148623
The temperature of source and sink of a heat engine are $127^{\circ} \mathrm{C}$ and $27^{\circ} \mathrm{C}$, respectively. An inventor claims its efficient to be $26 \%$, then
1 it is impossible
2 it is possible with high probability
3 it is possible with low probability
4 Data are insufficient
Explanation:
A Given, $\mathrm{T}_{1}=127^{\circ} \mathrm{C}+273=400 \mathrm{~K}$ $\mathrm{T}_{2}=27^{\circ} \mathrm{C}+273=300 \mathrm{~K}$ and $\eta=26 \%$ Maximum efficiency is obtained by- $\eta=\left(1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right) \times 100$ $=\left(1-\frac{300}{400}\right) \times 100$ $=25 \%$ $\therefore$ Efficiency $26 \%$, Hence, it is impossible for this heat engine.
JIPMER-2017
Thermodynamics
148624
In Carnot engine efficiency is $40 \%$ at hot reservoir temperature T. For efficiency $50 \%$, what will be temperature of hot reservoir?
1 $\frac{\mathrm{T}}{5}$
2 $\frac{2 \mathrm{~T}}{5}$
3 $6 \mathrm{~T}$
4 $\frac{6}{5} \mathrm{~T}$
Explanation:
D We know that, efficiency of a Carnot engine is, Efficiency $(\eta)=\frac{\text { Work done }}{\text { Work input }}=\frac{W}{Q}=1-\frac{T_{1}}{T}$ Where, $T_{1}=$ temperature of sink $\mathrm{T}=$ temperature of source (hot reservoir) Efficiency is $40 \%$ at hot reservoir temperature $\mathrm{T}$. $0.4=1-\frac{\mathrm{T}_{1}}{\mathrm{~T}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}}=0.6$ $\mathrm{~T}_{1}=0.6 \mathrm{~T}$ Let the hot reservoir temperature $\mathrm{T}$ ' for efficiency $50 \%$ $0.5=1-\frac{T_{1}}{T^{\prime}}$ $\frac{T_{1}}{T^{\prime}}=0.5$ Putting the value of $T_{1}$ from equation (i) $\frac{0.6 \mathrm{~T}}{\mathrm{~T}^{\prime}}=0.5$ $\mathrm{~T}^{\prime}=\frac{0.6}{0.5} \mathrm{~T}$ $\mathrm{~T}^{\prime}=\frac{6}{5} \mathrm{~T}$
UPSEE - 2015
Thermodynamics
148625
The maximum amount of work that a Carnot engine can perform per kilocalorie of heat input if it absorbs heat at $427^{\circ} \mathrm{C}$ and releases heat at $177^{\circ} \mathrm{C}$ is
1 $2.39 \mathrm{~kJ}$
2 $6.66 \mathrm{~kJ}$
3 $4.66 \mathrm{~kJ}$
4 $1.51 \mathrm{~kJ}$
Explanation:
D Given, $\mathrm{T}_{1}=427^{\circ} \mathrm{C}+273=700 \mathrm{~K}$ $\mathrm{~T}_{2}=177^{\circ} \mathrm{C}+273=450 \mathrm{~K}$ Heat input $(\mathrm{Q})=1 \mathrm{k}$ cal So, $\quad \eta=1-\frac{T_{2}}{T_{1}}$ $\eta=1-\frac{450}{700}=\frac{5}{14}$ Now, Work done $=\eta \times$ heat input $(\mathrm{Q})$ $=\frac{5}{14} \times 1 \mathrm{kcal}$ $=0.36 \mathrm{k} \mathrm{cal}$ $=0.36 \times 4.18 \mathrm{~kJ}$ $=1.51 \mathrm{~kJ}$
