Physics - Heat & Thermo Question with Solution | TestHub

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)
$SL=1350S$
$\frac{L}{S}=270$