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 = $4.2\,kJ/kg\cdot K$]

The temperature of a body is given by $T = 6t^2 + 4$. The rate of change of temperature with respect to time at $t = 10 \ s$ is ........ $K/s$. ($T$ is in $Kelvin$ and $t$ is in $seconds$)

The temperature of Bhakhra dam water at the ground level with respect to the temperature at a high level should be:

Water falls from a height of $500 \ m$. What is the rise in temperature of water at the bottom if the whole energy remains in the water? (Assume $g = 9.8 \ m/s^2$ and specific heat capacity of water $c = 4200 \ J/kg \cdot ^\circ C$)

$M \ kg$ 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 $= 540 \ cal \ g^{-1}$)

How much energy will be released by cooling $1\,kg$ of water up to a temperature of $10\,^{\circ}C$?