371401
A Carnot engine has efficiency \(25 \%\). It operates between reservoirs of constant temperatures with temperature difference of \(80\;K.\) What is the temperature of the low temperature reservoir?
1 \(-25^{\circ} C\)
2 \(25^{\circ} \mathrm{C}\)
3 \(-33^{\circ} C\)
4 \(33^{\circ} \mathrm{C}\)
Explanation:
Efficiency of a Carnot engine, i.e. \(\eta=1-\dfrac{T_{2}}{T_{1}}=\dfrac{T_{1}-T_{2}}{T_{1}}\) \(\Rightarrow \dfrac{25}{100}=\dfrac{T_{1}-\left(T_{1}-80\right)}{T_{1}}\) \(\Rightarrow \quad \dfrac{1}{4}=\dfrac{80}{T_{1}}\) \(\Rightarrow \quad T_{1}=320 \mathrm{~K}\) \(T_{1}=320-273\) \( = 47^\circ C\) Temperature of the low temperature reservoir \( = 47 - 80\) \( = - 33^\circ C\)
PHXI12:THERMODYNAMICS
371402
The temperature of food material in refrigerator is \(4^\circ C\) and temperature of environment is \(15^\circ C\). If Carnot cycle is used in its working gas, then find its Carnot efficiency.
1 0.038
2 0.028
3 0.053
4 0.072
Explanation:
Given, temperature of food material in refrigerator, \({T_2} = 4^\circ C = 273 + 4 = 277\;K\) Temperature of environment, \({T_1} = 15^\circ C = 273 + 15 = 288\;K\) \(\therefore\) Carnot efficiency, \(\eta=1-\dfrac{T_{2}}{T_{1}}\) \(=1-\dfrac{277}{288}=0.038\)
AIIMS - 2019
PHXI12:THERMODYNAMICS
371403
Two different adiabatic paths for the same gas intersect two isothermal curves as shown in \({P-V}\) diagram. The relation between the ratio \({\dfrac{V_{a}}{V_{d}}}\) and the ratio \({\dfrac{V_{b}}{V_{c}}}\) is
For an adiabatic process \({T V^{\gamma-1}=}\) constant \({\therefore T_{b} V_{b}^{\gamma-1}=T_{c} V_{c}^{\gamma-1} \quad}\) (1) (Adiabatic process) \({\therefore T_{a} V_{a}^{\gamma-1}=T_{d} V_{d}^{\gamma-1} \quad}\) (2) (Adiabatic process) Taking ratio of (1) and (2) \({\dfrac{T_{b} V_{b}^{\gamma-1}}{T_{a} V_{a}^{\gamma-1}}=\dfrac{T_{c} V_{c}^{\gamma-1}}{T_{a} V_{a}^{\gamma-1}}}\) For given isothermal curve, \({T_{b}=T_{c}}\) and \({T_{a}=T_{d} \cdot}\) So, \({\Rightarrow \dfrac{V_{b}}{V_{a}}=\dfrac{V_{c}}{V_{d}} \Rightarrow \dfrac{V_{a}}{V_{d}}=\dfrac{V_{b}}{V_{c}}}\) So, correct option is (2).
JEE - 2024
PHXI12:THERMODYNAMICS
371404
A carnot engine, whose efficiency is \(40{\rm{ }}\,\% \), takes in heat from a source maintained at a temperature of \(500\;K\). It is desired to have engine of efficiency \(60\,{\rm{ }}\% \). Then the intake temperature for the same exhaust (sink) temperature must be
1 \(1200\;K\)
2 Efficiency of carnot engine cannot be made larger than \(50 \%\)
3 \(600\;K\)
4 \(750\;K\)
Explanation:
For \(1^{s t}\) case efficiency \(=\eta=\left(1-\dfrac{T_{2}}{T_{1}}\right) \times 100\) \(\left(1-\dfrac{T_{2}}{500}\right) \times 100=40\) \(T_{2}=300 K\) for \(2^{\text {nd }}\) case \(\eta=\left(1-\dfrac{300}{T_{1}}\right) \times 100=60\) \({T_1} = 750\;K.\)
PHXI12:THERMODYNAMICS
371405
In Carnot engine efficiency is \(40\,\% \) at hot reservoir temperature T. For efficiency \(50\,\% \) what will be temperature of hot reservoir?
