Calculate the amount of heat (in calories) required to convert $5\,gm$ of ice at $0°C$ to steam at $100°C$

$500\, g$ 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 $\,= 540\, cal/ g$)

Find the quantity of heat required to convert $40\; gm$ of ice at $-20^{\circ} C$ into water at $20^{\circ} C$. Given $L _{\text {ice }}$ $=0.336 \times 10^6 J / kg$.

specific heat of ice $=2100 \;J / kg – K$ sp heat of water= $4200\; J / kg – K$

A heater supplying constant power $P$ watts is switched $ON$ at time $t=0 \,min$ to raise the temperature of a liquid kept in a calorimeter of negligible heat capacity. A student records the temperature of the liquid $T(t)$ at equal time intervals. A graph is plotted with $T(t)$ on the $Y$-axis versus $t$ on the $X$-axis. Assume that there is no heat loss to the surroundings during heating. Then,

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 :-