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 $= 336 \ kJ \ kg^{-1}$)

$A$ piece of metal weighing $100 \,g$ is heated to $80^{\circ} C$ and dropped into $1 \,kg$ of cold water in an insulated container at $15^{\circ} C$. If the final temperature of the water in the container is $15.69^{\circ} C$,the specific heat of the metal in $J / g^{\circ} C$ is:

$A$ small quantity of water of mass $m$ at temperature $\theta^{\circ} C$ is mixed with a large mass $M$ of ice which is at its melting point. If $s$ is the specific heat capacity of water and $L$ is the latent heat of fusion of ice,then the mass of ice melted is:

Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is held at a temperature of $100^{\circ} C$,while the other one is kept at $0^{\circ} C$. If the two are brought into contact,then assuming no heat loss to the environment,the final temperature that they will reach is

$A$ liquid of mass $m$ and specific heat $c$ is heated to a temperature $2T$. Another liquid of mass $m/2$ and specific heat $2c$ is heated to a temperature $T$. If these two liquids are mixed, the resulting temperature of the mixture is:

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