148535
A Carnot engine whose low temperature reservoir is at $7^{\circ} \mathrm{C}$ has an efficiency of $50 \%$. It is desired to increase the efficiency to $70 \%$. By how many degrees should the temperature of the high temperature reservoir be increased?
1 $840 \mathrm{~K}$
2 $280 \mathrm{~K}$
3 $560 \mathrm{~K}$
4 $380 \mathrm{~K}$
Explanation:
D Efficiency is defined as- $\eta=1-\frac{T_{2}}{T_{1}}$ Initially, $\frac{50}{100}=1-\frac{273+7}{T_{1}}$ $\frac{1}{2}=1-\frac{280}{T_{1}}$ $\frac{280}{T_{1}}=\frac{1}{2}$ $\mathrm{~T}_{1}=560 \mathrm{~K}$ Let new temperature of higher temperature reservoir is $\mathrm{T}_{1}^{\prime}-$ $\frac{70}{100}=1-\frac{280}{\mathrm{~T}_{1}^{\prime}}$ $\frac{280}{\mathrm{~T}_{1}^{\prime}}=\frac{3}{10}$ $\mathrm{~T}_{1}^{\prime}=\frac{280 \times 10}{3}=933 \mathrm{~K}$ $\therefore$ Increase in temperature $=933-560=373 \mathrm{~K}$ The nearest option is option (d). Hence, correct answer is $380 \mathrm{~K}$.
UPSEE - 2005
Thermodynamics
148536
A Carnot engine absorbs an amount $Q$ of heat from a reservoir at an absolute temperature $T$ and rejects heat to a sink at a temperature of $T / 3$. The amount of heat rejected is:
1 $\mathrm{Q} / 4$
2 $\mathrm{Q} / 3$
3 $\mathrm{Q} / 2$
4 $2 \mathrm{Q} / 3$
Explanation:
B We Know that, $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{\mathrm{W}}{\mathrm{Q}_{1}}=\frac{\mathrm{Q}_{1}-\mathrm{Q}_{2}}{\mathrm{Q}_{1}}$ Where, $\mathrm{Q}_{1}=$ heat absorbed, $\mathrm{Q}_{2}=$ heat rejected $1-\frac{\mathrm{T} / 3}{\mathrm{~T}}=\frac{\mathrm{W}}{\mathrm{Q}_{1}}$ $\frac{2}{3}=\frac{\mathrm{W}}{\mathrm{Q}_{1}}=\frac{\mathrm{Q}_{1}-\mathrm{Q}_{2}}{\mathrm{Q}_{1}}$ $\frac{2}{3}=1-\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}}$ $\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}}=\frac{1}{3}$ $\mathrm{Q}_{2}=\frac{\mathrm{Q}_{1}}{3}=\frac{\mathrm{Q}}{3}$
UPSEE - 2004
Thermodynamics
148537
The efficiency of an ideal Carnot engine working between temperature $T_{1}$ and $T_{2}$ is $1 / 3$. If the temperature of the sink is reduced by $\mathbf{4 0} \%$, then its efficiency will be
1 $50 \%$
2 $25 \%$
3 $60 \%$
4 $75 \%$
Explanation:
C Given, Efficiency of carnot engine $(\eta)=\frac{1}{3}$ $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ Where, $\mathrm{T}_{1}=\text { temperature of source }$ $\mathrm{T}_{2}=\text { temperature of sink }$ $\therefore \quad \frac{1}{3}=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}} \Rightarrow \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-\frac{1}{3} \Rightarrow \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{2}{3}$ When temperature of sink is reduced by $40 \%$ then its new temperature- $\mathrm{T}_{2}^{\prime} =\mathrm{T}_{2}-40 \% \text { of } \mathrm{T}_{2}$ $=\mathrm{T}_{2}-0.4 \mathrm{~T}_{2}$ $\mathrm{~T}_{2}^{\prime} =0.6 \mathrm{~T}_{2}$ Now, efficiency of Carnot's engine- $\eta^{\prime} =\left(1-\frac{\mathrm{T}_{2}^{\prime}}{\mathrm{T}_{1}}\right) \times 100=\left(1-0.6 \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right) \times 100$ $=\left(1-0.6 \times \frac{2}{3}\right) \times 100=0.6 \times 100=60 \%$
AP EAMCET (22.09.2020) Shift-II
Thermodynamics
148538
A Carnot's engine has an efficiency of $25 \%$ when its sink is at $27^{\circ} \mathrm{C}$. If it has to be increased to $40 \%$, what should be the temperature of the sink keeping the temperature of the source constant?
