If mass energy equivalence is taken into account, when water is cooled to form ice, the mass of water should

We have half a bucket $(6l)$ of water at $20\,^oC$  . If we want water at $40\,^oC$, how much steam at $100\,^oC$ should be added to it ?

Heat needed to convert $1$ kilogram ice at $0^o\,C$ to steam at $100^o\,C$ is

$1 \,kg$ of ice at $-20^{\circ} C$ is mixed with $2 \,kg$ of water at $90^{\circ} C$. Assuming that there is no loss of energy to the environment, the final temperature of the mixture is ………… $^{\circ} C$ (Assume, latent heat of ice $=334.4 \,kJ / kg$, specific heat of water and ice are $4.18 \,kJ kg ^{-1} K ^{-1}$ and $2.09 \,kJ kg ^{-1}- K ^{-1}$, respectively.)

A liquid of mass $M$ and specific heat $S$ is at a temperature $2t$. If another liquid of thermal capacity $1.5$ times, at a temperature of $\frac{t}{3}$ is added to it, the resultant temperature will be