$2 \ kg$ of ice at $-20^{\circ} C$ is mixed with $5 \ kg$ of water at $20^{\circ} C$. The final mass of water formed is: (in $kg$)

$A$ calorimeter contains $0.5 \,kg$ of water at $30^{\circ} C$. When $0.3 \,kg$ of water at $60^{\circ} C$ is added to it, the resulting temperature is found to be $40^{\circ} C$. The water equivalent of the calorimeter is (in $\,kg$)

One kilogram of ice at $0^{\circ}C$ is mixed with one kilogram of water at $80^{\circ}C$. The final temperature of the mixture is ........ $^{\circ}C$. (Take: specific heat of water $= 4200 \ J \ kg^{-1} \ K^{-1}$,latent heat of ice $= 336 \ kJ \ kg^{-1}$)

$A$ glass beaker contains $200 \,g$ of carbonated water initially at $20^{\circ} C$. How much ice should be added to obtain the final temperature of $0^{\circ} C$ with all ice melted, if the initial temperature of ice is $-10^{\circ} C$ (in $\,g$)? Neglect the heat capacity of the glass.
[Take, $C_{\text{water}} = 4190 \,J/kg^{\circ} C$, $C_{\text{ice}} = 2100 \,J/kg^{\circ} C$, $L_F = 3.34 \times 10^5 \,J/kg$]

If equal masses of $10$ liquids of specific heats $s, 2s, 3s, \ldots, 10s$ at temperatures $10^{\circ} C, 20^{\circ} C, 30^{\circ} C, \ldots, 100^{\circ} C$ respectively are mixed,the resultant temperature of the mixture is . . . . . . . (in $^{\circ} C$)

The temperatures of two liquids $A$ and $B$ are $32 \, ^\circ C$ and $24 \, ^\circ C$ respectively. When equal masses of both are mixed,the temperature of the mixture becomes $28 \, ^\circ C$. The ratio of their specific heats is .....