1 $320 \mathrm{~K}$
2 $375 \mathrm{~K}$
3 $240 \mathrm{~K}$
4 $300 \mathrm{~K}$
Explanation:
C Efficiency of carnot's engine, $\eta=25 \%=0.25$ Temperature of sink, $\mathrm{T}_{2}=27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K}$ Temperature of source $=\mathrm{T}_{1}$ $\because \quad \eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}} \Rightarrow 0.25=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-0.25=0.75$ $\mathrm{T}_{1}=\frac{\mathrm{T}_{2}}{0.75}=\frac{300}{0.75}=400 \mathrm{~K}$ $\mathrm{T}_{1}=400 \mathrm{~K}$ When efficiency, $\eta=40 \%=0.4$ $1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=0.4$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-0.4=0.6$ $\mathrm{~T}_{2}=0.6 \mathrm{~T}_{1}=0.6 \times 400$ $\quad=240 \mathrm{~K}$
AP EAMCET (22.09.2020) Shift-II
Thermodynamics
148539
A Carnot engine takes $3 \times 10^{6}$ calories of heat from reservoir at $627^{\circ} \mathrm{C}$ and gives it to a sink at $27^{\circ} \mathrm{C}$. The work done by the engine is
148535
A Carnot engine whose low temperature reservoir is at $7^{\circ} \mathrm{C}$ has an efficiency of $50 \%$. It is desired to increase the efficiency to $70 \%$. By how many degrees should the temperature of the high temperature reservoir be increased?
1 $840 \mathrm{~K}$
2 $280 \mathrm{~K}$
3 $560 \mathrm{~K}$
4 $380 \mathrm{~K}$
Explanation:
D Efficiency is defined as- $\eta=1-\frac{T_{2}}{T_{1}}$ Initially, $\frac{50}{100}=1-\frac{273+7}{T_{1}}$ $\frac{1}{2}=1-\frac{280}{T_{1}}$ $\frac{280}{T_{1}}=\frac{1}{2}$ $\mathrm{~T}_{1}=560 \mathrm{~K}$ Let new temperature of higher temperature reservoir is $\mathrm{T}_{1}^{\prime}-$ $\frac{70}{100}=1-\frac{280}{\mathrm{~T}_{1}^{\prime}}$ $\frac{280}{\mathrm{~T}_{1}^{\prime}}=\frac{3}{10}$ $\mathrm{~T}_{1}^{\prime}=\frac{280 \times 10}{3}=933 \mathrm{~K}$ $\therefore$ Increase in temperature $=933-560=373 \mathrm{~K}$ The nearest option is option (d). Hence, correct answer is $380 \mathrm{~K}$.
UPSEE - 2005
Thermodynamics
148536
A Carnot engine absorbs an amount $Q$ of heat from a reservoir at an absolute temperature $T$ and rejects heat to a sink at a temperature of $T / 3$. The amount of heat rejected is:
1 $\mathrm{Q} / 4$
2 $\mathrm{Q} / 3$
3 $\mathrm{Q} / 2$
4 $2 \mathrm{Q} / 3$
Explanation:
B We Know that, $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{\mathrm{W}}{\mathrm{Q}_{1}}=\frac{\mathrm{Q}_{1}-\mathrm{Q}_{2}}{\mathrm{Q}_{1}}$ Where, $\mathrm{Q}_{1}=$ heat absorbed, $\mathrm{Q}_{2}=$ heat rejected $1-\frac{\mathrm{T} / 3}{\mathrm{~T}}=\frac{\mathrm{W}}{\mathrm{Q}_{1}}$ $\frac{2}{3}=\frac{\mathrm{W}}{\mathrm{Q}_{1}}=\frac{\mathrm{Q}_{1}-\mathrm{Q}_{2}}{\mathrm{Q}_{1}}$ $\frac{2}{3}=1-\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}}$ $\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}}=\frac{1}{3}$ $\mathrm{Q}_{2}=\frac{\mathrm{Q}_{1}}{3}=\frac{\mathrm{Q}}{3}$
UPSEE - 2004
Thermodynamics
148537
The efficiency of an ideal Carnot engine working between temperature $T_{1}$ and $T_{2}$ is $1 / 3$. If the temperature of the sink is reduced by $\mathbf{4 0} \%$, then its efficiency will be
1 $50 \%$
2 $25 \%$
3 $60 \%$
4 $75 \%$
Explanation:
C Given, Efficiency of carnot engine $(\eta)=\frac{1}{3}$ $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ Where, $\mathrm{T}_{1}=\text { temperature of source }$ $\mathrm{T}_{2}=\text { temperature of sink }$ $\therefore \quad \frac{1}{3}=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}} \Rightarrow \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-\frac{1}{3} \Rightarrow \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{2}{3}$ When temperature of sink is reduced by $40 \%$ then its new temperature- $\mathrm{T}_{2}^{\prime} =\mathrm{T}_{2}-40 \% \text { of } \mathrm{T}_{2}$ $=\mathrm{T}_{2}-0.4 \mathrm{~T}_{2}$ $\mathrm{~T}_{2}^{\prime} =0.6 \mathrm{~T}_{2}$ Now, efficiency of Carnot's engine- $\eta^{\prime} =\left(1-\frac{\mathrm{T}_{2}^{\prime}}{\mathrm{T}_{1}}\right) \times 100=\left(1-0.6 \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right) \times 100$ $=\left(1-0.6 \times \frac{2}{3}\right) \times 100=0.6 \times 100=60 \%$
AP EAMCET (22.09.2020) Shift-II
Thermodynamics
148538
A Carnot's engine has an efficiency of $25 \%$ when its sink is at $27^{\circ} \mathrm{C}$. If it has to be increased to $40 \%$, what should be the temperature of the sink keeping the temperature of the source constant?
1 $320 \mathrm{~K}$
2 $375 \mathrm{~K}$
3 $240 \mathrm{~K}$
4 $300 \mathrm{~K}$
Explanation:
C Efficiency of carnot's engine, $\eta=25 \%=0.25$ Temperature of sink, $\mathrm{T}_{2}=27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K}$ Temperature of source $=\mathrm{T}_{1}$ $\because \quad \eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}} \Rightarrow 0.25=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-0.25=0.75$ $\mathrm{T}_{1}=\frac{\mathrm{T}_{2}}{0.75}=\frac{300}{0.75}=400 \mathrm{~K}$ $\mathrm{T}_{1}=400 \mathrm{~K}$ When efficiency, $\eta=40 \%=0.4$ $1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=0.4$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-0.4=0.6$ $\mathrm{~T}_{2}=0.6 \mathrm{~T}_{1}=0.6 \times 400$ $\quad=240 \mathrm{~K}$
AP EAMCET (22.09.2020) Shift-II
Thermodynamics
148539
A Carnot engine takes $3 \times 10^{6}$ calories of heat from reservoir at $627^{\circ} \mathrm{C}$ and gives it to a sink at $27^{\circ} \mathrm{C}$. The work done by the engine is
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Thermodynamics
148535
A Carnot engine whose low temperature reservoir is at $7^{\circ} \mathrm{C}$ has an efficiency of $50 \%$. It is desired to increase the efficiency to $70 \%$. By how many degrees should the temperature of the high temperature reservoir be increased?
