What amount of ice will remains when 52 g ice is added to 100 g of water at `40^(@)C`?
Specific heat of water is 1 cal/g and latent heat of fusion of ice is 80 cal/g.
Correct Answer - 2 g ice
At the stage of thermal equilibrium at `0^(@)C`
heat lost by water=Heat absorbed by ice to melt.
`msDeltaT=mL`
`100xx1xx40=mxx80`
`m=50g`
Remaining ice`=52-50=2g`
Since the density of hydrogen is `0.0899 g cm^(-3)` therefore volume occupied by `0.0899 `g of hydrogen at `NTP` is `1cm^(3)` . So, volume of `1` mole `(2.016g)` of gas...
Correct Answer - B
`DeltaQ_("top") = DeltaQ_("freezing")`
`m.(etaL) = M(L) implies M = etam`
L = latent heat of freezing
m = mass of vapour
`M` = mass of freezed
`therefore...
(i) `C_(P)-C_(V)=(R)/(M)` where, M=molecular of gas
`0.125-0.075=(1.987)/(M)`
`M=39.74~~40`
(ii) `(C_(P))/(C_(V))=gamma`
`therefore=(0.125)/(0.075)=1.66`
`therefore` 1.66 value of `gamma` shows that the gas is monoatomic.