Assam CEE-31.07.2022
Thermodynamics
148626
Temperature of a cold reservoir of a Carnot engine is $127^{\circ} \mathrm{C}$. If the efficiency of the Carnot engine is $20 \%$, then the temperature of the hot reservoir is-
1 $500^{\circ} \mathrm{C}$
2 $227^{\circ} \mathrm{C}$
3 $273^{\circ} \mathrm{C}$
4 $400^{\circ} \mathrm{C}$
Explanation:
B Given, Temperature of cold reservoir $\left(\mathrm{T}_{2}\right)=127^{\circ} \mathrm{C}+273=400 \mathrm{~K}$ and Efficiency $\quad(\eta)=20 \%=0.20$ $\mathrm{T}_{1}=\text { ? }$ We know, Efficiency of Carnot engine $(\eta)=1-\frac{T_{2}}{T_{1}}$ $0.20 =1-\frac{400}{T_{1}}$ $0.80 =\frac{400}{\mathrm{~T}_{1}}$ $\mathrm{~T}_{1} =\frac{400}{0.80}$ $\mathrm{~T}_{1} =500 \mathrm{~K}$ Then, $\mathrm{T}_{1}=500-273=227^{\circ} \mathrm{C}$
148623
The temperature of source and sink of a heat engine are $127^{\circ} \mathrm{C}$ and $27^{\circ} \mathrm{C}$, respectively. An inventor claims its efficient to be $26 \%$, then
1 it is impossible
2 it is possible with high probability
3 it is possible with low probability
4 Data are insufficient
Explanation:
A Given, $\mathrm{T}_{1}=127^{\circ} \mathrm{C}+273=400 \mathrm{~K}$ $\mathrm{T}_{2}=27^{\circ} \mathrm{C}+273=300 \mathrm{~K}$ and $\eta=26 \%$ Maximum efficiency is obtained by- $\eta=\left(1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right) \times 100$ $=\left(1-\frac{300}{400}\right) \times 100$ $=25 \%$ $\therefore$ Efficiency $26 \%$, Hence, it is impossible for this heat engine.
JIPMER-2017
Thermodynamics
148624
In Carnot engine efficiency is $40 \%$ at hot reservoir temperature T. For efficiency $50 \%$, what will be temperature of hot reservoir?
1 $\frac{\mathrm{T}}{5}$
2 $\frac{2 \mathrm{~T}}{5}$
3 $6 \mathrm{~T}$
4 $\frac{6}{5} \mathrm{~T}$
Explanation:
D We know that, efficiency of a Carnot engine is, Efficiency $(\eta)=\frac{\text { Work done }}{\text { Work input }}=\frac{W}{Q}=1-\frac{T_{1}}{T}$ Where, $T_{1}=$ temperature of sink $\mathrm{T}=$ temperature of source (hot reservoir) Efficiency is $40 \%$ at hot reservoir temperature $\mathrm{T}$. $0.4=1-\frac{\mathrm{T}_{1}}{\mathrm{~T}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}}=0.6$ $\mathrm{~T}_{1}=0.6 \mathrm{~T}$ Let the hot reservoir temperature $\mathrm{T}$ ' for efficiency $50 \%$ $0.5=1-\frac{T_{1}}{T^{\prime}}$ $\frac{T_{1}}{T^{\prime}}=0.5$ Putting the value of $T_{1}$ from equation (i) $\frac{0.6 \mathrm{~T}}{\mathrm{~T}^{\prime}}=0.5$ $\mathrm{~T}^{\prime}=\frac{0.6}{0.5} \mathrm{~T}$ $\mathrm{~T}^{\prime}=\frac{6}{5} \mathrm{~T}$
