NEET UG 2008 — Thermodynamics

If $Q, E$ and $W$ denote respectively the heat added, change in internal energy and the work done in a closed cycle process, then

An electric kettle takes $4 \mathrm{~A}$ current at $220 \mathrm{~V}$. How much time will it take to boil $1 \mathrm{~kg}$ of water from temperature $20^{\circ} \mathrm{C}$ ? The temperature of boiling water is $100^{\circ} \mathrm{C}$.

At $10^{\circ} \mathrm{C}$ the value of the density of a fixed mass of an ideal gas divided by its pressure is $\mathrm{x}$. At $110^{\circ} \mathrm{C}$ this ratio is

At $10^{\circ} \mathrm{C}$ the value of the density of a fixed mass of an ideal gas divided by its pressure is $x$. At $110^{\circ} \mathrm{C}$ this ratio is

An electric kettle takes $4 \mathrm{~A}$ current at $220 \mathrm{~V}$. How much time will it take to boil $1 \mathrm{~kg}$ of water from temperature $20^{\circ} \mathrm{C}$ ? The temperature of boiling water is $100^{\circ} \mathrm{C}$

If $\mathrm{Q}, \mathrm{E}$ and $\mathrm{W}$ denote respectively the heat added, change in internal energy and the work done in a closed cycle process, then

On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water are $39^{\circ} \mathrm{W}$ and $239^{\circ} \mathrm{W}$ respectively. What will be the temperature on the new scale, corresponding to a temperature of $39^{\circ} \mathrm{C}$ on the Celsius scale?

Оп a new scale of temperature (which is linear) and called the $W$ scale, the freezing and boiling points of water are $39^{\circ} \mathrm{W}$ and $239^{\circ} \mathrm{W}$ respectively. What will be the temperature on the new scale, corresponding to a temperature of $39^{\circ} \mathrm{C}$ on the Celsius scale?