The water equivalent of $20 \,g$ of aluminium (specific heat $0.2 \,cal \,g^{-1} {}^{\circ}C^{-1}$) is ......... $g$.

Three copper blocks of masses $M_1, M_2$ and $M_3$ $kg$ respectively are brought into thermal contact until they reach equilibrium. Before contact,they were at temperatures $T_1, T_2, T_3$ $(T_1 > T_3)$. Assuming there is no heat loss to the surroundings,the equilibrium temperature $T$ is ($s$ is the specific heat of copper).

$100 \, g$ of water is supercooled to $-10 \, ^\circ C$. At this point,due to some disturbance,some of it suddenly freezes to ice. What will be the temperature of the resultant mixture and how much mass would freeze? $[S_W = 1 \, cal \, g^{-1} \, ^\circ C^{-1}$ and $L_{fusion} = 80 \, cal \, g^{-1}]$

$10\,g$ of ice at $-20\,^{\circ}C$ is dropped into a calorimeter containing $10\,g$ of water at $10\,^{\circ}C$. The specific heat of water is twice that of ice. When equilibrium is reached,the calorimeter will contain

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

$10 \, g$ of ice at $0 \, ^oC$ is mixed with $m \, g$ of water at $50 \, ^oC$. What is the minimum value of $m$ (in $g$) so that the ice melts completely? (Given: $L_f = 80 \, cal/g$ and $S_W = 1 \, cal/g \cdot ^oC$)