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 $^{\circ} C$)

If an electric heater is rated at $1000\, W$, then the time required to heat one litre of water from $20\, ^oC$ to $60\, ^oC$ is

The water equivalent of a calorimeter is $10 \ g$ and it contains $50 \ g$ of water at $15^{\circ} C$. Some amount of ice, initially at $-10^{\circ} C$, is dropped in it and half of the ice melts till equilibrium is reached. What was the initial amount of ice that was dropped (given specific heat of ice $= 0.5 \ cal \ g^{-1} {}^{\circ} C^{-1}$, specific heat of water $= 1.0 \ cal \ g^{-1} {}^{\circ} C^{-1}$ and latent heat of melting of ice $= 80 \ cal \ g^{-1}$) (in $g$)?

$A$ 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:

$A$ liquid of mass $2m$ and specific heat $C$ is heated to a temperature $4T$. Another liquid of mass $m$ and specific heat $2C$ is heated to a temperature $T$. If these two liquids are mixed,the resulting temperature of the mixture 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: