'Stem correction' in platinum resistance thermometers is eliminated by the use of

An amount of ice of mass $10^{-3} \ kg$ and temperature $-10^{\circ} C$ is transformed to vapour of temperature $110^{\circ} C$ by applying heat. The total amount of heat required for this conversion is,(Take,specific heat of ice $= 2100 \ J \ kg^{-1} \ K^{-1}$,specific heat of water $= 4180 \ J \ kg^{-1} \ K^{-1}$,specific heat of steam $= 1920 \ J \ kg^{-1} \ K^{-1}$,Latent heat of ice $= 3.35 \times 10^5 \ J \ kg^{-1}$ and Latent heat of steam $= 2.25 \times 10^6 \ J \ kg^{-1}$) (in $J$)

$A$ beaker is filled with water at $4\,^{\circ}C$. At one time the temperature is increased by a few degrees above $4\,^{\circ}C$ and at another time it is decreased by a few degrees below $4\,^{\circ}C$. One shall observe that:

In a thermocouple,if the thermo $EMF$ is given by $E = 40\theta - \frac{\theta^2}{20}$,then the neutral temperature will be .......... $^oC$.

The height of a waterfall is $50 \ m$. If $g = 9.8 \ m/s^2$,the difference between the temperature at the top and the bottom of the waterfall is: (in $^{\circ} C$)

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 $; J$)