The heat required to convert $8 \ g$ of ice at a temperature of $-20^{\circ}C$ to steam at $100^{\circ}C$ is (Specific heat capacity of ice $= 2100 \ J \ kg^{-1} \ K^{-1}$, specific heat capacity of water $= 4200 \ J \ kg^{-1} \ K^{-1}$, latent heat of fusion of ice $= 336 \times 10^3 \ J \ kg^{-1}$ and latent heat of steam $= 2.268 \times 10^6 \ J \ kg^{-1}$) (in $kJ$)

$A$ steel pendulum clock manufactured at $32^{\circ} C$ and working at $47^{\circ} C$ is nearly (Coefficient of linear expansion of steel $= 12 \times 10^{-6} /{ }^{\circ} C$)

$A$ cylinder of mass $1\,kg$ is given heat of $20000\,J$ at atmospheric pressure. If the initial temperature of the cylinder is $20\,^oC$,then the work done by the cylinder will be .......$J$. (Given: Specific heat of cylinder $= 400\,J\,kg^{-1}\,^oC^{-1}$,Coefficient of volume expansion $= 9 \times 10^{-5}\,^oC^{-1}$,Atmospheric pressure $= 10^5\,N/m^2$,and density of cylinder $= 9000\,kg/m^3$)

$A$ mass $m$ of material exists in its solid state at its melting temperature $0^\circ C$. The following processes occur to the material:
Process-$1$: An amount of thermal energy $Q$ is added to the material, and $\frac{2}{3}$ of the material melts.
Process-$2$: An identical additional amount of thermal energy $Q$ is added to the material, and the material is now a liquid at $40^\circ C$.
What is the ratio of the latent heat of fusion $(L)$ to the specific heat of the liquid $(s)$ for this material?

$A$ lead bullet of mass $50 \ g$ at a temperature of $30^{\circ}C$ is fired vertically upwards with a speed of $840 \ m/s$. When the bullet returns to the point of departure,it strikes a large block of ice at $0^{\circ}C$. How much ice (in $g$) will melt? (Specific heat of lead is $0.02 \ cal/g^{\circ}C$,latent heat of fusion of ice is $80 \ cal/g$,and assume all energy is used to melt the ice.)

The height of a waterfall is $50 \ m$. If $g = 9.8 \ m/s^2$, the difference between the temperature at the top and the bottom of the waterfall is: (in $^{\circ} C$)