1 $840 \mathrm{~K}$
2 $280 \mathrm{~K}$
3 $560 \mathrm{~K}$
4 $380 \mathrm{~K}$
Explanation:
D Efficiency is defined as- $\eta=1-\frac{T_{2}}{T_{1}}$ Initially, $\frac{50}{100}=1-\frac{273+7}{T_{1}}$ $\frac{1}{2}=1-\frac{280}{T_{1}}$ $\frac{280}{T_{1}}=\frac{1}{2}$ $\mathrm{~T}_{1}=560 \mathrm{~K}$ Let new temperature of higher temperature reservoir is $\mathrm{T}_{1}^{\prime}-$ $\frac{70}{100}=1-\frac{280}{\mathrm{~T}_{1}^{\prime}}$ $\frac{280}{\mathrm{~T}_{1}^{\prime}}=\frac{3}{10}$ $\mathrm{~T}_{1}^{\prime}=\frac{280 \times 10}{3}=933 \mathrm{~K}$ $\therefore$ Increase in temperature $=933-560=373 \mathrm{~K}$ The nearest option is option (d). Hence, correct answer is $380 \mathrm{~K}$.
UPSEE - 2005
Thermodynamics
148536
A Carnot engine absorbs an amount $Q$ of heat from a reservoir at an absolute temperature $T$ and rejects heat to a sink at a temperature of $T / 3$. The amount of heat rejected is:
1 $\mathrm{Q} / 4$
2 $\mathrm{Q} / 3$
3 $\mathrm{Q} / 2$
4 $2 \mathrm{Q} / 3$
Explanation:
B We Know that, $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{\mathrm{W}}{\mathrm{Q}_{1}}=\frac{\mathrm{Q}_{1}-\mathrm{Q}_{2}}{\mathrm{Q}_{1}}$ Where, $\mathrm{Q}_{1}=$ heat absorbed, $\mathrm{Q}_{2}=$ heat rejected $1-\frac{\mathrm{T} / 3}{\mathrm{~T}}=\frac{\mathrm{W}}{\mathrm{Q}_{1}}$ $\frac{2}{3}=\frac{\mathrm{W}}{\mathrm{Q}_{1}}=\frac{\mathrm{Q}_{1}-\mathrm{Q}_{2}}{\mathrm{Q}_{1}}$ $\frac{2}{3}=1-\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}}$ $\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}}=\frac{1}{3}$ $\mathrm{Q}_{2}=\frac{\mathrm{Q}_{1}}{3}=\frac{\mathrm{Q}}{3}$
UPSEE - 2004
Thermodynamics
148537
The efficiency of an ideal Carnot engine working between temperature $T_{1}$ and $T_{2}$ is $1 / 3$. If the temperature of the sink is reduced by $\mathbf{4 0} \%$, then its efficiency will be
1 $50 \%$
2 $25 \%$
3 $60 \%$
4 $75 \%$
Explanation:
C Given, Efficiency of carnot engine $(\eta)=\frac{1}{3}$ $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ Where, $\mathrm{T}_{1}=\text { temperature of source }$ $\mathrm{T}_{2}=\text { temperature of sink }$ $\therefore \quad \frac{1}{3}=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}} \Rightarrow \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-\frac{1}{3} \Rightarrow \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{2}{3}$ When temperature of sink is reduced by $40 \%$ then its new temperature- $\mathrm{T}_{2}^{\prime} =\mathrm{T}_{2}-40 \% \text { of } \mathrm{T}_{2}$ $=\mathrm{T}_{2}-0.4 \mathrm{~T}_{2}$ $\mathrm{~T}_{2}^{\prime} =0.6 \mathrm{~T}_{2}$ Now, efficiency of Carnot's engine- $\eta^{\prime} =\left(1-\frac{\mathrm{T}_{2}^{\prime}}{\mathrm{T}_{1}}\right) \times 100=\left(1-0.6 \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right) \times 100$ $=\left(1-0.6 \times \frac{2}{3}\right) \times 100=0.6 \times 100=60 \%$
AP EAMCET (22.09.2020) Shift-II
Thermodynamics
148538
A Carnot's engine has an efficiency of $25 \%$ when its sink is at $27^{\circ} \mathrm{C}$. If it has to be increased to $40 \%$, what should be the temperature of the sink keeping the temperature of the source constant?
