A perfectly black body emits a radiation at temperature 'T$$_1$$' K. If it is to radiate at 16 times this power, its temperature 'T$$_2$$' K should be

One mole of an ideal gas expands adiabatically at constant pressure such that its temperature $$T \propto {1 \over {\sqrt V }}$$. The value of $$\gamma$$ for the gas is ($$\gamma = {{{C_p}} \over {{C_v}}},V = $$ Volume of the gas)

On an imaginary linear scale of temperature (called 'W' scale) the freezing and boiling points of water are 39$$^\circ$$ W and 239$$^\circ$$ W respectively. The temperature on the new scale corresponding to 39$$^\circ$$C temperature on Celsius scale will be

Specific heats of an ideal gas at constant pressure and volume are denoted by $$\mathrm{C}_{\mathrm{p}}$$ and $$\mathrm{C}_{\mathrm{v}}$$ respectively. If $$\gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}$$ and $$\mathrm{R}$$ it's the universal gas constant then $$\mathrm{C}_{\mathrm{v}}$$ is equal to