$540 \ g$ of ice at $0 \ ^\circ C$ is mixed with $540 \ g$ of water at $80 \ ^\circ C$. The final temperature of the mixture is

Water of volume $2\, L$ in a closed container is heated with a coil of $1\, kW$. While water is heated,the container loses energy at a rate of $160\, J/s$. In how much time will the temperature of water rise from $27\, ^\circ C$ to $77\, ^\circ C$? (Specific heat of water is $4.2\, kJ/kg\cdot K$ and that of the container is negligible)

$A$ thermos flask contains $250 \ g$ of coffee at $90^{\circ} C$. To this $20 \ g$ of milk at $5^{\circ} C$ is added. After equilibrium is established, the temperature of the liquid is (Assume no heat loss to the thermos bottle. Take specific heat of coffee and milk as $1.00 \ cal/g^{\circ} C$) (in $^{\circ} C$)

$A$ liquid of mass $m$ and specific heat $c$ is at a temperature of $2T$. Another liquid of mass $m/2$ and specific heat $2c$ is at a temperature of $T$. What will be the final temperature of the mixture?

The temperatures of equal masses of three different liquids $A, B$ and $C$ are $12^{\circ}C, 19^{\circ}C$ and $28^{\circ}C$ respectively. The temperature when $A$ and $B$ are mixed is $16^{\circ}C$ and when $B$ and $C$ are mixed is $23^{\circ}C$. The temperature when $A$ and $C$ are mixed is ......... $^{\circ}C$.

$5 \ kg$ of water at $20^{\circ} C$ is added to $10 \ kg$ of water at $60^{\circ} C$. Neglecting the heat capacity of the vessel and other losses,the resultant temperature will be nearly: (in $^{\circ} C$)