UPSEE - 2015
Thermodynamics
148625
The maximum amount of work that a Carnot engine can perform per kilocalorie of heat input if it absorbs heat at $427^{\circ} \mathrm{C}$ and releases heat at $177^{\circ} \mathrm{C}$ is
1 $2.39 \mathrm{~kJ}$
2 $6.66 \mathrm{~kJ}$
3 $4.66 \mathrm{~kJ}$
4 $1.51 \mathrm{~kJ}$
Explanation:
D Given, $\mathrm{T}_{1}=427^{\circ} \mathrm{C}+273=700 \mathrm{~K}$ $\mathrm{~T}_{2}=177^{\circ} \mathrm{C}+273=450 \mathrm{~K}$ Heat input $(\mathrm{Q})=1 \mathrm{k}$ cal So, $\quad \eta=1-\frac{T_{2}}{T_{1}}$ $\eta=1-\frac{450}{700}=\frac{5}{14}$ Now, Work done $=\eta \times$ heat input $(\mathrm{Q})$ $=\frac{5}{14} \times 1 \mathrm{kcal}$ $=0.36 \mathrm{k} \mathrm{cal}$ $=0.36 \times 4.18 \mathrm{~kJ}$ $=1.51 \mathrm{~kJ}$
Assam CEE-31.07.2022
Thermodynamics
148626
Temperature of a cold reservoir of a Carnot engine is $127^{\circ} \mathrm{C}$. If the efficiency of the Carnot engine is $20 \%$, then the temperature of the hot reservoir is-
1 $500^{\circ} \mathrm{C}$
2 $227^{\circ} \mathrm{C}$
3 $273^{\circ} \mathrm{C}$
4 $400^{\circ} \mathrm{C}$
Explanation:
B Given, Temperature of cold reservoir $\left(\mathrm{T}_{2}\right)=127^{\circ} \mathrm{C}+273=400 \mathrm{~K}$ and Efficiency $\quad(\eta)=20 \%=0.20$ $\mathrm{T}_{1}=\text { ? }$ We know, Efficiency of Carnot engine $(\eta)=1-\frac{T_{2}}{T_{1}}$ $0.20 =1-\frac{400}{T_{1}}$ $0.80 =\frac{400}{\mathrm{~T}_{1}}$ $\mathrm{~T}_{1} =\frac{400}{0.80}$ $\mathrm{~T}_{1} =500 \mathrm{~K}$ Then, $\mathrm{T}_{1}=500-273=227^{\circ} \mathrm{C}$
148623
The temperature of source and sink of a heat engine are $127^{\circ} \mathrm{C}$ and $27^{\circ} \mathrm{C}$, respectively. An inventor claims its efficient to be $26 \%$, then
1 it is impossible
2 it is possible with high probability
3 it is possible with low probability
4 Data are insufficient
Explanation:
A Given, $\mathrm{T}_{1}=127^{\circ} \mathrm{C}+273=400 \mathrm{~K}$ $\mathrm{T}_{2}=27^{\circ} \mathrm{C}+273=300 \mathrm{~K}$ and $\eta=26 \%$ Maximum efficiency is obtained by- $\eta=\left(1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right) \times 100$ $=\left(1-\frac{300}{400}\right) \times 100$ $=25 \%$ $\therefore$ Efficiency $26 \%$, Hence, it is impossible for this heat engine.
JIPMER-2017
Thermodynamics
148624
In Carnot engine efficiency is $40 \%$ at hot reservoir temperature T. For efficiency $50 \%$, what will be temperature of hot reservoir?