1 \(\dfrac{T}{5}\)
2 \(\dfrac{2 T}{5}\)
3 \(6\;T\)
4 \(\dfrac{6 T}{5}\)
Explanation:
\(\eta=1-\dfrac{T_{2}}{T_{1}} \quad T_{1}=T\) (Temperature of hot reservoir) \(\begin{aligned}& \eta=40 \%, \\& \dfrac{40}{100}=1-\dfrac{T_{2}}{T} \Rightarrow \dfrac{T_{2}}{T}=\dfrac{3}{5} \Rightarrow T_{2}=\dfrac{3}{5} T\end{aligned}\) For \(\eta = 50\,\% \), \(\dfrac{50}{100}=1-\dfrac{5}{T_{1}} \Rightarrow T_{1}=\dfrac{6 T}{5}\)
371401
A Carnot engine has efficiency \(25 \%\). It operates between reservoirs of constant temperatures with temperature difference of \(80\;K.\) What is the temperature of the low temperature reservoir?
1 \(-25^{\circ} C\)
2 \(25^{\circ} \mathrm{C}\)
3 \(-33^{\circ} C\)
4 \(33^{\circ} \mathrm{C}\)
Explanation:
Efficiency of a Carnot engine, i.e. \(\eta=1-\dfrac{T_{2}}{T_{1}}=\dfrac{T_{1}-T_{2}}{T_{1}}\) \(\Rightarrow \dfrac{25}{100}=\dfrac{T_{1}-\left(T_{1}-80\right)}{T_{1}}\) \(\Rightarrow \quad \dfrac{1}{4}=\dfrac{80}{T_{1}}\) \(\Rightarrow \quad T_{1}=320 \mathrm{~K}\) \(T_{1}=320-273\) \( = 47^\circ C\) Temperature of the low temperature reservoir \( = 47 - 80\) \( = - 33^\circ C\)
PHXI12:THERMODYNAMICS
371402
The temperature of food material in refrigerator is \(4^\circ C\) and temperature of environment is \(15^\circ C\). If Carnot cycle is used in its working gas, then find its Carnot efficiency.
1 0.038
2 0.028
3 0.053
4 0.072
Explanation:
Given, temperature of food material in refrigerator, \({T_2} = 4^\circ C = 273 + 4 = 277\;K\) Temperature of environment, \({T_1} = 15^\circ C = 273 + 15 = 288\;K\) \(\therefore\) Carnot efficiency, \(\eta=1-\dfrac{T_{2}}{T_{1}}\) \(=1-\dfrac{277}{288}=0.038\)
AIIMS - 2019
PHXI12:THERMODYNAMICS
371403
Two different adiabatic paths for the same gas intersect two isothermal curves as shown in \({P-V}\) diagram. The relation between the ratio \({\dfrac{V_{a}}{V_{d}}}\) and the ratio \({\dfrac{V_{b}}{V_{c}}}\) is
For an adiabatic process \({T V^{\gamma-1}=}\) constant \({\therefore T_{b} V_{b}^{\gamma-1}=T_{c} V_{c}^{\gamma-1} \quad}\) (1) (Adiabatic process) \({\therefore T_{a} V_{a}^{\gamma-1}=T_{d} V_{d}^{\gamma-1} \quad}\) (2) (Adiabatic process) Taking ratio of (1) and (2) \({\dfrac{T_{b} V_{b}^{\gamma-1}}{T_{a} V_{a}^{\gamma-1}}=\dfrac{T_{c} V_{c}^{\gamma-1}}{T_{a} V_{a}^{\gamma-1}}}\) For given isothermal curve, \({T_{b}=T_{c}}\) and \({T_{a}=T_{d} \cdot}\) So, \({\Rightarrow \dfrac{V_{b}}{V_{a}}=\dfrac{V_{c}}{V_{d}} \Rightarrow \dfrac{V_{a}}{V_{d}}=\dfrac{V_{b}}{V_{c}}}\) So, correct option is (2).
