In the definition of 'calorie' one calorie is the heat required to raise the temperature of $1\  gram$ of water through $1\  ^oC$ in a certain interval of temperature. The temperature interval is

One end of a $2.35\,\,m$ long and $2.0\,\,cm$ radius aluminium rod$(K = 235 \,\,W.m^{-1}K^{-1})$ is held at $20^0\,\,C$. The other end of the rod is in contact with a block of ice at its melting point. The rate in $kg.s^{-1}$ at which ice melts is

[Take latent heat of fusion for ice as $\frac{{10}}{3} ×10^5 J.kg^{-1} $]

$50\, gm$ of copper is heated to increase its temperature by $10^oC$. If the same quantity of heat is given to $10\; gm$ of water, the rise in its temperature is ........ $^oC$ (Specific heat of copper $= 420 \;Joule-kg^{-1} {°C^{-1}}$)

$100\,g$ of water is supercooled to $-\,10\,^oC$. At this point, due to some disturbance mechanised or otherwise some of it suddenly freezes to ice. What will be the temperature of the resultant mixture and how much mass would freeze ? $[S_W = 1\,cal\,g^{-1}\,^oC^{-1}$ and ${L^W}_{{\text{fussion}}}$ $= 80\,cal\,g^{-1}]$

The specific heat of alcohol is about half that of water. Suppose you have identical masses of alcohol and water. The alcohol is initially at temperature $T_A$ . The water is initially at a different temperature $T_W$ . Now the two fluids are mixed in the same container and allowed to come into thermal equilibrium, with no loss of heat to the surroundings. The final temperature of the mixture will be 

$10\, gm$ of ice at $-20°C$ is dropped into a calorimeter containing $10\, gm$ of water at $10°C;$ the specific heat of water is twice that of ice. When equilibrium is reached, the calorimeter will contain