$A$ 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 $\,kg$)

$A$ $2 \, kg$ block of ice at $-20^{\circ}C$ is added to $5 \, kg$ of water at $20^{\circ}C$. What will be the total mass of water in $kg$? (Specific heat of water = $1 \, kcal/kg/^{\circ}C$,specific heat of ice = $0.5 \, kcal/kg/^{\circ}C$,latent heat of fusion of ice = $80 \, kcal/kg$)

$A$ 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, $C_{\text{water}} = 4190 \,J/kg^{\circ} C$, $C_{\text{ice}} = 2100 \,J/kg^{\circ} C$, $L_F = 3.34 \times 10^5 \,J/kg$]

When $100\,g$ of a liquid $A$ at $100\,^oC$ is added to $50\,g$ of a liquid $B$ at temperature $75\,^oC$,the temperature of the mixture becomes $90\,^oC$. The temperature of the mixture,if $100\,g$ of liquid $A$ at $100\,^oC$ is added to $50\,g$ of liquid $B$ at $50\,^oC$,will be ........$^oC$

$A$ 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

$A$ metal ball of mass $100 \ g$ at $20^{\circ} C$ is dropped in $200 \ ml$ of water at $80^{\circ} C$. If the resultant temperature is $70^{\circ} C$,then the ratio of specific heat of the metal to that of water is