JEE - 2024
PHXI12:THERMODYNAMICS
371404
A carnot engine, whose efficiency is \(40{\rm{ }}\,\% \), takes in heat from a source maintained at a temperature of \(500\;K\). It is desired to have engine of efficiency \(60\,{\rm{ }}\% \). Then the intake temperature for the same exhaust (sink) temperature must be
1 \(1200\;K\)
2 Efficiency of carnot engine cannot be made larger than \(50 \%\)
3 \(600\;K\)
4 \(750\;K\)
Explanation:
For \(1^{s t}\) case efficiency \(=\eta=\left(1-\dfrac{T_{2}}{T_{1}}\right) \times 100\) \(\left(1-\dfrac{T_{2}}{500}\right) \times 100=40\) \(T_{2}=300 K\) for \(2^{\text {nd }}\) case \(\eta=\left(1-\dfrac{300}{T_{1}}\right) \times 100=60\) \({T_1} = 750\;K.\)
PHXI12:THERMODYNAMICS
371405
In Carnot engine efficiency is \(40\,\% \) at hot reservoir temperature T. For efficiency \(50\,\% \) what will be temperature of hot reservoir?
1 \(\dfrac{T}{5}\)
2 \(\dfrac{2 T}{5}\)
3 \(6\;T\)
4 \(\dfrac{6 T}{5}\)
Explanation:
\(\eta=1-\dfrac{T_{2}}{T_{1}} \quad T_{1}=T\) (Temperature of hot reservoir) \(\begin{aligned}& \eta=40 \%, \\& \dfrac{40}{100}=1-\dfrac{T_{2}}{T} \Rightarrow \dfrac{T_{2}}{T}=\dfrac{3}{5} \Rightarrow T_{2}=\dfrac{3}{5} T\end{aligned}\) For \(\eta = 50\,\% \), \(\dfrac{50}{100}=1-\dfrac{5}{T_{1}} \Rightarrow T_{1}=\dfrac{6 T}{5}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXI12:THERMODYNAMICS
371401
A Carnot engine has efficiency \(25 \%\). It operates between reservoirs of constant temperatures with temperature difference of \(80\;K.\) What is the temperature of the low temperature reservoir?
1 \(-25^{\circ} C\)
2 \(25^{\circ} \mathrm{C}\)
3 \(-33^{\circ} C\)
4 \(33^{\circ} \mathrm{C}\)
Explanation:
Efficiency of a Carnot engine, i.e. \(\eta=1-\dfrac{T_{2}}{T_{1}}=\dfrac{T_{1}-T_{2}}{T_{1}}\) \(\Rightarrow \dfrac{25}{100}=\dfrac{T_{1}-\left(T_{1}-80\right)}{T_{1}}\) \(\Rightarrow \quad \dfrac{1}{4}=\dfrac{80}{T_{1}}\) \(\Rightarrow \quad T_{1}=320 \mathrm{~K}\) \(T_{1}=320-273\) \( = 47^\circ C\) Temperature of the low temperature reservoir \( = 47 - 80\) \( = - 33^\circ C\)
PHXI12:THERMODYNAMICS
371402
The temperature of food material in refrigerator is \(4^\circ C\) and temperature of environment is \(15^\circ C\). If Carnot cycle is used in its working gas, then find its Carnot efficiency.
1 0.038
2 0.028
3 0.053
4 0.072
Explanation:
Given, temperature of food material in refrigerator, \({T_2} = 4^\circ C = 273 + 4 = 277\;K\) Temperature of environment, \({T_1} = 15^\circ C = 273 + 15 = 288\;K\) \(\therefore\) Carnot efficiency, \(\eta=1-\dfrac{T_{2}}{T_{1}}\) \(=1-\dfrac{277}{288}=0.038\)
AIIMS - 2019
PHXI12:THERMODYNAMICS
371403
Two different adiabatic paths for the same gas intersect two isothermal curves as shown in \({P-V}\) diagram. The relation between the ratio \({\dfrac{V_{a}}{V_{d}}}\) and the ratio \({\dfrac{V_{b}}{V_{c}}}\) is
For an adiabatic process \({T V^{\gamma-1}=}\) constant \({\therefore T_{b} V_{b}^{\gamma-1}=T_{c} V_{c}^{\gamma-1} \quad}\) (1) (Adiabatic process) \({\therefore T_{a} V_{a}^{\gamma-1}=T_{d} V_{d}^{\gamma-1} \quad}\) (2) (Adiabatic process) Taking ratio of (1) and (2) \({\dfrac{T_{b} V_{b}^{\gamma-1}}{T_{a} V_{a}^{\gamma-1}}=\dfrac{T_{c} V_{c}^{\gamma-1}}{T_{a} V_{a}^{\gamma-1}}}\) For given isothermal curve, \({T_{b}=T_{c}}\) and \({T_{a}=T_{d} \cdot}\) So, \({\Rightarrow \dfrac{V_{b}}{V_{a}}=\dfrac{V_{c}}{V_{d}} \Rightarrow \dfrac{V_{a}}{V_{d}}=\dfrac{V_{b}}{V_{c}}}\) So, correct option is (2).