1 $320 \mathrm{~K}$
2 $375 \mathrm{~K}$
3 $240 \mathrm{~K}$
4 $300 \mathrm{~K}$
Explanation:
C Efficiency of carnot's engine, $\eta=25 \%=0.25$ Temperature of sink, $\mathrm{T}_{2}=27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K}$ Temperature of source $=\mathrm{T}_{1}$ $\because \quad \eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}} \Rightarrow 0.25=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-0.25=0.75$ $\mathrm{T}_{1}=\frac{\mathrm{T}_{2}}{0.75}=\frac{300}{0.75}=400 \mathrm{~K}$ $\mathrm{T}_{1}=400 \mathrm{~K}$ When efficiency, $\eta=40 \%=0.4$ $1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=0.4$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-0.4=0.6$ $\mathrm{~T}_{2}=0.6 \mathrm{~T}_{1}=0.6 \times 400$ $\quad=240 \mathrm{~K}$
AP EAMCET (22.09.2020) Shift-II
Thermodynamics
148539
A Carnot engine takes $3 \times 10^{6}$ calories of heat from reservoir at $627^{\circ} \mathrm{C}$ and gives it to a sink at $27^{\circ} \mathrm{C}$. The work done by the engine is
148535
A Carnot engine whose low temperature reservoir is at $7^{\circ} \mathrm{C}$ has an efficiency of $50 \%$. It is desired to increase the efficiency to $70 \%$. By how many degrees should the temperature of the high temperature reservoir be increased?
1 $840 \mathrm{~K}$
2 $280 \mathrm{~K}$
3 $560 \mathrm{~K}$
4 $380 \mathrm{~K}$
Explanation:
D Efficiency is defined as- $\eta=1-\frac{T_{2}}{T_{1}}$ Initially, $\frac{50}{100}=1-\frac{273+7}{T_{1}}$ $\frac{1}{2}=1-\frac{280}{T_{1}}$ $\frac{280}{T_{1}}=\frac{1}{2}$ $\mathrm{~T}_{1}=560 \mathrm{~K}$ Let new temperature of higher temperature reservoir is $\mathrm{T}_{1}^{\prime}-$ $\frac{70}{100}=1-\frac{280}{\mathrm{~T}_{1}^{\prime}}$ $\frac{280}{\mathrm{~T}_{1}^{\prime}}=\frac{3}{10}$ $\mathrm{~T}_{1}^{\prime}=\frac{280 \times 10}{3}=933 \mathrm{~K}$ $\therefore$ Increase in temperature $=933-560=373 \mathrm{~K}$ The nearest option is option (d). Hence, correct answer is $380 \mathrm{~K}$.
UPSEE - 2005
Thermodynamics
148536
A Carnot engine absorbs an amount $Q$ of heat from a reservoir at an absolute temperature $T$ and rejects heat to a sink at a temperature of $T / 3$. The amount of heat rejected is:
1 $\mathrm{Q} / 4$
2 $\mathrm{Q} / 3$
3 $\mathrm{Q} / 2$
4 $2 \mathrm{Q} / 3$
Explanation:
B We Know that, $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{\mathrm{W}}{\mathrm{Q}_{1}}=\frac{\mathrm{Q}_{1}-\mathrm{Q}_{2}}{\mathrm{Q}_{1}}$ Where, $\mathrm{Q}_{1}=$ heat absorbed, $\mathrm{Q}_{2}=$ heat rejected $1-\frac{\mathrm{T} / 3}{\mathrm{~T}}=\frac{\mathrm{W}}{\mathrm{Q}_{1}}$ $\frac{2}{3}=\frac{\mathrm{W}}{\mathrm{Q}_{1}}=\frac{\mathrm{Q}_{1}-\mathrm{Q}_{2}}{\mathrm{Q}_{1}}$ $\frac{2}{3}=1-\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}}$ $\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}}=\frac{1}{3}$ $\mathrm{Q}_{2}=\frac{\mathrm{Q}_{1}}{3}=\frac{\mathrm{Q}}{3}$
UPSEE - 2004
Thermodynamics
148537
The efficiency of an ideal Carnot engine working between temperature $T_{1}$ and $T_{2}$ is $1 / 3$. If the temperature of the sink is reduced by $\mathbf{4 0} \%$, then its efficiency will be
1 $50 \%$
2 $25 \%$
3 $60 \%$
4 $75 \%$
Explanation:
