Heat and Thermodynamics

Efficiency is maximum in Carnot engine which is an ideal engine.

$ \because \eta = {{400 - 300} \over {400}} \times 100\% = 25\% $

$ \therefore $ efficiency 26% is impossible for his heat engine.

$\gamma = 1 + {2 \over f}$

where f is the degree of freedom

$ \therefore $ ${2 \over f} = \gamma - 1 \Rightarrow f = {2 \over {\gamma - 1}}$

Initially the efficiency of the engine was ${1 \over 6}$ which increases to ${1 \over 3}$ when the sink temperature reduces by 62º C.

$\eta = {1 \over 6} = 1 - {{{T_2}} \over {{T_1}}}$ when T₂ = sink temperature , T₁ = source temperature

$ \Rightarrow {T_2} = {5 \over 6}{T_1}$

Secondly,
${1 \over 3} = 1 - {{{T_2} - 62} \over {{T_1}}} = 1 - {{{T_2}} \over {{T_1}}} + {{62} \over {{T_1}}}$

$ = 1 - {5 \over 6} + {{62} \over {{T_1}}}$

$ \Rightarrow {T_1} = 62 \times 6 = 372K = 372 - 273 = {99^o}C$

& ${T_2} = {5 \over 6} \times 372 = 310K = 310 - 273 = {37^o}C$