Heat Engine, Second Law of Thermodynamics and Carnot Engine — NEET Physics PYQ MCQs with Solutions

Solution: Efficiency of Carnot engine is given as $\eta=1-\dfrac{T_2}{T_1}$, where $T_2=$ temperature of sink and $T_1=$ temperature of source. Hence, $\eta$ depends upon the temperature of source and sink both.

Solution: Efficiency of an ideal heat engine is given as $\eta=1-\dfrac{T_2}{T_1}$, where $T_1$ is the temperature of the source and $T_2$ is the temperature of the sink. Here, $T_1=100+273=373$ K and $T_2=0+273=273$ K. $\Rightarrow \eta=1-\dfrac{273}{373}=\dfrac{373-273}{373}=\dfrac{100}{373}=0.268$. $\therefore \eta\%=0.268\times100=26.8\%$.

Solution: In case of engine, engine efficiency $=\dfrac{\text{work}}{\text{heat absorbed}}=\dfrac{W}{q_1}$. $\therefore \dfrac{W}{q_1}=\dfrac{1}{10} \Rightarrow \dfrac{10\text{ J}}{q_1}=\dfrac{1}{10}$ or $q_1=100$ J. When this engine is reversed, it takes in work $W$ and heat $q_2$ from cold reservoir and ejects 100 J of heat to hot reservoir. $\therefore W+q_2=q_1 \Rightarrow 10+q_2=100$ or $q_2=90$ J.

Solution: Given, temperature of source, $T=30^{\circ}$C $=30+273 \Rightarrow T_1=303$ K. Temperature of sink, $T_2=4^{\circ}$C $=4+273=277$ K. As we know that $\dfrac{Q_1}{Q_2}=\dfrac{T_1}{T_2} \Rightarrow \dfrac{Q_2+W}{Q_2}=\dfrac{T_1}{T_2}$ (since $W=Q_1-Q_2$). $\Rightarrow WT_2+T_2Q_2=T_1Q_2 \Rightarrow WT_2=Q_2(T_1-T_2) \Rightarrow W=Q_2\left(\dfrac{T_1}{T_2}-1\right)=600\times4.2\times\left(\dfrac{303}{277}-1\right)=600\times4.2\times\left(\dfrac{26}{277}\right)=236.5$ Joules. Power $=\dfrac{\text{Work done}}{\text{Time}}=\dfrac{W}{t}=\dfrac{236.5}{1}=236.5$ W.

Solution: For a refrigerator, we know that $\dfrac{Q_1}{W}=\dfrac{Q_1}{Q_1-Q_2}=\dfrac{T_1}{T_1-T_2}$, where $Q_1=$ amount of heat delivered to the room, $W=$ electrical energy consumed, $T_1=$ room temperature $=t_1+273$ and $T_2=$ temperature of sink $=t_2+273$. $\therefore \dfrac{Q_1}{1}=\dfrac{t_1+273}{t_1+273-(t_2+273)} \Rightarrow Q_1=\dfrac{t_1+273}{t_1-t_2}$.

Solution: As, $Q_1+W=Q_2$. Given, $\eta=\dfrac{1}{10}$. Now using $\eta=1-\dfrac{T_1}{T_2}$, so $\dfrac{1}{10}=1-\dfrac{T_1}{T_2} \Rightarrow \dfrac{T_1}{T_2}=\dfrac{9}{10}$. Now $\dfrac{Q_1}{Q_2}=\dfrac{T_1}{T_2} \Rightarrow \dfrac{Q_1}{Q_1+W}=\dfrac{9}{10} \Rightarrow 10Q_1=9Q_1+9W \Rightarrow Q_1=9W=9\times10=90$ J.

Solution: Coefficient of performance $(\beta)$ of a refrigerator is defined as the ratio of quantity of heat removed per cycle $(Q_2)$ to the work done on the working substance per cycle to remove this heat. Given, $\beta=5$. Temperature inside freezer, $T_2=-20^{\circ}$C $=-20+273=253$ K. Temperature of surrounding, $T_1=?$. So $\beta=\dfrac{T_2}{T_1-T_2} \Rightarrow 5=\dfrac{253}{T_1-253} \Rightarrow 5T_1-1265=253 \Rightarrow 5T_1=1518 \Rightarrow T_1=\dfrac{1518}{5}=303.6$ K $=303.6-273=31^{\circ}$C.

