$A$ steel ball of mass $0.1 \, kg$ falls freely from a height of $10 \, m$ and bounces to a height of $5.4 \, m$ from the ground. If the dissipated energy in this process is absorbed by the ball,the rise in its temperature is ........... $^\circ C$ (Specific heat of steel $= 460 \, J \, kg^{-1} \, ^\circ C^{-1}, \; g = 10 \, m \, s^{-2}$)

Two different liquids of same mass are kept in two identical vessels,which are placed in a freezer that extracts heat from them at the same rate,causing each liquid to transform into a solid. The schematic figure below shows the temperature $T$ versus time $t$ plot for the two materials. We denote the specific heat of materials in the liquid (solid) states to be $C_{L1}$ $(C_{S1})$ and $C_{L2}$ $(C_{S2})$,respectively. Choose the correct option given below.

Water falls from a height of $210 \, m$. Assuming the whole of the energy due to the fall is converted into heat,the rise in temperature of water would be .......... $^\circ C$ $(J = 4.3 \, J/cal)$.

$A$ water cooler of storage capacity $120$ litres can cool water at a constant rate of $P$ watts. In a closed circulation system (as shown schematically in the figure),the water from the cooler is used to cool an external device that generates constantly $3 \text{ kW}$ of heat (thermal load). The temperature of water fed into the device cannot exceed $30^{\circ}\text{C}$ and the entire stored $120$ litres of water is initially cooled to $10^{\circ}\text{C}$. The entire system is thermally insulated. The minimum value of $P$ (in watts) for which the device can be operated for $3$ hours is (Specific heat of water is $4.2 \text{ kJ kg}^{-1} \text{K}^{-1}$ and the density of water is $1000 \text{ kg m}^{-3}$)

Calculate the heat required to convert $3 \; kg$ of ice at $-12 \; ^{\circ}C$ kept in a calorimeter to steam at $100 \; ^{\circ}C$ at atmospheric pressure. Given: specific heat capacity of ice $= 2100 \; J \; kg^{-1} \; K^{-1}$,specific heat capacity of water $= 4186 \; J \; kg^{-1} \; K^{-1}$,latent heat of fusion of ice $= 3.35 \times 10^{5} \; J \; kg^{-1}$,and latent heat of steam $= 2.256 \times 10^{6} \; J \; kg^{-1}$.

$A$ 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 $^{\circ} C$)