C Given, Efficiency of carnot engine $(\eta)=\frac{1}{3}$ $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ Where, $\mathrm{T}_{1}=\text { temperature of source }$ $\mathrm{T}_{2}=\text { temperature of sink }$ $\therefore \quad \frac{1}{3}=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}} \Rightarrow \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-\frac{1}{3} \Rightarrow \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{2}{3}$ When temperature of sink is reduced by $40 \%$ then its new temperature- $\mathrm{T}_{2}^{\prime} =\mathrm{T}_{2}-40 \% \text { of } \mathrm{T}_{2}$ $=\mathrm{T}_{2}-0.4 \mathrm{~T}_{2}$ $\mathrm{~T}_{2}^{\prime} =0.6 \mathrm{~T}_{2}$ Now, efficiency of Carnot's engine- $\eta^{\prime} =\left(1-\frac{\mathrm{T}_{2}^{\prime}}{\mathrm{T}_{1}}\right) \times 100=\left(1-0.6 \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right) \times 100$ $=\left(1-0.6 \times \frac{2}{3}\right) \times 100=0.6 \times 100=60 \%$
AP EAMCET (22.09.2020) Shift-II
Thermodynamics
148538
A Carnot's engine has an efficiency of $25 \%$ when its sink is at $27^{\circ} \mathrm{C}$. If it has to be increased to $40 \%$, what should be the temperature of the sink keeping the temperature of the source constant?
1 $320 \mathrm{~K}$
2 $375 \mathrm{~K}$
3 $240 \mathrm{~K}$
4 $300 \mathrm{~K}$
Explanation:
C Efficiency of carnot's engine, $\eta=25 \%=0.25$ Temperature of sink, $\mathrm{T}_{2}=27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K}$ Temperature of source $=\mathrm{T}_{1}$ $\because \quad \eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}} \Rightarrow 0.25=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-0.25=0.75$ $\mathrm{T}_{1}=\frac{\mathrm{T}_{2}}{0.75}=\frac{300}{0.75}=400 \mathrm{~K}$ $\mathrm{T}_{1}=400 \mathrm{~K}$ When efficiency, $\eta=40 \%=0.4$ $1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=0.4$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-0.4=0.6$ $\mathrm{~T}_{2}=0.6 \mathrm{~T}_{1}=0.6 \times 400$ $\quad=240 \mathrm{~K}$
AP EAMCET (22.09.2020) Shift-II
Thermodynamics
148539
A Carnot engine takes $3 \times 10^{6}$ calories of heat from reservoir at $627^{\circ} \mathrm{C}$ and gives it to a sink at $27^{\circ} \mathrm{C}$. The work done by the engine is
148535
A Carnot engine whose low temperature reservoir is at $7^{\circ} \mathrm{C}$ has an efficiency of $50 \%$. It is desired to increase the efficiency to $70 \%$. By how many degrees should the temperature of the high temperature reservoir be increased?
1 $840 \mathrm{~K}$
2 $280 \mathrm{~K}$
3 $560 \mathrm{~K}$
4 $380 \mathrm{~K}$
Explanation:
D Efficiency is defined as- $\eta=1-\frac{T_{2}}{T_{1}}$ Initially, $\frac{50}{100}=1-\frac{273+7}{T_{1}}$ $\frac{1}{2}=1-\frac{280}{T_{1}}$ $\frac{280}{T_{1}}=\frac{1}{2}$ $\mathrm{~T}_{1}=560 \mathrm{~K}$ Let new temperature of higher temperature reservoir is $\mathrm{T}_{1}^{\prime}-$ $\frac{70}{100}=1-\frac{280}{\mathrm{~T}_{1}^{\prime}}$ $\frac{280}{\mathrm{~T}_{1}^{\prime}}=\frac{3}{10}$ $\mathrm{~T}_{1}^{\prime}=\frac{280 \times 10}{3}=933 \mathrm{~K}$ $\therefore$ Increase in temperature $=933-560=373 \mathrm{~K}$ The nearest option is option (d). Hence, correct answer is $380 \mathrm{~K}$.
UPSEE - 2005
Thermodynamics
148536
A Carnot engine absorbs an amount $Q$ of heat from a reservoir at an absolute temperature $T$ and rejects heat to a sink at a temperature of $T / 3$. The amount of heat rejected is:
1 $\mathrm{Q} / 4$
2 $\mathrm{Q} / 3$
3 $\mathrm{Q} / 2$
4 $2 \mathrm{Q} / 3$
Explanation:
B We Know that, $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{\mathrm{W}}{\mathrm{Q}_{1}}=\frac{\mathrm{Q}_{1}-\mathrm{Q}_{2}}{\mathrm{Q}_{1}}$ Where, $\mathrm{Q}_{1}=$ heat absorbed, $\mathrm{Q}_{2}=$ heat rejected $1-\frac{\mathrm{T} / 3}{\mathrm{~T}}=\frac{\mathrm{W}}{\mathrm{Q}_{1}}$ $\frac{2}{3}=\frac{\mathrm{W}}{\mathrm{Q}_{1}}=\frac{\mathrm{Q}_{1}-\mathrm{Q}_{2}}{\mathrm{Q}_{1}}$ $\frac{2}{3}=1-\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}}$ $\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}}=\frac{1}{3}$ $\mathrm{Q}_{2}=\frac{\mathrm{Q}_{1}}{3}=\frac{\mathrm{Q}}{3}$
UPSEE - 2004
Thermodynamics
148537
The efficiency of an ideal Carnot engine working between temperature $T_{1}$ and $T_{2}$ is $1 / 3$. If the temperature of the sink is reduced by $\mathbf{4 0} \%$, then its efficiency will be
1 $50 \%$
2 $25 \%$
3 $60 \%$
4 $75 \%$
Explanation:
C Given, Efficiency of carnot engine $(\eta)=\frac{1}{3}$ $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ Where, $\mathrm{T}_{1}=\text { temperature of source }$ $\mathrm{T}_{2}=\text { temperature of sink }$ $\therefore \quad \frac{1}{3}=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}} \Rightarrow \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-\frac{1}{3} \Rightarrow \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{2}{3}$ When temperature of sink is reduced by $40 \%$ then its new temperature- $\mathrm{T}_{2}^{\prime} =\mathrm{T}_{2}-40 \% \text { of } \mathrm{T}_{2}$ $=\mathrm{T}_{2}-0.4 \mathrm{~T}_{2}$ $\mathrm{~T}_{2}^{\prime} =0.6 \mathrm{~T}_{2}$ Now, efficiency of Carnot's engine- $\eta^{\prime} =\left(1-\frac{\mathrm{T}_{2}^{\prime}}{\mathrm{T}_{1}}\right) \times 100=\left(1-0.6 \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right) \times 100$ $=\left(1-0.6 \times \frac{2}{3}\right) \times 100=0.6 \times 100=60 \%$
AP EAMCET (22.09.2020) Shift-II
Thermodynamics
148538
A Carnot's engine has an efficiency of $25 \%$ when its sink is at $27^{\circ} \mathrm{C}$. If it has to be increased to $40 \%$, what should be the temperature of the sink keeping the temperature of the source constant?
1 $320 \mathrm{~K}$
2 $375 \mathrm{~K}$
3 $240 \mathrm{~K}$
4 $300 \mathrm{~K}$
Explanation:
C Efficiency of carnot's engine, $\eta=25 \%=0.25$ Temperature of sink, $\mathrm{T}_{2}=27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K}$ Temperature of source $=\mathrm{T}_{1}$ $\because \quad \eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}} \Rightarrow 0.25=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-0.25=0.75$ $\mathrm{T}_{1}=\frac{\mathrm{T}_{2}}{0.75}=\frac{300}{0.75}=400 \mathrm{~K}$ $\mathrm{T}_{1}=400 \mathrm{~K}$ When efficiency, $\eta=40 \%=0.4$ $1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=0.4$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-0.4=0.6$ $\mathrm{~T}_{2}=0.6 \mathrm{~T}_{1}=0.6 \times 400$ $\quad=240 \mathrm{~K}$
AP EAMCET (22.09.2020) Shift-II
Thermodynamics
148539
A Carnot engine takes $3 \times 10^{6}$ calories of heat from reservoir at $627^{\circ} \mathrm{C}$ and gives it to a sink at $27^{\circ} \mathrm{C}$. The work done by the engine is