149590
If the Wien's constant $b=0.3 \mathrm{~cm}-K$, then the temperature of the sun having maximum intensity of radiation at $6000 \mathrm{~A}^{\circ}$ wavelength is
1 $2000 \mathrm{~K}$
2 $5000 \mathrm{~K}$
3 $6000 \mathrm{~K}$
4 $7000 \mathrm{~K}$
Explanation:
B Given, Wein's constant $b=0.3 \mathrm{~cm}-\mathrm{K}$ Wavelength of radiation $=6000 \AA$ We know that, Wien's displacement Law $\lambda \mathrm{T}=$ constant $\therefore \quad 6000 \times \mathrm{T}=0.3$ $\mathrm{T}=\frac{0.3}{6000 \AA} \mathrm{cmK}$ $=\frac{0.3 \times 10^{-2}}{6000 \times 10^{-10}} \mathrm{cmK}$ $\mathrm{T}=5000 \mathrm{~K}$
UPSEE 2020
Heat Transfer
149595
Two bodies of same shape, same size and same radiating power have emissivity's 0.2 and 0.8 . The ratio of their temperatures is:
1 $\sqrt{3}: 1$
2 $\sqrt{2}: 1$
3 $1: \sqrt{5}$
4 $1: \sqrt{8}$
Explanation:
B From Stefan - Boltzmann equation, $\mathrm{P}=\sigma \mathrm{e} \cdot \mathrm{AT}^{4}$ Where, $\mathrm{P}=$ Radioactive power $\sigma=$ Stefan-Boltzmann constant $=5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}^{4}$ $\mathrm{e}=$ Emissivity of the object $A=$ Surface area of the object $\mathrm{T}=$ Temperature According to the question, Two bodies of same shape, same size and same radiating power. $\frac{\mathrm{P}}{\mathrm{P}}=\frac{\sigma \mathrm{e}_{1} \mathrm{AT}_{1}^{4}}{\sigma \mathrm{e}_{2} \mathrm{AT}_{2}^{4}}$ $\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}=\frac{\mathrm{e}_{2}}{\mathrm{e}_{1}}=\frac{0.8}{0.2}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\left(\frac{4}{1}\right)^{1 / 4}=\frac{\sqrt{2}}{1}$ $\mathrm{~T}_{1}: \mathrm{T}_{2}=\sqrt{2}: 1$
AP EAMCET(Medical)-2005
Heat Transfer
149596
If a black body emits $0.5 \mathrm{~J}$ of energy per second when it is at $27^{\circ} \mathrm{C}$, then the amount of energy emitted by it when it is at $627^{\circ} \mathrm{C}$ will be:
1 $40.5 \mathrm{~J}$
2 $162 \mathrm{~J}$
3 $13.5 \mathrm{~J}$
4 $135 \mathrm{~J}$
Explanation:
A Given, Energy emitted by black body $\mathrm{E}_{2}=0.5 \mathrm{~J} /$ second Initial temperature, $\mathrm{T}_{2}=27^{\circ} \mathrm{C}$ Final temperature, $\mathrm{T}_{1}=627^{\circ} \mathrm{C}$ According to Stefan's law, $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}_{1}}{0.5}=\left(\frac{273+627}{273+27}\right)^{4}$ $\mathrm{E}_{1}=0.5\left(\frac{900}{300}\right)^{4}=0.5 \times$ $\Rightarrow \quad \mathrm{E}_{1}=40.5 \mathrm{~J}$
Karnataka CET-2008
Heat Transfer
149598
The emissive power of a body at temperature $T(C)$ is $E$. Then the graph between $\log _{e} E$ and $\log _{\mathrm{e}} \mathrm{T}$ will be :
1 a
2 b
3 c
4 d
Explanation:
C According to the Stefan's law. The power emitted by radiation is proportional to fourth power of temperature $\mathrm{E}=\sigma \mathrm{AT}^{4}$ Taking $\log$ on both side, $\log _{e} \mathrm{E}=4 \log _{\mathrm{e}} \mathrm{T}+\log \sigma$ $\log _{e} E=4 \log _{e} T+C$ This equation show a linear relation between $\log _{\mathrm{e}} \mathrm{E}$ and $\log _{\mathrm{e}} \mathrm{T}$ with 4 as a gradient and $\mathrm{C}$ is the $\mathrm{y}$ intercept.
149590
If the Wien's constant $b=0.3 \mathrm{~cm}-K$, then the temperature of the sun having maximum intensity of radiation at $6000 \mathrm{~A}^{\circ}$ wavelength is
1 $2000 \mathrm{~K}$
2 $5000 \mathrm{~K}$
3 $6000 \mathrm{~K}$
4 $7000 \mathrm{~K}$
Explanation:
B Given, Wein's constant $b=0.3 \mathrm{~cm}-\mathrm{K}$ Wavelength of radiation $=6000 \AA$ We know that, Wien's displacement Law $\lambda \mathrm{T}=$ constant $\therefore \quad 6000 \times \mathrm{T}=0.3$ $\mathrm{T}=\frac{0.3}{6000 \AA} \mathrm{cmK}$ $=\frac{0.3 \times 10^{-2}}{6000 \times 10^{-10}} \mathrm{cmK}$ $\mathrm{T}=5000 \mathrm{~K}$
UPSEE 2020
Heat Transfer
149595
Two bodies of same shape, same size and same radiating power have emissivity's 0.2 and 0.8 . The ratio of their temperatures is:
1 $\sqrt{3}: 1$
2 $\sqrt{2}: 1$
3 $1: \sqrt{5}$
4 $1: \sqrt{8}$
Explanation:
B From Stefan - Boltzmann equation, $\mathrm{P}=\sigma \mathrm{e} \cdot \mathrm{AT}^{4}$ Where, $\mathrm{P}=$ Radioactive power $\sigma=$ Stefan-Boltzmann constant $=5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}^{4}$ $\mathrm{e}=$ Emissivity of the object $A=$ Surface area of the object $\mathrm{T}=$ Temperature According to the question, Two bodies of same shape, same size and same radiating power. $\frac{\mathrm{P}}{\mathrm{P}}=\frac{\sigma \mathrm{e}_{1} \mathrm{AT}_{1}^{4}}{\sigma \mathrm{e}_{2} \mathrm{AT}_{2}^{4}}$ $\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}=\frac{\mathrm{e}_{2}}{\mathrm{e}_{1}}=\frac{0.8}{0.2}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\left(\frac{4}{1}\right)^{1 / 4}=\frac{\sqrt{2}}{1}$ $\mathrm{~T}_{1}: \mathrm{T}_{2}=\sqrt{2}: 1$
AP EAMCET(Medical)-2005
Heat Transfer
149596
If a black body emits $0.5 \mathrm{~J}$ of energy per second when it is at $27^{\circ} \mathrm{C}$, then the amount of energy emitted by it when it is at $627^{\circ} \mathrm{C}$ will be:
1 $40.5 \mathrm{~J}$
2 $162 \mathrm{~J}$
3 $13.5 \mathrm{~J}$
4 $135 \mathrm{~J}$
Explanation:
A Given, Energy emitted by black body $\mathrm{E}_{2}=0.5 \mathrm{~J} /$ second Initial temperature, $\mathrm{T}_{2}=27^{\circ} \mathrm{C}$ Final temperature, $\mathrm{T}_{1}=627^{\circ} \mathrm{C}$ According to Stefan's law, $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}_{1}}{0.5}=\left(\frac{273+627}{273+27}\right)^{4}$ $\mathrm{E}_{1}=0.5\left(\frac{900}{300}\right)^{4}=0.5 \times$ $\Rightarrow \quad \mathrm{E}_{1}=40.5 \mathrm{~J}$
Karnataka CET-2008
Heat Transfer
149598
The emissive power of a body at temperature $T(C)$ is $E$. Then the graph between $\log _{e} E$ and $\log _{\mathrm{e}} \mathrm{T}$ will be :
1 a
2 b
3 c
4 d
Explanation:
C According to the Stefan's law. The power emitted by radiation is proportional to fourth power of temperature $\mathrm{E}=\sigma \mathrm{AT}^{4}$ Taking $\log$ on both side, $\log _{e} \mathrm{E}=4 \log _{\mathrm{e}} \mathrm{T}+\log \sigma$ $\log _{e} E=4 \log _{e} T+C$ This equation show a linear relation between $\log _{\mathrm{e}} \mathrm{E}$ and $\log _{\mathrm{e}} \mathrm{T}$ with 4 as a gradient and $\mathrm{C}$ is the $\mathrm{y}$ intercept.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Heat Transfer
