$A$ block of steel of mass $2 \,kg$ slides down a rough inclined plane of inclination $\sin ^{-1}\left(\frac{3}{5}\right)$ at a constant speed. Assuming that the mechanical energy lost due to friction is used to increase the temperature of the block, calculate the rise in temperature of the block as it slides through $80 \,cm$. (Specific heat capacity of steel $= 420 \,J kg^{-1} K^{-1}$ and acceleration due to gravity $= 10 \,ms^{-2}$) (in $^{\circ} C$)

The amount of heat required to change $1 \, g$ of ice at $0^\circ C$ into water at $100^\circ C$ is ............ $cal$.

Heat energy of $184\,kJ$ is given to ice of mass $600\,g$ at $-12^{\circ}\,C$. The specific heat of ice is $2222.3\,J\,kg^{-1\circ}C^{-1}$ and the latent heat of fusion of ice is $336\,kJ\,kg^{-1}$.
$(A)$ Final temperature of the system will be $0^{\circ}C$.
$(B)$ Final temperature of the system will be greater than $0^{\circ}C$.
$(C)$ The final system will have a mixture of ice and water in the ratio of $5:1$.
$(D)$ The final system will have a mixture of ice and water in the ratio of $1:5$.
$(E)$ The final system will have water only.
Choose the correct answer from the options given below:

$250\,g$ of water and an equal volume of alcohol of mass $200\,g$ are placed successively in the same calorimeter and cool from $60^{\circ}C$ to $55^{\circ}C$ in $130\,s$ and $67\,s$ respectively. If the water equivalent of the calorimeter is $10\,g$,then the specific heat of alcohol in $\text{cal/g}^{\circ}C$ is:

$A$ piece of ice (heat capacity = $2100 \text{ J kg}^{-1} \text{ }^\circ\text{C}^{-1}$ and latent heat = $3.36 \times 10^5 \ J kg^{-1}$) of mass $m$ grams is at $-5^oC$ 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 $1 \ gm$ of ice has melted. Assuming there is no other heat exchange in the process, the value of $m$ is ...... $gm$.

Find the amount of heat supplied to decrease the volume of an ice-water mixture by $1 \, cm^3$ without any change in temperature. (Given: $\rho_{ice} = 0.9 \rho_{water}$,$L_{ice} = 80 \, cal/g$).