When $54 \ g$ of ice at $-20^{\circ} C$ is mixed with $25 \ g$ of steam at $100^{\circ} C$, then the final mixture at thermal equilibrium contains

$A$ metal ball immersed in alcohol weighs $W_1$ at $0^{\circ}C$ and $W_2$ at $59^{\circ}C$. The coefficient of cubical expansion of the metal is less than that of alcohol. Assuming that the density of metal is large compared to that of alcohol,it can be shown that

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

'Stem correction' in platinum resistance thermometers is eliminated by the use of

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 $^{\circ} C$)

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 $= 2.256 \times 10^{6} \; J \; kg^{-1}$.