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

Three containers $C_{1}, C_{2}$ and $C_{3}$ have water at different temperatures. The table below shows the final temperature $T$ when different amounts of water (given in litres) are taken from each container and mixed (assume no loss of heat during the process).

$C_{1}$	$C_{2}$	$C_{3}$	$T$
$1 \ l$	$2 \ l$	$-$	$60^{\circ} C$
$-$	$1 \ l$	$2 \ l$	$30^{\circ} C$
$2 \ l$	$-$	$1 \ l$	$60^{\circ} C$
$1 \ l$	$1 \ l$	$1 \ l$	$\theta$

The value of $\theta$ (in $^{\circ} C$ to the nearest integer) is

$15 \text{ g}$ of ice at $0^{\circ}\text{C}$ is added to a vessel containing water at $40^{\circ}\text{C}$. The mass of water and water equivalent of the vessel is $60 \text{ g}$. Assuming that negligible heat is taken from the surroundings, the final temperature of the mixture will be $[L_{\text{ice}} = 80 \text{ cal/g}, S_{\text{water}} = 1 \text{ cal/g}]$ (in $^{\circ}\text{C}$)

Three bodies of the same material and having masses $m, m$ and $3m$ are at temperatures $40^{\circ} C, 50^{\circ} C$ and $60^{\circ} C$ respectively. If the bodies are brought in thermal contact, the final temperature will be (in $^{\circ} C$)

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

The temperature of water of mass $100 \ g$ is raised from $24^{\circ} C$ to $90^{\circ} C$ by adding steam to it. The mass of the steam added is (Latent heat of steam $= 540 \ cal \ g^{-1}$ and specific heat capacity of water $= 1 \ cal \ g^{-1} {}^{\circ} C^{-1}$) (in $g$)