When $M_1$ gram of ice at $-10\,^oC$ (specific heat $= 0.5\, cal\, g^{-1}\,^oC^{-1}$) is added to $M_2$ gram of water at $50\,^oC$, finally no ice is left and the water is at $0\,^oC$. The value of latent heat of ice, in $cal\, g^{-1}$ is
$\frac{{50{M_2}}}{{{M_1}}} - 5$
$\frac{{5{M_2}}}{{{M_1}}} - 5$
$\frac{{50{M_2}}}{{{M_1}}}$
$\frac{{5{M_1}}}{{{M_2}}} - 50$
What is isolated system ?
The saturation vapour pressure of water at $100°C$ is ........ $mm$ of mercury
A metal rod $\mathrm{AB}$ of length $10 x$ has its one end $\mathrm{A}$ in ice at $0^{\circ} \mathrm{C}$ and the other end $\mathrm{B}$ in water at $100^{\circ} \mathrm{C}$. If a point $\mathrm{P}$ on the rod is maintained at $400^{\circ} \mathrm{C}$, then it is found that equal amounts of water and ice evaporate and melt per unit time. The latent heat of evaporation of water is $540 \ \mathrm{cal} / \mathrm{g}$ and latent heat of melting of ice is $80 \ \mathrm{cal} / \mathrm{g}$. If the point $\mathrm{P}$ is at a distance of $\lambda x$ from the ice end $\mathrm{A}$, find the value of $\lambda$.
[Neglect any heat loss to the surrounding.|
The temperature at which the vapour pressure of a liquid becomes equals to the external (atmospheric) pressure is its