JEE - 2024
PHXI12:THERMODYNAMICS
371404
A carnot engine, whose efficiency is \(40{\rm{ }}\,\% \), takes in heat from a source maintained at a temperature of \(500\;K\). It is desired to have engine of efficiency \(60\,{\rm{ }}\% \). Then the intake temperature for the same exhaust (sink) temperature must be
1 \(1200\;K\)
2 Efficiency of carnot engine cannot be made larger than \(50 \%\)
3 \(600\;K\)
4 \(750\;K\)
Explanation:
For \(1^{s t}\) case efficiency \(=\eta=\left(1-\dfrac{T_{2}}{T_{1}}\right) \times 100\) \(\left(1-\dfrac{T_{2}}{500}\right) \times 100=40\) \(T_{2}=300 K\) for \(2^{\text {nd }}\) case \(\eta=\left(1-\dfrac{300}{T_{1}}\right) \times 100=60\) \({T_1} = 750\;K.\)
PHXI12:THERMODYNAMICS
371405
In Carnot engine efficiency is \(40\,\% \) at hot reservoir temperature T. For efficiency \(50\,\% \) what will be temperature of hot reservoir?
1 \(\dfrac{T}{5}\)
2 \(\dfrac{2 T}{5}\)
3 \(6\;T\)
4 \(\dfrac{6 T}{5}\)
Explanation:
\(\eta=1-\dfrac{T_{2}}{T_{1}} \quad T_{1}=T\) (Temperature of hot reservoir) \(\begin{aligned}& \eta=40 \%, \\& \dfrac{40}{100}=1-\dfrac{T_{2}}{T} \Rightarrow \dfrac{T_{2}}{T}=\dfrac{3}{5} \Rightarrow T_{2}=\dfrac{3}{5} T\end{aligned}\) For \(\eta = 50\,\% \), \(\dfrac{50}{100}=1-\dfrac{5}{T_{1}} \Rightarrow T_{1}=\dfrac{6 T}{5}\)
371401
A Carnot engine has efficiency \(25 \%\). It operates between reservoirs of constant temperatures with temperature difference of \(80\;K.\) What is the temperature of the low temperature reservoir?
1 \(-25^{\circ} C\)
2 \(25^{\circ} \mathrm{C}\)
3 \(-33^{\circ} C\)
4 \(33^{\circ} \mathrm{C}\)
Explanation:
Efficiency of a Carnot engine, i.e. \(\eta=1-\dfrac{T_{2}}{T_{1}}=\dfrac{T_{1}-T_{2}}{T_{1}}\) \(\Rightarrow \dfrac{25}{100}=\dfrac{T_{1}-\left(T_{1}-80\right)}{T_{1}}\) \(\Rightarrow \quad \dfrac{1}{4}=\dfrac{80}{T_{1}}\) \(\Rightarrow \quad T_{1}=320 \mathrm{~K}\) \(T_{1}=320-273\) \( = 47^\circ C\) Temperature of the low temperature reservoir \( = 47 - 80\) \( = - 33^\circ C\)
PHXI12:THERMODYNAMICS
371402
The temperature of food material in refrigerator is \(4^\circ C\) and temperature of environment is \(15^\circ C\). If Carnot cycle is used in its working gas, then find its Carnot efficiency.