1 $\frac{\mathrm{T}}{5}$
2 $\frac{2 \mathrm{~T}}{5}$
3 $6 \mathrm{~T}$
4 $\frac{6}{5} \mathrm{~T}$
Explanation:
D We know that, efficiency of a Carnot engine is, Efficiency $(\eta)=\frac{\text { Work done }}{\text { Work input }}=\frac{W}{Q}=1-\frac{T_{1}}{T}$ Where, $T_{1}=$ temperature of sink $\mathrm{T}=$ temperature of source (hot reservoir) Efficiency is $40 \%$ at hot reservoir temperature $\mathrm{T}$. $0.4=1-\frac{\mathrm{T}_{1}}{\mathrm{~T}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}}=0.6$ $\mathrm{~T}_{1}=0.6 \mathrm{~T}$ Let the hot reservoir temperature $\mathrm{T}$ ' for efficiency $50 \%$ $0.5=1-\frac{T_{1}}{T^{\prime}}$ $\frac{T_{1}}{T^{\prime}}=0.5$ Putting the value of $T_{1}$ from equation (i) $\frac{0.6 \mathrm{~T}}{\mathrm{~T}^{\prime}}=0.5$ $\mathrm{~T}^{\prime}=\frac{0.6}{0.5} \mathrm{~T}$ $\mathrm{~T}^{\prime}=\frac{6}{5} \mathrm{~T}$
UPSEE - 2015
Thermodynamics
148625
The maximum amount of work that a Carnot engine can perform per kilocalorie of heat input if it absorbs heat at $427^{\circ} \mathrm{C}$ and releases heat at $177^{\circ} \mathrm{C}$ is
1 $2.39 \mathrm{~kJ}$
2 $6.66 \mathrm{~kJ}$
3 $4.66 \mathrm{~kJ}$
4 $1.51 \mathrm{~kJ}$
Explanation:
D Given, $\mathrm{T}_{1}=427^{\circ} \mathrm{C}+273=700 \mathrm{~K}$ $\mathrm{~T}_{2}=177^{\circ} \mathrm{C}+273=450 \mathrm{~K}$ Heat input $(\mathrm{Q})=1 \mathrm{k}$ cal So, $\quad \eta=1-\frac{T_{2}}{T_{1}}$ $\eta=1-\frac{450}{700}=\frac{5}{14}$ Now, Work done $=\eta \times$ heat input $(\mathrm{Q})$ $=\frac{5}{14} \times 1 \mathrm{kcal}$ $=0.36 \mathrm{k} \mathrm{cal}$ $=0.36 \times 4.18 \mathrm{~kJ}$ $=1.51 \mathrm{~kJ}$
Assam CEE-31.07.2022
Thermodynamics
148626
Temperature of a cold reservoir of a Carnot engine is $127^{\circ} \mathrm{C}$. If the efficiency of the Carnot engine is $20 \%$, then the temperature of the hot reservoir is-
1 $500^{\circ} \mathrm{C}$
2 $227^{\circ} \mathrm{C}$
3 $273^{\circ} \mathrm{C}$
4 $400^{\circ} \mathrm{C}$
Explanation:
B Given, Temperature of cold reservoir $\left(\mathrm{T}_{2}\right)=127^{\circ} \mathrm{C}+273=400 \mathrm{~K}$ and Efficiency $\quad(\eta)=20 \%=0.20$ $\mathrm{T}_{1}=\text { ? }$ We know, Efficiency of Carnot engine $(\eta)=1-\frac{T_{2}}{T_{1}}$ $0.20 =1-\frac{400}{T_{1}}$ $0.80 =\frac{400}{\mathrm{~T}_{1}}$ $\mathrm{~T}_{1} =\frac{400}{0.80}$ $\mathrm{~T}_{1} =500 \mathrm{~K}$ Then, $\mathrm{T}_{1}=500-273=227^{\circ} \mathrm{C}$
148623
The temperature of source and sink of a heat engine are $127^{\circ} \mathrm{C}$ and $27^{\circ} \mathrm{C}$, respectively. An inventor claims its efficient to be $26 \%$, then
1 it is impossible
2 it is possible with high probability
3 it is possible with low probability
4 Data are insufficient
Explanation:
A Given, $\mathrm{T}_{1}=127^{\circ} \mathrm{C}+273=400 \mathrm{~K}$ $\mathrm{T}_{2}=27^{\circ} \mathrm{C}+273=300 \mathrm{~K}$ and $\eta=26 \%$ Maximum efficiency is obtained by- $\eta=\left(1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right) \times 100$ $=\left(1-\frac{300}{400}\right) \times 100$ $=25 \%$ $\therefore$ Efficiency $26 \%$, Hence, it is impossible for this heat engine.
JIPMER-2017
Thermodynamics
148624
In Carnot engine efficiency is $40 \%$ at hot reservoir temperature T. For efficiency $50 \%$, what will be temperature of hot reservoir?