149590
If the Wien's constant $b=0.3 \mathrm{~cm}-K$, then the temperature of the sun having maximum intensity of radiation at $6000 \mathrm{~A}^{\circ}$ wavelength is
1 $2000 \mathrm{~K}$
2 $5000 \mathrm{~K}$
3 $6000 \mathrm{~K}$
4 $7000 \mathrm{~K}$
Explanation:
B Given, Wein's constant $b=0.3 \mathrm{~cm}-\mathrm{K}$ Wavelength of radiation $=6000 \AA$ We know that, Wien's displacement Law $\lambda \mathrm{T}=$ constant $\therefore \quad 6000 \times \mathrm{T}=0.3$ $\mathrm{T}=\frac{0.3}{6000 \AA} \mathrm{cmK}$ $=\frac{0.3 \times 10^{-2}}{6000 \times 10^{-10}} \mathrm{cmK}$ $\mathrm{T}=5000 \mathrm{~K}$
UPSEE 2020
Heat Transfer
149595
Two bodies of same shape, same size and same radiating power have emissivity's 0.2 and 0.8 . The ratio of their temperatures is:
1 $\sqrt{3}: 1$
2 $\sqrt{2}: 1$
3 $1: \sqrt{5}$
4 $1: \sqrt{8}$
Explanation:
B From Stefan - Boltzmann equation, $\mathrm{P}=\sigma \mathrm{e} \cdot \mathrm{AT}^{4}$ Where, $\mathrm{P}=$ Radioactive power $\sigma=$ Stefan-Boltzmann constant $=5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}^{4}$ $\mathrm{e}=$ Emissivity of the object $A=$ Surface area of the object $\mathrm{T}=$ Temperature According to the question, Two bodies of same shape, same size and same radiating power. $\frac{\mathrm{P}}{\mathrm{P}}=\frac{\sigma \mathrm{e}_{1} \mathrm{AT}_{1}^{4}}{\sigma \mathrm{e}_{2} \mathrm{AT}_{2}^{4}}$ $\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}=\frac{\mathrm{e}_{2}}{\mathrm{e}_{1}}=\frac{0.8}{0.2}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\left(\frac{4}{1}\right)^{1 / 4}=\frac{\sqrt{2}}{1}$ $\mathrm{~T}_{1}: \mathrm{T}_{2}=\sqrt{2}: 1$
AP EAMCET(Medical)-2005
Heat Transfer
149596
If a black body emits $0.5 \mathrm{~J}$ of energy per second when it is at $27^{\circ} \mathrm{C}$, then the amount of energy emitted by it when it is at $627^{\circ} \mathrm{C}$ will be:
1 $40.5 \mathrm{~J}$
2 $162 \mathrm{~J}$
3 $13.5 \mathrm{~J}$
4 $135 \mathrm{~J}$
Explanation:
A Given, Energy emitted by black body $\mathrm{E}_{2}=0.5 \mathrm{~J} /$ second Initial temperature, $\mathrm{T}_{2}=27^{\circ} \mathrm{C}$ Final temperature, $\mathrm{T}_{1}=627^{\circ} \mathrm{C}$ According to Stefan's law, $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}_{1}}{0.5}=\left(\frac{273+627}{273+27}\right)^{4}$ $\mathrm{E}_{1}=0.5\left(\frac{900}{300}\right)^{4}=0.5 \times$ $\Rightarrow \quad \mathrm{E}_{1}=40.5 \mathrm{~J}$
Karnataka CET-2008
Heat Transfer
149598
The emissive power of a body at temperature $T(C)$ is $E$. Then the graph between $\log _{e} E$ and $\log _{\mathrm{e}} \mathrm{T}$ will be :
1 a
2 b
3 c
4 d
Explanation:
C According to the Stefan's law. The power emitted by radiation is proportional to fourth power of temperature $\mathrm{E}=\sigma \mathrm{AT}^{4}$ Taking $\log$ on both side, $\log _{e} \mathrm{E}=4 \log _{\mathrm{e}} \mathrm{T}+\log \sigma$ $\log _{e} E=4 \log _{e} T+C$ This equation show a linear relation between $\log _{\mathrm{e}} \mathrm{E}$ and $\log _{\mathrm{e}} \mathrm{T}$ with 4 as a gradient and $\mathrm{C}$ is the $\mathrm{y}$ intercept.