1 0.038
2 0.028
3 0.053
4 0.072
Explanation:
Given, temperature of food material in refrigerator, \({T_2} = 4^\circ C = 273 + 4 = 277\;K\) Temperature of environment, \({T_1} = 15^\circ C = 273 + 15 = 288\;K\) \(\therefore\) Carnot efficiency, \(\eta=1-\dfrac{T_{2}}{T_{1}}\) \(=1-\dfrac{277}{288}=0.038\)
AIIMS - 2019
PHXI12:THERMODYNAMICS
371403
Two different adiabatic paths for the same gas intersect two isothermal curves as shown in \({P-V}\) diagram. The relation between the ratio \({\dfrac{V_{a}}{V_{d}}}\) and the ratio \({\dfrac{V_{b}}{V_{c}}}\) is
For an adiabatic process \({T V^{\gamma-1}=}\) constant \({\therefore T_{b} V_{b}^{\gamma-1}=T_{c} V_{c}^{\gamma-1} \quad}\) (1) (Adiabatic process) \({\therefore T_{a} V_{a}^{\gamma-1}=T_{d} V_{d}^{\gamma-1} \quad}\) (2) (Adiabatic process) Taking ratio of (1) and (2) \({\dfrac{T_{b} V_{b}^{\gamma-1}}{T_{a} V_{a}^{\gamma-1}}=\dfrac{T_{c} V_{c}^{\gamma-1}}{T_{a} V_{a}^{\gamma-1}}}\) For given isothermal curve, \({T_{b}=T_{c}}\) and \({T_{a}=T_{d} \cdot}\) So, \({\Rightarrow \dfrac{V_{b}}{V_{a}}=\dfrac{V_{c}}{V_{d}} \Rightarrow \dfrac{V_{a}}{V_{d}}=\dfrac{V_{b}}{V_{c}}}\) So, correct option is (2).
JEE - 2024
PHXI12:THERMODYNAMICS
371404
A carnot engine, whose efficiency is \(40{\rm{ }}\,\% \), takes in heat from a source maintained at a temperature of \(500\;K\). It is desired to have engine of efficiency \(60\,{\rm{ }}\% \). Then the intake temperature for the same exhaust (sink) temperature must be
1 \(1200\;K\)
2 Efficiency of carnot engine cannot be made larger than \(50 \%\)
3 \(600\;K\)
4 \(750\;K\)
Explanation:
For \(1^{s t}\) case efficiency \(=\eta=\left(1-\dfrac{T_{2}}{T_{1}}\right) \times 100\) \(\left(1-\dfrac{T_{2}}{500}\right) \times 100=40\) \(T_{2}=300 K\) for \(2^{\text {nd }}\) case \(\eta=\left(1-\dfrac{300}{T_{1}}\right) \times 100=60\) \({T_1} = 750\;K.\)
PHXI12:THERMODYNAMICS
371405
In Carnot engine efficiency is \(40\,\% \) at hot reservoir temperature T. For efficiency \(50\,\% \) what will be temperature of hot reservoir?
1 \(\dfrac{T}{5}\)
2 \(\dfrac{2 T}{5}\)
3 \(6\;T\)
4 \(\dfrac{6 T}{5}\)
Explanation:
\(\eta=1-\dfrac{T_{2}}{T_{1}} \quad T_{1}=T\) (Temperature of hot reservoir) \(\begin{aligned}& \eta=40 \%, \\& \dfrac{40}{100}=1-\dfrac{T_{2}}{T} \Rightarrow \dfrac{T_{2}}{T}=\dfrac{3}{5} \Rightarrow T_{2}=\dfrac{3}{5} T\end{aligned}\) For \(\eta = 50\,\% \), \(\dfrac{50}{100}=1-\dfrac{5}{T_{1}} \Rightarrow T_{1}=\dfrac{6 T}{5}\)
371401
A Carnot engine has efficiency \(25 \%\). It operates between reservoirs of constant temperatures with temperature difference of \(80\;K.\) What is the temperature of the low temperature reservoir?
1 \(-25^{\circ} C\)
2 \(25^{\circ} \mathrm{C}\)
3 \(-33^{\circ} C\)
4 \(33^{\circ} \mathrm{C}\)
Explanation:
Efficiency of a Carnot engine, i.e. \(\eta=1-\dfrac{T_{2}}{T_{1}}=\dfrac{T_{1}-T_{2}}{T_{1}}\) \(\Rightarrow \dfrac{25}{100}=\dfrac{T_{1}-\left(T_{1}-80\right)}{T_{1}}\) \(\Rightarrow \quad \dfrac{1}{4}=\dfrac{80}{T_{1}}\) \(\Rightarrow \quad T_{1}=320 \mathrm{~K}\) \(T_{1}=320-273\) \( = 47^\circ C\) Temperature of the low temperature reservoir \( = 47 - 80\) \( = - 33^\circ C\)
PHXI12:THERMODYNAMICS
371402
The temperature of food material in refrigerator is \(4^\circ C\) and temperature of environment is \(15^\circ C\). If Carnot cycle is used in its working gas, then find its Carnot efficiency.