1 $\frac{\mathrm{T}}{5}$
2 $\frac{2 \mathrm{~T}}{5}$
3 $6 \mathrm{~T}$
4 $\frac{6}{5} \mathrm{~T}$
Explanation:
D We know that, efficiency of a Carnot engine is, Efficiency $(\eta)=\frac{\text { Work done }}{\text { Work input }}=\frac{W}{Q}=1-\frac{T_{1}}{T}$ Where, $T_{1}=$ temperature of sink $\mathrm{T}=$ temperature of source (hot reservoir) Efficiency is $40 \%$ at hot reservoir temperature $\mathrm{T}$. $0.4=1-\frac{\mathrm{T}_{1}}{\mathrm{~T}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}}=0.6$ $\mathrm{~T}_{1}=0.6 \mathrm{~T}$ Let the hot reservoir temperature $\mathrm{T}$ ' for efficiency $50 \%$ $0.5=1-\frac{T_{1}}{T^{\prime}}$ $\frac{T_{1}}{T^{\prime}}=0.5$ Putting the value of $T_{1}$ from equation (i) $\frac{0.6 \mathrm{~T}}{\mathrm{~T}^{\prime}}=0.5$ $\mathrm{~T}^{\prime}=\frac{0.6}{0.5} \mathrm{~T}$ $\mathrm{~T}^{\prime}=\frac{6}{5} \mathrm{~T}$
UPSEE - 2015
Thermodynamics
148625
The maximum amount of work that a Carnot engine can perform per kilocalorie of heat input if it absorbs heat at $427^{\circ} \mathrm{C}$ and releases heat at $177^{\circ} \mathrm{C}$ is
1 $2.39 \mathrm{~kJ}$
2 $6.66 \mathrm{~kJ}$
3 $4.66 \mathrm{~kJ}$
4 $1.51 \mathrm{~kJ}$
Explanation:
D Given, $\mathrm{T}_{1}=427^{\circ} \mathrm{C}+273=700 \mathrm{~K}$ $\mathrm{~T}_{2}=177^{\circ} \mathrm{C}+273=450 \mathrm{~K}$ Heat input $(\mathrm{Q})=1 \mathrm{k}$ cal So, $\quad \eta=1-\frac{T_{2}}{T_{1}}$ $\eta=1-\frac{450}{700}=\frac{5}{14}$ Now, Work done $=\eta \times$ heat input $(\mathrm{Q})$ $=\frac{5}{14} \times 1 \mathrm{kcal}$ $=0.36 \mathrm{k} \mathrm{cal}$ $=0.36 \times 4.18 \mathrm{~kJ}$ $=1.51 \mathrm{~kJ}$
Assam CEE-31.07.2022
Thermodynamics
148626
Temperature of a cold reservoir of a Carnot engine is $127^{\circ} \mathrm{C}$. If the efficiency of the Carnot engine is $20 \%$, then the temperature of the hot reservoir is-
1 $500^{\circ} \mathrm{C}$
2 $227^{\circ} \mathrm{C}$
3 $273^{\circ} \mathrm{C}$
4 $400^{\circ} \mathrm{C}$
Explanation:
B Given, Temperature of cold reservoir $\left(\mathrm{T}_{2}\right)=127^{\circ} \mathrm{C}+273=400 \mathrm{~K}$ and Efficiency $\quad(\eta)=20 \%=0.20$ $\mathrm{T}_{1}=\text { ? }$ We know, Efficiency of Carnot engine $(\eta)=1-\frac{T_{2}}{T_{1}}$ $0.20 =1-\frac{400}{T_{1}}$ $0.80 =\frac{400}{\mathrm{~T}_{1}}$ $\mathrm{~T}_{1} =\frac{400}{0.80}$ $\mathrm{~T}_{1} =500 \mathrm{~K}$ Then, $\mathrm{T}_{1}=500-273=227^{\circ} \mathrm{C}$
