Heat and Thermodynamics

$1 - {{{T_2}} \over {{T_1}}} = 1 - {{{Q_2}} \over {{Q_1}}}$

$ \Rightarrow 1 - {{400} \over {500}} = 1 - {{{Q_2}} \over {6 \times {{10}^4}}}$

$ \Rightarrow {4 \over 5} = {{{Q_2}} \over {6 \times {{10}^4}}} \Rightarrow {Q_2} = 4.8 \times {10^4}cal$

Net heat converted into work

$ = 6.0 \times {10^4} - 4.8 \times {10^4} = 1.2 \times {10^4}cal$

As PV = nRT

$n = {m \over {molecular\,mass}} = {5 \over {32}}$

$ \Rightarrow PV = \left( {{5 \over {32}}} \right)RT$

${T_1} = T,W = 6R\,joules,\gamma = {5 \over 3}$

$W = {{{P_1}{V_1} - {P_2}{V_2}} \over {\gamma - 1}} = {{nR{T_1} - nR{T_2}} \over {\gamma - 1}}$

$ = {{nR\left( {{T_1} - {T_2}} \right)} \over {\gamma - 1}}$

$n = 1,{T_1} = T \Rightarrow {{R\left( {T - {T_2}} \right)} \over {5/3 - 1}} = 6R$

$ \Rightarrow {T_2} = \left( {T - 4} \right)K$

Efficiency of Carnot engine

$ = {W \over {{Q_1}}} = 1 - {{{T_2}} \over {{T_1}}}$

${W \over 6} = 1 - {{400} \over {500}} = {1 \over 5} \Rightarrow W = {6 \over 5} = 1.2\,kcal$

Efficiency ($\eta$) of a carnot engine is given by $\eta = 1 - {{{T_2}} \over {{T_1}}}$, where T₁ is the temperature of the source and T₂ is the temperature of the sink.

Here, T₂ = 500 K

$ \therefore $ $0.5 = 1 - {{500} \over {{T_1}}} \Rightarrow {T_1} = 1000K$

Now, $\eta = 0.6 = 1 - {{{T_2}'} \over {1000}}$ (T₂' is the new sink temperature)

$ \Rightarrow {T_2}' = 400\,K$

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$

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