148674
A Carnot engine has an efficiency of only $15 \%$. If it operates between constant temperature reservoirs differing in temperatures by $55^{\circ} \mathrm{C}$ the temperature of higher temperature reservoir is
1 $367^{\circ} \mathrm{K}$
2 $382^{\circ} \mathrm{K}$
3 $418^{\circ} \mathrm{K}$
4 $421^{\circ} \mathrm{K}$
Explanation:
A Efficiency of Carnot cycle, $\eta=\frac{T_{H}-T_{L}}{T_{H}}$ $\eta=15 \%=0.15$ Difference in temperature of two reservoirs, $\Delta \mathrm{T}=\mathrm{T}_{\mathrm{H}}-\mathrm{T}_{\mathrm{L}}=55^{\circ} \mathrm{C}=55 \mathrm{~K} .$ $\therefore \quad 0.15=\frac{55}{\mathrm{~T}_{\mathrm{H}}}$ $\mathrm{T}_{\mathrm{H}} =\frac{55}{0.15}=366.67 \simeq 367 \mathrm{~K}$
AMU-2019
Thermodynamics
148675
A quarter horse power motor runs at a speed of $600 \mathrm{rpm}$. Assuming $40 \%$ efficiency, the work done by the motor in one rotation will be
148676
The efficiency of an ideal heat engine working between the freezing point and boiling point of water, is
1 $6.25 \%$
2 $20 \%$
3 $26.8 \%$
4 $12.5 \%$
Explanation:
C Given, Freezing point of water $=0^{\circ}=273+0^{\circ}=273 \mathrm{~K}$ Boiling point of water $=100^{\circ} \mathrm{C}=273+100=373 \mathrm{~K}$ Efficiency of engine, $\eta=1-\frac{T_{L}}{T_{H}}$ $=\frac{373-273}{373}$ $=\frac{100}{373} \times 100$ $=26.80$
NEET-2018
Thermodynamics
148677
A Carnot engine having an efficiency of $\frac{1}{10}$ as heat engine, is used as a refrigerator. If the work done on the system is $10 \mathrm{~J}$, the amount of energy absorbed from the reservoir at lower temperature is
1 $1 \mathrm{~J}$
2 $90 \mathrm{~J}$
3 $99 \mathrm{~J}$
4 $100 \mathrm{~J}$
Explanation:
B For a heat engine, $\eta=\frac{\text { Work output }}{\text { Heat input }}$ $\eta=\frac{W}{Q} \quad\left\{\because \eta=\frac{1}{10}\right.$ $\frac{1}{10}=\frac{10}{Q_{1}}$ $Q_{1}=100 \mathrm{~J}$ So, energy released at lower temperature, $\mathrm{Q}_{2}=\mathrm{Q}_{1}-\mathrm{W}$ $\mathrm{Q}_{2}=100-10$ $\mathrm{Q}_{2}=90 \mathrm{~J}$
148674
A Carnot engine has an efficiency of only $15 \%$. If it operates between constant temperature reservoirs differing in temperatures by $55^{\circ} \mathrm{C}$ the temperature of higher temperature reservoir is
1 $367^{\circ} \mathrm{K}$
2 $382^{\circ} \mathrm{K}$
3 $418^{\circ} \mathrm{K}$
4 $421^{\circ} \mathrm{K}$
Explanation:
A Efficiency of Carnot cycle, $\eta=\frac{T_{H}-T_{L}}{T_{H}}$ $\eta=15 \%=0.15$ Difference in temperature of two reservoirs, $\Delta \mathrm{T}=\mathrm{T}_{\mathrm{H}}-\mathrm{T}_{\mathrm{L}}=55^{\circ} \mathrm{C}=55 \mathrm{~K} .$ $\therefore \quad 0.15=\frac{55}{\mathrm{~T}_{\mathrm{H}}}$ $\mathrm{T}_{\mathrm{H}} =\frac{55}{0.15}=366.67 \simeq 367 \mathrm{~K}$
AMU-2019
Thermodynamics
148675
A quarter horse power motor runs at a speed of $600 \mathrm{rpm}$. Assuming $40 \%$ efficiency, the work done by the motor in one rotation will be
148676
The efficiency of an ideal heat engine working between the freezing point and boiling point of water, is
1 $6.25 \%$
2 $20 \%$
3 $26.8 \%$
4 $12.5 \%$
Explanation:
C Given, Freezing point of water $=0^{\circ}=273+0^{\circ}=273 \mathrm{~K}$ Boiling point of water $=100^{\circ} \mathrm{C}=273+100=373 \mathrm{~K}$ Efficiency of engine, $\eta=1-\frac{T_{L}}{T_{H}}$ $=\frac{373-273}{373}$ $=\frac{100}{373} \times 100$ $=26.80$
NEET-2018
Thermodynamics
148677
A Carnot engine having an efficiency of $\frac{1}{10}$ as heat engine, is used as a refrigerator. If the work done on the system is $10 \mathrm{~J}$, the amount of energy absorbed from the reservoir at lower temperature is
1 $1 \mathrm{~J}$
2 $90 \mathrm{~J}$
3 $99 \mathrm{~J}$
4 $100 \mathrm{~J}$
Explanation:
B For a heat engine, $\eta=\frac{\text { Work output }}{\text { Heat input }}$ $\eta=\frac{W}{Q} \quad\left\{\because \eta=\frac{1}{10}\right.$ $\frac{1}{10}=\frac{10}{Q_{1}}$ $Q_{1}=100 \mathrm{~J}$ So, energy released at lower temperature, $\mathrm{Q}_{2}=\mathrm{Q}_{1}-\mathrm{W}$ $\mathrm{Q}_{2}=100-10$ $\mathrm{Q}_{2}=90 \mathrm{~J}$
