Work done (heat energy required) in converting $1\,g$ of ice at $-10\,^{\circ}C$ into steam at $100\,^{\circ}C$ is ......... $J$. (Take specific heat of ice $= 0.5\,cal/g^{\circ}C$,latent heat of fusion $= 80\,cal/g$,specific heat of water $= 1\,cal/g^{\circ}C$,latent heat of vaporization $= 540\,cal/g$,and $1\,cal = 4.2\,J$)

$A$ steel ball of mass $200 \,g$ falls freely from a height of $20 \,m$ and bounces to a height of $10.8 \,m$ from the ground. If the energy lost in this process is absorbed by the ball, the rise in its temperature is ($g = 10 \,ms^{-2}$, specific heat capacity of steel is $460 \,Jkg^{-1} K^{-1}$). (in $^{\circ} C$)

$22320 \ cal$ 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:

$M \ kg$ 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 in the winter season the surface temperature of a lake is $1^{\circ} C$,the temperature at the bottom of the lake will be ............

During summers in India,one of the common practices to keep cool is to make ice balls of crushed ice,dip them in flavoured sugar syrup,and sip them. For this,a stick is inserted into crushed ice,and it is squeezed in the palm to make it into a ball. Equivalently in winter,in those areas where it snows,people make snowballs and throw them around. Explain the formation of a ball out of crushed ice or snow in the light of the $p-T$ diagram of water.