1 0.038
2 0.028
3 0.053
4 0.072
Explanation:
Given, temperature of food material in refrigerator, \({T_2} = 4^\circ C = 273 + 4 = 277\;K\) Temperature of environment, \({T_1} = 15^\circ C = 273 + 15 = 288\;K\) \(\therefore\) Carnot efficiency, \(\eta=1-\dfrac{T_{2}}{T_{1}}\) \(=1-\dfrac{277}{288}=0.038\)
AIIMS - 2019
PHXI12:THERMODYNAMICS
371403
Two different adiabatic paths for the same gas intersect two isothermal curves as shown in \({P-V}\) diagram. The relation between the ratio \({\dfrac{V_{a}}{V_{d}}}\) and the ratio \({\dfrac{V_{b}}{V_{c}}}\) is
For an adiabatic process \({T V^{\gamma-1}=}\) constant \({\therefore T_{b} V_{b}^{\gamma-1}=T_{c} V_{c}^{\gamma-1} \quad}\) (1) (Adiabatic process) \({\therefore T_{a} V_{a}^{\gamma-1}=T_{d} V_{d}^{\gamma-1} \quad}\) (2) (Adiabatic process) Taking ratio of (1) and (2) \({\dfrac{T_{b} V_{b}^{\gamma-1}}{T_{a} V_{a}^{\gamma-1}}=\dfrac{T_{c} V_{c}^{\gamma-1}}{T_{a} V_{a}^{\gamma-1}}}\) For given isothermal curve, \({T_{b}=T_{c}}\) and \({T_{a}=T_{d} \cdot}\) So, \({\Rightarrow \dfrac{V_{b}}{V_{a}}=\dfrac{V_{c}}{V_{d}} \Rightarrow \dfrac{V_{a}}{V_{d}}=\dfrac{V_{b}}{V_{c}}}\) So, correct option is (2).
JEE - 2024
PHXI12:THERMODYNAMICS
371404
A carnot engine, whose efficiency is \(40{\rm{ }}\,\% \), takes in heat from a source maintained at a temperature of \(500\;K\). It is desired to have engine of efficiency \(60\,{\rm{ }}\% \). Then the intake temperature for the same exhaust (sink) temperature must be
1 \(1200\;K\)
2 Efficiency of carnot engine cannot be made larger than \(50 \%\)
3 \(600\;K\)
4 \(750\;K\)
Explanation:
For \(1^{s t}\) case efficiency \(=\eta=\left(1-\dfrac{T_{2}}{T_{1}}\right) \times 100\) \(\left(1-\dfrac{T_{2}}{500}\right) \times 100=40\) \(T_{2}=300 K\) for \(2^{\text {nd }}\) case \(\eta=\left(1-\dfrac{300}{T_{1}}\right) \times 100=60\) \({T_1} = 750\;K.\)
PHXI12:THERMODYNAMICS
371405
In Carnot engine efficiency is \(40\,\% \) at hot reservoir temperature T. For efficiency \(50\,\% \) what will be temperature of hot reservoir?
1 \(\dfrac{T}{5}\)
2 \(\dfrac{2 T}{5}\)
3 \(6\;T\)
4 \(\dfrac{6 T}{5}\)
Explanation:
\(\eta=1-\dfrac{T_{2}}{T_{1}} \quad T_{1}=T\) (Temperature of hot reservoir) \(\begin{aligned}& \eta=40 \%, \\& \dfrac{40}{100}=1-\dfrac{T_{2}}{T} \Rightarrow \dfrac{T_{2}}{T}=\dfrac{3}{5} \Rightarrow T_{2}=\dfrac{3}{5} T\end{aligned}\) For \(\eta = 50\,\% \), \(\dfrac{50}{100}=1-\dfrac{5}{T_{1}} \Rightarrow T_{1}=\dfrac{6 T}{5}\)