148674
A Carnot engine has an efficiency of only $15 \%$. If it operates between constant temperature reservoirs differing in temperatures by $55^{\circ} \mathrm{C}$ the temperature of higher temperature reservoir is
1 $367^{\circ} \mathrm{K}$
2 $382^{\circ} \mathrm{K}$
3 $418^{\circ} \mathrm{K}$
4 $421^{\circ} \mathrm{K}$
Explanation:
A Efficiency of Carnot cycle, $\eta=\frac{T_{H}-T_{L}}{T_{H}}$ $\eta=15 \%=0.15$ Difference in temperature of two reservoirs, $\Delta \mathrm{T}=\mathrm{T}_{\mathrm{H}}-\mathrm{T}_{\mathrm{L}}=55^{\circ} \mathrm{C}=55 \mathrm{~K} .$ $\therefore \quad 0.15=\frac{55}{\mathrm{~T}_{\mathrm{H}}}$ $\mathrm{T}_{\mathrm{H}} =\frac{55}{0.15}=366.67 \simeq 367 \mathrm{~K}$
AMU-2019
Thermodynamics
148675
A quarter horse power motor runs at a speed of $600 \mathrm{rpm}$. Assuming $40 \%$ efficiency, the work done by the motor in one rotation will be
148676
The efficiency of an ideal heat engine working between the freezing point and boiling point of water, is
1 $6.25 \%$
2 $20 \%$
3 $26.8 \%$
4 $12.5 \%$
Explanation:
C Given, Freezing point of water $=0^{\circ}=273+0^{\circ}=273 \mathrm{~K}$ Boiling point of water $=100^{\circ} \mathrm{C}=273+100=373 \mathrm{~K}$ Efficiency of engine, $\eta=1-\frac{T_{L}}{T_{H}}$ $=\frac{373-273}{373}$ $=\frac{100}{373} \times 100$ $=26.80$
NEET-2018
Thermodynamics
148677
A Carnot engine having an efficiency of $\frac{1}{10}$ as heat engine, is used as a refrigerator. If the work done on the system is $10 \mathrm{~J}$, the amount of energy absorbed from the reservoir at lower temperature is
1 $1 \mathrm{~J}$
2 $90 \mathrm{~J}$
3 $99 \mathrm{~J}$
4 $100 \mathrm{~J}$
Explanation:
B For a heat engine, $\eta=\frac{\text { Work output }}{\text { Heat input }}$ $\eta=\frac{W}{Q} \quad\left\{\because \eta=\frac{1}{10}\right.$ $\frac{1}{10}=\frac{10}{Q_{1}}$ $Q_{1}=100 \mathrm{~J}$ So, energy released at lower temperature, $\mathrm{Q}_{2}=\mathrm{Q}_{1}-\mathrm{W}$ $\mathrm{Q}_{2}=100-10$ $\mathrm{Q}_{2}=90 \mathrm{~J}$
148674
A Carnot engine has an efficiency of only $15 \%$. If it operates between constant temperature reservoirs differing in temperatures by $55^{\circ} \mathrm{C}$ the temperature of higher temperature reservoir is
1 $367^{\circ} \mathrm{K}$
2 $382^{\circ} \mathrm{K}$
3 $418^{\circ} \mathrm{K}$
4 $421^{\circ} \mathrm{K}$
Explanation:
A Efficiency of Carnot cycle, $\eta=\frac{T_{H}-T_{L}}{T_{H}}$ $\eta=15 \%=0.15$ Difference in temperature of two reservoirs, $\Delta \mathrm{T}=\mathrm{T}_{\mathrm{H}}-\mathrm{T}_{\mathrm{L}}=55^{\circ} \mathrm{C}=55 \mathrm{~K} .$ $\therefore \quad 0.15=\frac{55}{\mathrm{~T}_{\mathrm{H}}}$ $\mathrm{T}_{\mathrm{H}} =\frac{55}{0.15}=366.67 \simeq 367 \mathrm{~K}$
AMU-2019
Thermodynamics
148675
A quarter horse power motor runs at a speed of $600 \mathrm{rpm}$. Assuming $40 \%$ efficiency, the work done by the motor in one rotation will be
148676
The efficiency of an ideal heat engine working between the freezing point and boiling point of water, is
1 $6.25 \%$
2 $20 \%$
3 $26.8 \%$
4 $12.5 \%$
Explanation:
C Given, Freezing point of water $=0^{\circ}=273+0^{\circ}=273 \mathrm{~K}$ Boiling point of water $=100^{\circ} \mathrm{C}=273+100=373 \mathrm{~K}$ Efficiency of engine, $\eta=1-\frac{T_{L}}{T_{H}}$ $=\frac{373-273}{373}$ $=\frac{100}{373} \times 100$ $=26.80$
NEET-2018
Thermodynamics
148677
A Carnot engine having an efficiency of $\frac{1}{10}$ as heat engine, is used as a refrigerator. If the work done on the system is $10 \mathrm{~J}$, the amount of energy absorbed from the reservoir at lower temperature is
1 $1 \mathrm{~J}$
2 $90 \mathrm{~J}$
3 $99 \mathrm{~J}$
4 $100 \mathrm{~J}$
Explanation:
B For a heat engine, $\eta=\frac{\text { Work output }}{\text { Heat input }}$ $\eta=\frac{W}{Q} \quad\left\{\because \eta=\frac{1}{10}\right.$ $\frac{1}{10}=\frac{10}{Q_{1}}$ $Q_{1}=100 \mathrm{~J}$ So, energy released at lower temperature, $\mathrm{Q}_{2}=\mathrm{Q}_{1}-\mathrm{W}$ $\mathrm{Q}_{2}=100-10$ $\mathrm{Q}_{2}=90 \mathrm{~J}$
