$500\, g$ of water and $100\, g$ of ice at $0\,^oC$ are in a calorimeter whose water equivalent is $40\, g$. $10\, g$ of steam at $100\,^oC$ is added to it. Then water in the calorimeter is ....... $g$ (Latent heat of ice $\,= 80\, cal/g$, Latent heat of steam $\,= 540\, cal/ g$)

A $2\,kg$ copper block is heated to $500^o\,C$ and then it is placed on a large block of ice  at $0^o\,C$. If the specific heat capacity of copper is $400\, J/kg/ ^o\,C$ and latent heat of  fusion of water is $3.5 \times 10^5\, J/kg$, the amount of ice, that can melt is :-

A bullet of mass $10 \,g$ moving with a speed of $20 \,m / s$ hits an ice block of mass $990 \,g$ kept on a frictionless floor and gets stuck in it. How much ice will melt if $50 \%$ of the lost KE goes to ice is .......... $g$ (initial temperature of the ice block and bullet $=0^{\circ} C$ )

A calorimeter of water equivalent $20\, g$ contains $180\, g$ of water at $25^{\circ} C$. '$m$' grams of steam at $100^{\circ} C$ is mixed in it till the temperature of the mixure is $31^{\circ} C$. The value of $'m'$ is close to

(Latent heat of water $=540$ cal $g ^{-1}$, specific heat of water $=1$ cal $g^{-1}{ }^{\circ} C ^{-1}$ )

$10\; gm$ of ice cubes at $0\;^{\circ} C$ are released in a tumbler (water equivalent $55\; g$ ) at $40\;^{\circ} C$. Assuming that negligible heat is taken from the surroundings, the temperature(in $^o C$) of water in the tumbler becomes nearely $(L_f=80\; cal / g )$

$100 \,gm$ of ice at $0°C$ is mixed with $100\, g$ of water at $100°C.$ What will be the final temperature of the mixture .......... $^oC$