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 $^{\circ} C$)

$200 \, g$ of a solid ball at $20 \, ^\circ C$ is dropped in an equal amount of water at $80 \, ^\circ C$. The resulting temperature is $60 \, ^\circ C$. This means that the specific heat of the solid 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:

$50\, g$ of ice at $0\,^{\circ}C$ is dropped into a calorimeter containing $100\, g$ of water at $30\,^{\circ}C$. If the thermal capacity of the calorimeter is zero,then the amount of ice left in the mixture at equilibrium is ........ $g$.

$300 \text{ g}$ of water at $25^{\circ}C$ is added to $100 \text{ g}$ of ice at $0^{\circ}C$. The final temperature of the mixture is (in $^{\circ}C$)

$A$ piece of ice (heat capacity $= 2100 \ J \ kg^{-1} \ ^{\circ}C^{-1}$ and latent heat $= 3.36 \times 10^5 \ J \ kg^{-1}$) of mass $m$ grams is at $-5^{\circ}C$ at atmospheric pressure. It is given $420 \ J$ of heat so that the ice starts melting. Finally,when the ice-water mixture is in equilibrium,it is found that $1 \ g$ of ice has melted. Assuming there is no other heat exchange in the process,the value of $m$ is: