It is possible for a substance to coexist in all three phases in equilibrium,when the substance is at

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 .......... $^\circ C$ $(J = 4.3 \, J/cal)$.

$A$ clock pendulum having a coefficient of linear expansion $\alpha = 9 \times 10^{-7} /^{\circ}C$ has a period of $0.5 \ s$ at $20^{\circ}C$. If the clock is used in a climate where the temperature is $30^{\circ}C$,how much time does the clock lose in each oscillation? (Assume $g$ is constant)

$M \ kg$ of water at $t^{\circ} C$ is divided into two parts so that one part of mass $m \ kg$ when converted into ice at $0^{\circ} C$ would release enough heat to vaporize the other part. Then $\frac{m}{M}$ is equal to (Specific heat of water $= 1 \ cal \ g^{-1} {}^{\circ} C^{-1}$,Latent heat of fusion of ice $= 80 \ cal \ g^{-1}$,Latent heat of steam $= 540 \ cal \ g^{-1}$)

If $60 \%$ of the kinetic energy of water falling from a $210 \ m$ high waterfall is converted into heat,the rise in temperature of the water at the bottom of the fall is nearly (specific heat of water $= 4.2 \times 10^3 \ J \ kg^{-1} \ K^{-1}$ and $g = 10 \ m/s^2$):

Heat required to convert $5\ kg$ of ice at $0\ ^oC$ into water at $100\ ^oC$ is: