371354
The freezer in a refrigerator is located at the top section, so that
1 the entire chamber of the refrigerator is cooled quickly due to convection
2 the motor is not heated
3 the heat gained from the environment is high
4 the heat gained from the environment is low
Explanation:
The freezer in a refrigerator is located at the top section, so that the entire chamber of the refrigerator is cooled quickly due to convection. In convection, transfer of heat takes place by transport of matter in the form of motion of particles.
PHXI12:THERMODYNAMICS
371355
If \(\alpha\) is the coefficient of performance of a refrigerator and ' \(Q_{1}\) ' is heat released to the hot reservoir, then the heat extracted from the cold reservoir ' \(Q_{2}\) ' is
1 \(\dfrac{\alpha Q_{1}}{\alpha-1}\)
2 \(\dfrac{\alpha-1}{\alpha} Q_{1}\)
3 \(\dfrac{\alpha Q_{1}}{1+\alpha}\)
4 \(\dfrac{1+\alpha}{\alpha} Q_{1}\)
Explanation:
As we know that the coefficient of performance of a refrigerator, \(\alpha=\dfrac{Q_{2}}{W}\left[\therefore W=Q_{1}-Q_{2}\right]\) \(\Rightarrow \alpha=\dfrac{Q_{2}}{Q_{1}-Q_{2}}\) where, \(Q_{2}=\) heat extracted from the cold reservoir, \(Q_{1}=\) heat released to the hot reservoir, \(W = \) work done by heat pump and \(\alpha=\) coefficient of performance So, \(\alpha\left(Q_{1}-Q_{2}\right)=Q_{2} \quad\) [from eq.(1)] \(\Rightarrow Q_{2}=\dfrac{\alpha Q_{1}}{1+\alpha}\)
PHXI12:THERMODYNAMICS
371356
A refrigerator is to remove heat from the eatable kept inside at \(9^\circ C\). Calculate the coefficient of performance, if the room temperature is \(36^\circ C.\)
1 10.4
2 11.5
3 9.8
4 None of these
Explanation:
Given, temperature of source \(\left(T_{1}\right)=(36+273) K=309 K\) Temperature of sink \(\left(T_{1}\right)=(9+273) K=282 K\) Coefficient of performance of a refrigerator \(\beta=\dfrac{T_{2}}{T_{1}-T_{2}}=\dfrac{282}{309-282}=10.4\)
NCERT Exemplar
PHXI12:THERMODYNAMICS
371357
In a mechanical refrigerator, the low temperature coils are at a temperature of \( - 23^\circ C\) and the compressed gas in the condenser has a temperature of \(27^\circ C\). The theoretical coefficient of performance is
1 8
2 5
3 6.5
4 6
Explanation:
Coefficient of performance \(\beta=\dfrac{T_{2}}{T_{1}-T_{2}}\) \(=\dfrac{(273-23)}{(273+27)-(273-23)}=\dfrac{250}{300-250}=\dfrac{250}{50}=5 .\)
371354
The freezer in a refrigerator is located at the top section, so that
1 the entire chamber of the refrigerator is cooled quickly due to convection
2 the motor is not heated
3 the heat gained from the environment is high
4 the heat gained from the environment is low
Explanation:
The freezer in a refrigerator is located at the top section, so that the entire chamber of the refrigerator is cooled quickly due to convection. In convection, transfer of heat takes place by transport of matter in the form of motion of particles.
PHXI12:THERMODYNAMICS
371355
If \(\alpha\) is the coefficient of performance of a refrigerator and ' \(Q_{1}\) ' is heat released to the hot reservoir, then the heat extracted from the cold reservoir ' \(Q_{2}\) ' is
1 \(\dfrac{\alpha Q_{1}}{\alpha-1}\)
2 \(\dfrac{\alpha-1}{\alpha} Q_{1}\)
3 \(\dfrac{\alpha Q_{1}}{1+\alpha}\)
4 \(\dfrac{1+\alpha}{\alpha} Q_{1}\)
Explanation:
As we know that the coefficient of performance of a refrigerator, \(\alpha=\dfrac{Q_{2}}{W}\left[\therefore W=Q_{1}-Q_{2}\right]\) \(\Rightarrow \alpha=\dfrac{Q_{2}}{Q_{1}-Q_{2}}\) where, \(Q_{2}=\) heat extracted from the cold reservoir, \(Q_{1}=\) heat released to the hot reservoir, \(W = \) work done by heat pump and \(\alpha=\) coefficient of performance So, \(\alpha\left(Q_{1}-Q_{2}\right)=Q_{2} \quad\) [from eq.(1)] \(\Rightarrow Q_{2}=\dfrac{\alpha Q_{1}}{1+\alpha}\)
PHXI12:THERMODYNAMICS
371356
A refrigerator is to remove heat from the eatable kept inside at \(9^\circ C\). Calculate the coefficient of performance, if the room temperature is \(36^\circ C.\)
1 10.4
2 11.5
3 9.8
4 None of these
Explanation:
Given, temperature of source \(\left(T_{1}\right)=(36+273) K=309 K\) Temperature of sink \(\left(T_{1}\right)=(9+273) K=282 K\) Coefficient of performance of a refrigerator \(\beta=\dfrac{T_{2}}{T_{1}-T_{2}}=\dfrac{282}{309-282}=10.4\)
NCERT Exemplar
PHXI12:THERMODYNAMICS
371357
In a mechanical refrigerator, the low temperature coils are at a temperature of \( - 23^\circ C\) and the compressed gas in the condenser has a temperature of \(27^\circ C\). The theoretical coefficient of performance is
1 8
2 5
3 6.5
4 6
Explanation:
Coefficient of performance \(\beta=\dfrac{T_{2}}{T_{1}-T_{2}}\) \(=\dfrac{(273-23)}{(273+27)-(273-23)}=\dfrac{250}{300-250}=\dfrac{250}{50}=5 .\)
371354
The freezer in a refrigerator is located at the top section, so that
1 the entire chamber of the refrigerator is cooled quickly due to convection
2 the motor is not heated
3 the heat gained from the environment is high
4 the heat gained from the environment is low
Explanation:
The freezer in a refrigerator is located at the top section, so that the entire chamber of the refrigerator is cooled quickly due to convection. In convection, transfer of heat takes place by transport of matter in the form of motion of particles.
PHXI12:THERMODYNAMICS
371355
If \(\alpha\) is the coefficient of performance of a refrigerator and ' \(Q_{1}\) ' is heat released to the hot reservoir, then the heat extracted from the cold reservoir ' \(Q_{2}\) ' is
1 \(\dfrac{\alpha Q_{1}}{\alpha-1}\)
2 \(\dfrac{\alpha-1}{\alpha} Q_{1}\)
3 \(\dfrac{\alpha Q_{1}}{1+\alpha}\)
4 \(\dfrac{1+\alpha}{\alpha} Q_{1}\)
Explanation:
As we know that the coefficient of performance of a refrigerator, \(\alpha=\dfrac{Q_{2}}{W}\left[\therefore W=Q_{1}-Q_{2}\right]\) \(\Rightarrow \alpha=\dfrac{Q_{2}}{Q_{1}-Q_{2}}\) where, \(Q_{2}=\) heat extracted from the cold reservoir, \(Q_{1}=\) heat released to the hot reservoir, \(W = \) work done by heat pump and \(\alpha=\) coefficient of performance So, \(\alpha\left(Q_{1}-Q_{2}\right)=Q_{2} \quad\) [from eq.(1)] \(\Rightarrow Q_{2}=\dfrac{\alpha Q_{1}}{1+\alpha}\)
PHXI12:THERMODYNAMICS
371356
A refrigerator is to remove heat from the eatable kept inside at \(9^\circ C\). Calculate the coefficient of performance, if the room temperature is \(36^\circ C.\)
1 10.4
2 11.5
3 9.8
4 None of these
Explanation:
Given, temperature of source \(\left(T_{1}\right)=(36+273) K=309 K\) Temperature of sink \(\left(T_{1}\right)=(9+273) K=282 K\) Coefficient of performance of a refrigerator \(\beta=\dfrac{T_{2}}{T_{1}-T_{2}}=\dfrac{282}{309-282}=10.4\)
NCERT Exemplar
PHXI12:THERMODYNAMICS
371357
In a mechanical refrigerator, the low temperature coils are at a temperature of \( - 23^\circ C\) and the compressed gas in the condenser has a temperature of \(27^\circ C\). The theoretical coefficient of performance is
1 8
2 5
3 6.5
4 6
Explanation:
Coefficient of performance \(\beta=\dfrac{T_{2}}{T_{1}-T_{2}}\) \(=\dfrac{(273-23)}{(273+27)-(273-23)}=\dfrac{250}{300-250}=\dfrac{250}{50}=5 .\)
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PHXI12:THERMODYNAMICS
371354
The freezer in a refrigerator is located at the top section, so that
1 the entire chamber of the refrigerator is cooled quickly due to convection
2 the motor is not heated
3 the heat gained from the environment is high
4 the heat gained from the environment is low
Explanation:
The freezer in a refrigerator is located at the top section, so that the entire chamber of the refrigerator is cooled quickly due to convection. In convection, transfer of heat takes place by transport of matter in the form of motion of particles.
PHXI12:THERMODYNAMICS
371355
If \(\alpha\) is the coefficient of performance of a refrigerator and ' \(Q_{1}\) ' is heat released to the hot reservoir, then the heat extracted from the cold reservoir ' \(Q_{2}\) ' is
1 \(\dfrac{\alpha Q_{1}}{\alpha-1}\)
2 \(\dfrac{\alpha-1}{\alpha} Q_{1}\)
3 \(\dfrac{\alpha Q_{1}}{1+\alpha}\)
4 \(\dfrac{1+\alpha}{\alpha} Q_{1}\)
Explanation:
As we know that the coefficient of performance of a refrigerator, \(\alpha=\dfrac{Q_{2}}{W}\left[\therefore W=Q_{1}-Q_{2}\right]\) \(\Rightarrow \alpha=\dfrac{Q_{2}}{Q_{1}-Q_{2}}\) where, \(Q_{2}=\) heat extracted from the cold reservoir, \(Q_{1}=\) heat released to the hot reservoir, \(W = \) work done by heat pump and \(\alpha=\) coefficient of performance So, \(\alpha\left(Q_{1}-Q_{2}\right)=Q_{2} \quad\) [from eq.(1)] \(\Rightarrow Q_{2}=\dfrac{\alpha Q_{1}}{1+\alpha}\)
PHXI12:THERMODYNAMICS
371356
A refrigerator is to remove heat from the eatable kept inside at \(9^\circ C\). Calculate the coefficient of performance, if the room temperature is \(36^\circ C.\)
1 10.4
2 11.5
3 9.8
4 None of these
Explanation:
Given, temperature of source \(\left(T_{1}\right)=(36+273) K=309 K\) Temperature of sink \(\left(T_{1}\right)=(9+273) K=282 K\) Coefficient of performance of a refrigerator \(\beta=\dfrac{T_{2}}{T_{1}-T_{2}}=\dfrac{282}{309-282}=10.4\)
NCERT Exemplar
PHXI12:THERMODYNAMICS
371357
In a mechanical refrigerator, the low temperature coils are at a temperature of \( - 23^\circ C\) and the compressed gas in the condenser has a temperature of \(27^\circ C\). The theoretical coefficient of performance is
1 8
2 5
3 6.5
4 6
Explanation:
Coefficient of performance \(\beta=\dfrac{T_{2}}{T_{1}-T_{2}}\) \(=\dfrac{(273-23)}{(273+27)-(273-23)}=\dfrac{250}{300-250}=\dfrac{250}{50}=5 .\)
