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 $g = 10 \ m/s^2$):

$A$ $100 \, g$ of iron nail is hit by a $1.5 \, kg$ hammer striking at a velocity of $60 \, ms^{-1}$. What will be the rise in the temperature of the nail in $^{\circ}C$ if one-fourth of the energy of the hammer goes into heating the nail? [Specific heat capacity of iron $= 0.42 \, Jg^{-1} {}^{\circ}C^{-1}$]

$A$ thermoelectric emf of $200 \, \mu V$ is generated between $0 \, ^oC$ and $100 \, ^oC$. The emf generated between $(0 \, ^oC - 32 \, ^oC)$ and $(32 \, ^oC - 70 \, ^oC)$ are $64 \, \mu V$ and $76 \, \mu V$ respectively. What is the thermo $emf$ generated between $(70 \, ^oC - 100 \, ^oC)$ in $\mu V$?

An amount of ice of mass $10^{-3} \ kg$ and temperature $-10^{\circ} C$ is transformed to vapour of temperature $110^{\circ} C$ by applying heat. The total amount of heat required for this conversion is,(Take,specific heat of ice $= 2100 \ J \ kg^{-1} \ K^{-1}$,specific heat of water $= 4180 \ J \ kg^{-1} \ K^{-1}$,specific heat of steam $= 1920 \ J \ kg^{-1} \ K^{-1}$,Latent heat of ice $= 3.35 \times 10^5 \ J \ kg^{-1}$ and Latent heat of steam $= 2.25 \times 10^6 \ J \ kg^{-1}$) (in $J$)

Water falls from a height of $500\,m$. The rise in temperature of the water at the bottom,assuming all energy remains in the water,will be ........ $^oC$ [Specific heat of water = $4.2\,kJ/kg\cdot K$]

$22320 \ cal$ of heat is supplied to $100 \ g$ of ice at $0^{\circ} C$. If the latent heat of fusion of ice is $80 \ cal \ g^{-1}$ and latent heat of vaporization of water is $540 \ cal \ g^{-1}$, the final amount of water thus obtained and its temperature respectively are: