Work done in converting one gram of ice at $-10^{\circ}C$ into steam at $100^{\circ}C$ is ....... $J$

$A$ thermoelectric emf of $200 \, \mu V$ is generated between $0 \, ^oC$ and $100 \, ^oC$. The emf generated between $(0 \, ^oC - 32 \, ^oC)$ and $(32 \, ^oC - 70 \, ^oC)$ are $64 \, \mu V$ and $76 \, \mu V$ respectively. What is the thermo $emf$ generated between $(70 \, ^oC - 100 \, ^oC)$ in $\mu V$?

$A$ mass $m$ of material exists in its solid state at its melting temperature $0^\circ C$. The following processes occur to the material:
Process-$1$: An amount of thermal energy $Q$ is added to the material, and $\frac{2}{3}$ of the material melts.
Process-$2$: An identical additional amount of thermal energy $Q$ is added to the material, and the material is now a liquid at $40^\circ C$.
What is the ratio of the latent heat of fusion $(L)$ to the specific heat of the liquid $(s)$ for this material?

The thermo $e.m.f.$ of a thermocouple varies with the temperature $\theta$ of the hot junction as $E = a\theta + b\theta^2$ in volts,where the ratio $a/b$ is $700\,^{\circ}C$. If the cold junction is kept at $0\,^{\circ}C$,then the neutral temperature is:

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 $J$)

$Assertion :$ In the pressure-temperature $(P-T)$ phase diagram of water,the slope of the melting curve is found to be negative.
$Reason :$ Ice contracts on melting to water.