149590
If the Wien's constant $b=0.3 \mathrm{~cm}-K$, then the temperature of the sun having maximum intensity of radiation at $6000 \mathrm{~A}^{\circ}$ wavelength is
1 $2000 \mathrm{~K}$
2 $5000 \mathrm{~K}$
3 $6000 \mathrm{~K}$
4 $7000 \mathrm{~K}$
Explanation:
B Given, Wein's constant $b=0.3 \mathrm{~cm}-\mathrm{K}$ Wavelength of radiation $=6000 \AA$ We know that, Wien's displacement Law $\lambda \mathrm{T}=$ constant $\therefore \quad 6000 \times \mathrm{T}=0.3$ $\mathrm{T}=\frac{0.3}{6000 \AA} \mathrm{cmK}$ $=\frac{0.3 \times 10^{-2}}{6000 \times 10^{-10}} \mathrm{cmK}$ $\mathrm{T}=5000 \mathrm{~K}$
UPSEE 2020
Heat Transfer
149595
Two bodies of same shape, same size and same radiating power have emissivity's 0.2 and 0.8 . The ratio of their temperatures is:
1 $\sqrt{3}: 1$
2 $\sqrt{2}: 1$
3 $1: \sqrt{5}$
4 $1: \sqrt{8}$
Explanation:
B From Stefan - Boltzmann equation, $\mathrm{P}=\sigma \mathrm{e} \cdot \mathrm{AT}^{4}$ Where, $\mathrm{P}=$ Radioactive power $\sigma=$ Stefan-Boltzmann constant $=5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}^{4}$ $\mathrm{e}=$ Emissivity of the object $A=$ Surface area of the object $\mathrm{T}=$ Temperature According to the question, Two bodies of same shape, same size and same radiating power. $\frac{\mathrm{P}}{\mathrm{P}}=\frac{\sigma \mathrm{e}_{1} \mathrm{AT}_{1}^{4}}{\sigma \mathrm{e}_{2} \mathrm{AT}_{2}^{4}}$ $\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}=\frac{\mathrm{e}_{2}}{\mathrm{e}_{1}}=\frac{0.8}{0.2}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\left(\frac{4}{1}\right)^{1 / 4}=\frac{\sqrt{2}}{1}$ $\mathrm{~T}_{1}: \mathrm{T}_{2}=\sqrt{2}: 1$
AP EAMCET(Medical)-2005
Heat Transfer
149596
If a black body emits $0.5 \mathrm{~J}$ of energy per second when it is at $27^{\circ} \mathrm{C}$, then the amount of energy emitted by it when it is at $627^{\circ} \mathrm{C}$ will be:
1 $40.5 \mathrm{~J}$
2 $162 \mathrm{~J}$
3 $13.5 \mathrm{~J}$
4 $135 \mathrm{~J}$
Explanation:
A Given, Energy emitted by black body $\mathrm{E}_{2}=0.5 \mathrm{~J} /$ second Initial temperature, $\mathrm{T}_{2}=27^{\circ} \mathrm{C}$ Final temperature, $\mathrm{T}_{1}=627^{\circ} \mathrm{C}$ According to Stefan's law, $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}_{1}}{0.5}=\left(\frac{273+627}{273+27}\right)^{4}$ $\mathrm{E}_{1}=0.5\left(\frac{900}{300}\right)^{4}=0.5 \times$ $\Rightarrow \quad \mathrm{E}_{1}=40.5 \mathrm{~J}$
Karnataka CET-2008
Heat Transfer
149598
The emissive power of a body at temperature $T(C)$ is $E$. Then the graph between $\log _{e} E$ and $\log _{\mathrm{e}} \mathrm{T}$ will be :
1 a
2 b
3 c
4 d
Explanation:
C According to the Stefan's law. The power emitted by radiation is proportional to fourth power of temperature $\mathrm{E}=\sigma \mathrm{AT}^{4}$ Taking $\log$ on both side, $\log _{e} \mathrm{E}=4 \log _{\mathrm{e}} \mathrm{T}+\log \sigma$ $\log _{e} E=4 \log _{e} T+C$ This equation show a linear relation between $\log _{\mathrm{e}} \mathrm{E}$ and $\log _{\mathrm{e}} \mathrm{T}$ with 4 as a gradient and $\mathrm{C}$ is the $\mathrm{y}$ intercept.
