$A$ hammer of mass $200 \ kg$ strikes a steel block of mass $200 \ g$ with a velocity $8 \ ms^{-1}$. If $23 \%$ of the energy is utilized to heat the steel block,the rise in temperature of the block is (specific heat capacity of steel,$s = 460 \ J \ kg^{-1} \ K^{-1}$) (in $K$)

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

$A$ $100 \, g$ 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 $= 0.42 \, Jg^{-1} {}^{\circ}C^{-1}$]

How much thermal energy is required to change a $40 \ g$ ice cube from solid at $-10^{\circ} C$ to steam at $110^{\circ} C$ (in $kcal$)?
[Assume, latent heat of fusion for water $= 80 \ kcal/kg$, specific heat of water $= 1 \ kcal/kg^{\circ} C$, specific heat of ice $= 0.5 \ kcal/kg^{\circ} C$, specific heat of steam $= 0.48 \ kcal/kg^{\circ} C$, latent heat of vaporisation of water $= 540 \ kcal/kg$]

$A$ thermocouple develops $40\,\mu V/K$. If the hot and cold junctions are at $40\,^{\circ}C$ and $20\,^{\circ}C$ respectively,then the emf developed by a thermopile using $150$ such thermocouples in series shall be ............... $mV$.

Calculate the heat required to convert $3 \; kg$ of ice at $-12 \; ^{\circ}C$ kept in a calorimeter to steam at $100 \; ^{\circ}C$ at atmospheric pressure. Given: specific heat capacity of ice $= 2100 \; J \; kg^{-1} \; K^{-1}$,specific heat capacity of water $= 4186 \; J \; kg^{-1} \; K^{-1}$,latent heat of fusion of ice $= 3.35 \times 10^{5} \; J \; kg^{-1}$,and latent heat of steam $= 2.256 \times 10^{6} \; J \; kg^{-1}$.