(1)

Gravitational potential energy of a piece of ice at a height $\left(h\right)=mgh$

Heat absorbed by the ice to melt completely

           $\Delta Q=\frac{1}{4}mgh$                                                                                 ...(i)

Also, $\Delta Q=mL$                                                                                         ...(ii)

From eqns. (i) and (ii), $mL=\frac{1}{4}mgh \text{or,} h=\frac{4L}{g}$

Here $L=3.4\times {10}^5{\text{J kg}}^{-1}\text{,}$ $g=10{\text{N kg}}^{-1}$

$\therefore$ $h=\frac{4\times 3.4\times {10}^5}{10}=4\times 34\times {10}^3=136$ $\text{km}$

(4)

Here,

Specific heat of water, $s_w=1$ $\text{cal}$ $g^{-1} °{\text{C}}^{-1}$

Latent heat of steam, $L_s=540$ $\text{cal}$ $g^{-1}$

Heat lost by $m$ g of steam at ${100}^{\mathrm{o}}C$ to change into water at $80°$C is

$Q_1=m{L}_s+m{s}_w\Delta T_w$

      $=m\times 540+m\times 1\times (100-80)$

       $=540m+20m=560m$

Heat gained by 20 g of water to change its temperature from $10°$C to $80°$C is

         $Q_2=m_w{s}_w\Delta T_w=20\times 1\times (80-10)=1400$

According to the principle of calorimetry, $Q_1=Q_2$

$\therefore$ $560m=1400 \Rightarrow m=2.5$ $\text{g}$

Total mass of water present

          $=(20+m)\text{g}=(20+2.5)\text{g}=22.5$ $\text{g}$