Solution: Efficiency of engine is given by $\eta=1-\dfrac{T_2}{T_1}$, where $T_2=$ temperature of sink and $T_1=$ temperature of source. $\therefore \dfrac{T_2}{T_1}=1-\eta=1-\dfrac{1}{6}=\dfrac{5}{6}$ ...(i). In other case, $\dfrac{T_2-62}{T_1}=1-\eta'=1-\dfrac{2}{6}=\dfrac{2}{3}$ ...(ii). or $T_2-62=\dfrac{2}{3}T_1=\dfrac{2}{3}\times\dfrac{6}{5}T_2$ [Using Eq. (i)]. or $\dfrac{1}{5}T_2=62 \Rightarrow T_2=310$ K $=310-273=37^{\circ}$C. Here, $T_1=\dfrac{6}{5}T_2=\dfrac{6}{5}\times310=372$ K $=372-273=99^{\circ}$C.

Solution: The efficiency of Carnot engine is $\eta=\dfrac{\text{Work done}}{\text{Heat supplied}}=\dfrac{W}{Q_1}=\dfrac{Q_1-Q_2}{Q_1}=1-\dfrac{T_2}{T_1}$. As given, $\eta=40\%=\dfrac{40}{100}=0.4$ and $T_2=300$ K. So $0.4=1-\dfrac{300}{T_1} \Rightarrow T_1=\dfrac{300}{1-0.4}=\dfrac{300}{0.6}=500$ K. Let temperature of the source be increased by $x$ K, then efficiency becomes $\eta'=40\%+50\%$ of $\eta=\dfrac{40}{100}+\dfrac{50}{100}\times0.4=0.4+0.5\times0.4=0.6$. Hence, $0.6=1-\dfrac{300}{500+x} \Rightarrow \dfrac{300}{500+x}=0.4 \Rightarrow 500+x=\dfrac{300}{0.4}=750 \Rightarrow x=750-500=250$ K.

Solution: According to the Carnot cycle in heat engine $\dfrac{Q_2}{Q_1}=\dfrac{T_2}{T_1}$. Given, heat absorbed $Q_1=6\times10^{4}$ cal, temperature of source $T_1=227+273=500$ K, temperature of sink $T_2=127+273=400$ K. $\therefore \dfrac{Q_2}{6\times10^{4}}=\dfrac{400}{500} \Rightarrow$ heat rejected $Q_2=\dfrac{4}{5}\times6\times10^{4}=4.8\times10^{4}$ cal. Now, heat converted to work $W=Q_1-Q_2=6.0\times10^{4}-4.8\times10^{4}=1.2\times10^{4}$ cal.

Solution: $\eta=1-\dfrac{T_2}{T_1}$ or $\dfrac{W}{Q_1}=1-\dfrac{T_2}{T_1}$. Given $Q_1=6$ kcal, $T_1=227+273=500$ K, $T_2=127+273=400$ K. Hence $\dfrac{W}{6}=1-\dfrac{400}{500}=\dfrac{100}{500}\Rightarrow W=1.2$ kcal.

Solution: $\eta=1-\dfrac{T_2}{T_1}$ ...(i). Given $\eta=0.5$, $T_2=500$ K, so $0.5=1-\dfrac{500}{T_1}\Rightarrow T_1=\dfrac{500}{0.5}=1000$ K. Keeping the source constant and raising efficiency to $0.6$: $0.6=1-\dfrac{T_2'}{1000}\Rightarrow\dfrac{T_2'}{1000}=0.4\Rightarrow T_2'=400$ K.

Solution: $\eta=1-\dfrac{T_2}{T_1}=\dfrac{T_1-T_2}{T_1}$. Given $T_1=273+127=400$ K, $T_2=273+27=300$ K. So $\eta=\dfrac{400-300}{400}=\dfrac{100}{400}=0.25=25\%$. Hence a claimed efficiency of 26% is impossible for this heat engine.

Solution: $\eta=\dfrac{W}{Q_1}=1-\dfrac{T_2}{T_1}=\dfrac{1}{6}$ ...(i). When sink is reduced by $62^{\circ}$C, $\eta'=1-\dfrac{T_2-62}{T_1}=2\eta=\dfrac{1}{3}$ ...(ii). From (i) $\dfrac{T_2}{T_1}=\dfrac{5}{6}$; from (ii) $\dfrac{T_2-62}{T_1}=\dfrac{2}{3}$. Dividing, $\dfrac{T_2}{T_2-62}=\dfrac{5}{4}\Rightarrow4T_2=5T_2-310\Rightarrow T_2=310$ K. Then $\dfrac{310}{T_1}=\dfrac{5}{6}\Rightarrow T_1=372$ K. Hence $T_1=372-273=99^{\circ}$C and $T_2=310-273=37^{\circ}$C.

Solution: $\eta=1-\dfrac{T_2}{T_1}=\dfrac{T_1-T_2}{T_1}$. Given $T_1=100+273=373$ K and $T_2=-23+273=250$ K. So $\eta=\dfrac{373-250}{373}=\dfrac{123}{373}=\dfrac{100+23}{373}$.