A liquid at ${30\mathrm{ }}^{\mathrm{o}}\mathrm{C}$ is poured very slowly into a Calorimeter that is at temperature of ${110\mathrm{ }}^{\mathrm{o}}\mathrm{C}.$ The boiling temperature of the liquid is ${80\mathrm{ }}^{\mathrm{o}}\mathrm{C}.$ It is found that the first $5\mathrm{ }\mathrm{g}\mathrm{m}$ of the liquid completely evaporates. After pouring another $80\mathrm{ }\mathrm{g}\mathrm{m}$ of the liquid the equilibrium temperature is found to be ${50\mathrm{ }}^{\mathrm{o}}\mathrm{C}.$ The ratio of the Latent heat of the liquid to its specific heat will be ______ ${\mathrm{ }}^{\mathrm{o}}\mathrm{C}.$
[Neglect the heat exchange with surrounding]

Note: The information about condition of calorimeter is not given in question, so following two cases arise -
Case $1$ : If calorimeter is closed (vapour not allowed to escape)
Heat gain $=$ Heat loss
$5S(80-30)+5L=W(110-80)\to$ (as first $5 \mathrm{g}\mathrm{m}$ liquid is evaporated)
$S=$ Specific heat of liquid
$L=$ Latent heat of liquid
W=Water equivalent of calorimeter
$\Rightarrow 250S+5L=W\times 30$ ...(i)
Now $80\mathrm{ }\mathrm{g}\mathrm{m}$ liquid is poured,
Heat gain $=$ Heat loss
Here final temperature $=50°C$
$\therefore 80\times S\times 20=5L+5S\times 30+W\times 30$ ...(ii)
From (i) and (ii)
$\frac{L}{S}=120\mathrm{A}\mathrm{n}\mathrm{s}\mathrm{w}\mathrm{e}\mathrm{r}$
Case $2$ : If calorimeter is open and after evaporation liquid escapes
$5\times S\times 50+5L=W\times 30$ ...(i)
(as first $5\mathrm{ }\mathrm{g}\mathrm{m}$ liquid is evaporated)
$80\times S\times 20=W\times 30$ ...(ii)
(after pouring $80\mathrm{ }\mathrm{g}\mathrm{m}\mathrm{ }$ liquid, the equilibrium temperature is $50°C$
$\Rightarrow 80\times S\times 20=S\times S\times 50+5L$ (using (i) and (ii)
$SL=1350S$
$\frac{L}{S}=270$