If there are no heat losses,the heat released by the condensation of $x \, g$ of steam at $100^\circ C$ into water at $100^\circ C$ can be used to convert $y \, g$ of ice at $0^\circ C$ into water at $100^\circ C$. Then the ratio $y : x$ is nearly (in $:1$)

The amount of heat needed to heat $200 \ g$ of ice at $-10^{\circ}C$ to convert it into water at $30^{\circ}C$ is:
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 $= 3.35 \times 10^5 \ J \ kg^{-1}$ (in $J$)

$A$ 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 ............... $mV$.

Find the quantity of heat required to convert $40 \; g$ of ice at $-20^{\circ} C$ into water at $20^{\circ} C$. Given $L_{\text{ice}} = 0.336 \times 10^6 \; J/kg$,specific heat of ice $= 2100 \; J/kg \cdot K$,and specific heat of water $= 4200 \; J/kg \cdot K$. (in $; J$)

Water falls from a height of $200 \ m$ into a pool. Calculate the rise in temperature of the water assuming no heat dissipation from the water in the pool. ($g = 10 \ m/s^2$,specific heat of water $s = 4200 \ J/(kg \ K)$) (in $K$)

$A$ given mass $m$ of a hypothetical solid is supplied with heat continuously at a constant rate and the graph shown in the figure is plotted. If $L_f$ and $L_v$ are latent heats of fusion and vaporization,and $S_l$ and $S_s$ are specific heats of liquid and solid respectively,it can be concluded that: