Find the difference in temperature between the water at the top and the bottom of a $20 \ m$ high waterfall,assuming $10 \%$ of the energy of the fall is spent in heating the water. [Use specific heat capacity of water $= 4000 \ J \ kg^{-1} K^{-1}$ and $g = 10 \ m/s^2$]. (in $^{\circ} C$)

$A$ 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 $1000 \text{ kg m}^{-3}$)

If there are no heat losses,the heat released by the condensation of $x \, g$ of steam at $100^\circ C$ into water at $100^\circ C$ can be used to convert $y \, g$ of ice at $0^\circ C$ into water at $100^\circ C$. Then the ratio $y : x$ is nearly (in $:1$)

$A$ metal sphere immersed in water weighs $w_1$ at $0^{\circ} C$ and $w_2$ at $50^{\circ} C$. The coefficient of cubical expansion of the metal is less than that of water. Then

How many $J$ of heat energy is required to convert $1\,g$ of ice at $-10^{\circ}C$ into steam at $100^{\circ}C$?

An aluminium container of mass $100 \, g$ contains $200 \, g$ of ice at $-20^{\circ} C$. Heat is added to the system at the rate of $100 \, cal/s$. The temperature of the system after $4$ minutes will be ....... $^{\circ} C$ (specific heat of ice $= 0.5 \, cal/g^{\circ} C$,latent heat of fusion $L = 80 \, cal/g$,specific heat of $Al = 0.2 \, cal/g^{\circ} C$,specific heat of water $= 1 \, cal/g^{\circ} C$)