xx540=2160 cal` Heat required to raise temperature of water from `46^(@)C` to `100^(@)C` is `Q_(2)=mS Deltatheta=20xx1xx54=1080 J` `Q_(1) gt Q_(2)` and `(Q_(1))/(Q_(2))=2` Hence all steam is not converted to water only half steam shall be converted to water `:.` final mass of water =20+2=22 gm |
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