Heat and Thermodynamics

$P \propto {T^3} \Rightarrow P{T^{ - 3}} = $ constant ....$(i)$
But for an adiabatic process, the pressure temperature relationship is given by
${P^{1 - \gamma }}\,\,{T^\gamma } = $ constant $ \Rightarrow P{T^{{\gamma \over {1 - \gamma }}}} = $ constant. ....$(ii)$
From $(i)$ and $(ii)$ ${\gamma \over {1 - \gamma }} = - 3 \Rightarrow \gamma = - 3 + 3\gamma \Rightarrow \gamma = {3 \over 2}$

All reversible engines working for the same temperature of source and sink have same efficiencies. If the temperatures are different, the efficiency is different.

If ${n_1}$ moles of adiabatic exponent ${\gamma _1}$ is mixed with ${n_2}$ moles of adiabatic exponent ${\gamma _2}$ then the adiabatic component of the resulting mixture is given by

${{{n_1} + {n_2}} \over {\gamma - 1}} = {{{n_1}} \over {{\gamma _1} - 1}} + {{{n_2}} \over {{\gamma _2} - 1}}$
${{1 + 1} \over {\gamma - 1}} = {1 \over {{7 \over 5} - 1}} + {1 \over {{5 \over 3} - 1}}$

$\therefore$ ${2 \over {\gamma - 1}} = {5 \over 2} + {3 \over 2} = 4$

$\therefore$ $2 = 4\gamma - 4 \Rightarrow \gamma = {6 \over 4} = {3 \over 2}$

Pyrometer is used to detect infra-red radiation.

Black board paint is quite approximately equal to black bodies.

Heat required for raising the temperature of the whole body by ${1^ \circ }C$ is called the thermal capacity of the body.

${v_{rms}} = $$\sqrt {{{RT} \over M}} $

For ${v_{rms}}$ to be equal ${{{T_{{H_2}}}} \over {{M_{{H_2}}}}} = {{{T_{{O_2}}}} \over {{M_{{O_2}}}}}$

Here ${M_{{H_2}}} = 2;\,\,{M_{{O_2}}} = 32;$

${T_{{O_2}}} = 47 + 273 = 320K$

$\therefore$ ${{{T_{{H_2}}}} \over 2} = {{320} \over {32}} $

$\Rightarrow {T_{{H_2}}} = 20K$

When water is cooled to form ice, energy is released from water in the form of heat. As energy is equivalent to mass therefore when water is cooled to ice, its mass decreases.

$\eta = 1 - {{{T_2}} \over {{T_1}}}$
For $\eta = 1$ or $100\% $, ${T_2} = 0K.$
The temperature of $0$ $K$ (absolute zero) can not be obtained.

The energy radiated per second is given by $E = e\sigma {T^4}A$
For same material $e$ is same. $\sigma $ is stefan's constant
$\therefore$ ${{{E_1}} \over {{E_2}}} = {{T_1^4{A_1}} \over {T_2^4{A_2}}} = {{T_1^44\pi r_1^2} \over {T_2^44\pi r_2^2}}$
$ = {{{{\left( {4000} \right)}^4} \times {1^2}} \over {{{\left( {2000} \right)}^4} \times {4^2}}} = {1 \over 1}$

Since pressure and volume are not changing, so temperature remains same.