149460
The rate of radiation of a black body at $0^{\circ} \mathrm{C}$ is $\mathrm{E} \mathrm{Js} \mathrm{s}^{-1}$. The rate of radiation of the black body at $273^{\circ} \mathrm{C}$ will be
1 $\mathrm{E} \mathrm{Js}^{-1}$
2 $4 \mathrm{E} \mathrm{Js}^{-1}$
3 $\frac{\mathrm{E}}{2} \mathrm{Js}^{-1}$
4 $16 \mathrm{E} \mathrm{Js}^{-1}$
Explanation:
D Given that, $\mathrm{E}_{1}=\mathrm{E} \mathrm{J} / \mathrm{sec}$. $\mathrm{T}_{1}=0^{\circ}+273=273 \mathrm{~K} .$ $\mathrm{T}_{2}=273+273=546 \mathrm{~K} \text { and } \mathrm{E}_{2}=?$ According to Stefan-Boltzmann's law, $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{273}{546}\right)^{4}$ $\frac{\mathrm{E}}{\mathrm{E}_{2}}=\left(\frac{1}{2}\right)^{4}$ $\mathrm{E}_{2}=16 \mathrm{E} \mathrm{J/sec}$
AP EAMCET (21.09.2020) Shift-I
Heat Transfer
149463
The temperature of spherical black body is inversely proportional to its radius. If its radius is doubled, then the power radiating from it will be
1 Doubled
2 $\frac{1}{4}$ times of initial value
3 Halved
4 four times of initial value
Explanation:
B Given that, $\mathrm{T} \propto \frac{1}{\mathrm{R}}$ Where, $\mathrm{T}=$ Temperature of the body $\mathrm{R}=$ Radius of the spherical black body $\mathrm{T}=\frac{\mathrm{C}}{\mathrm{R}}$ Power radiating from the spherical black body given by. $\rho=\sigma \mathrm{AT}^{4}$ Area of spherical body, $\mathrm{A}=4 \pi \mathrm{R}^{2}$ From equation (i), (ii) \& (iii), $\mathrm{P}=\sigma \times\left(4 \pi \mathrm{R}^{2}\right) \times\left(\frac{\mathrm{C}}{\mathrm{R}}\right)^{4}$ $\mathrm{P} \propto \frac{1}{\mathrm{R}^{2}}$ Hence, when radius is doubled, radiating power of sphere become one fourth.
AP EAMCET (20.04.2019) Shift-1
Heat Transfer
149464
Two bodies $A$ and $B$ are placed in an evacuated vessel maintained at a temperature of $27^{\circ} \mathrm{C}$. The temperature of $A$ is $327^{\circ} \mathrm{C}$ and that of $B$ is $227^{\circ} \mathrm{C}$. The ratio of heat loss from $A$ and $B$ is about
1 $2: 1$
2 $1: 2$
3 $4: 1$
4 $1: 4$
Explanation:
A Given that, $\mathrm{T}_{1}=327^{\circ} \mathrm{C}=327+273=600 \mathrm{~K}$ $\mathrm{T}_{2}=227^{\circ} \mathrm{C}=227+273=500 \mathrm{~K}$ $\mathrm{~T}_{0}=27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K}$ From the Stefan's law of heat loss is given by $\mathrm{E}_{1}=\sigma\left[\mathrm{T}_{1}^{4}-\mathrm{T}_{0}^{4}\right]$ $\mathrm{E}_{2}=\sigma\left[\mathrm{T}_{2}^{4}-\mathrm{T}_{0}^{4}\right]$ From dividing equation (i) \& (ii) $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\sigma\left[(600)^{4}-(300)^{4}\right]}{\sigma\left[(500)^{4}-(300)^{4}\right]}$ $\begin{aligned} \frac{\mathrm{E}_1}{\mathrm{E}_2} =\frac{\left[6^4-3^4\right]}{\left[5^4-3^4\right]} \\ \frac{\mathrm{E}_1}{\mathrm{E}_2} =\frac{1296-81}{625-81} \\ \frac{\mathrm{E}_1}{\mathrm{E}_2} =\frac{1215}{544} \\ \frac{\mathrm{E}_1}{\mathrm{E}_2} =\frac{2.23}{1} \\ \mathrm{E}_1: \mathrm{E}_2 =2.23: 1 \approx 2: 1\end{aligned}$
BITSAT-2019
Heat Transfer
149466
If the radius of a star is $R$ and it acts as a black body, what would be the temperature of the star, in which the rate of energy production is Q?
D Given that, Rate of energy production is $\mathrm{Q}$. From the Stefan's law, $\mathrm{Q}=\sigma \mathrm{AT}^{4}$ $\mathrm{~T}^{4}=\frac{\mathrm{Q}}{\sigma \mathrm{A}}$ $\mathrm{T}=\left[\frac{\mathrm{Q}}{\sigma 4 \pi \mathrm{R}^{2}}\right]^{1 / 4} \quad\left[\because=4 \pi \mathrm{R}^{2}\right]$ $\mathrm{T}=\left[\frac{\mathrm{Q}}{\sigma\left(4 \pi \mathrm{R}^{2}\right)}\right]^{1 / 4}$
BITSAT-2013
Heat Transfer
149468
If the temperature of black body increases from $300 \mathrm{~K}$ to $900 \mathrm{~K}$, then the rate of energy radiation increases by how much times?
1 81
2 3
3 9
4 2
Explanation:
A Given that, $\mathrm{T}_{1}=300 \mathrm{~K}, \mathrm{~T}_{2}=900 \mathrm{~K}$ According to Stefan's law, $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{4}=\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\left(\frac{900}{300}\right)^{4}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=(3)^{4}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=81$ Hence, increase in radiation energy is 81 times.
149460
The rate of radiation of a black body at $0^{\circ} \mathrm{C}$ is $\mathrm{E} \mathrm{Js} \mathrm{s}^{-1}$. The rate of radiation of the black body at $273^{\circ} \mathrm{C}$ will be
1 $\mathrm{E} \mathrm{Js}^{-1}$
2 $4 \mathrm{E} \mathrm{Js}^{-1}$
3 $\frac{\mathrm{E}}{2} \mathrm{Js}^{-1}$
4 $16 \mathrm{E} \mathrm{Js}^{-1}$
Explanation:
D Given that, $\mathrm{E}_{1}=\mathrm{E} \mathrm{J} / \mathrm{sec}$. $\mathrm{T}_{1}=0^{\circ}+273=273 \mathrm{~K} .$ $\mathrm{T}_{2}=273+273=546 \mathrm{~K} \text { and } \mathrm{E}_{2}=?$ According to Stefan-Boltzmann's law, $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{273}{546}\right)^{4}$ $\frac{\mathrm{E}}{\mathrm{E}_{2}}=\left(\frac{1}{2}\right)^{4}$ $\mathrm{E}_{2}=16 \mathrm{E} \mathrm{J/sec}$
AP EAMCET (21.09.2020) Shift-I
Heat Transfer
149463
The temperature of spherical black body is inversely proportional to its radius. If its radius is doubled, then the power radiating from it will be
1 Doubled
2 $\frac{1}{4}$ times of initial value
3 Halved
4 four times of initial value
Explanation:
B Given that, $\mathrm{T} \propto \frac{1}{\mathrm{R}}$ Where, $\mathrm{T}=$ Temperature of the body $\mathrm{R}=$ Radius of the spherical black body $\mathrm{T}=\frac{\mathrm{C}}{\mathrm{R}}$ Power radiating from the spherical black body given by. $\rho=\sigma \mathrm{AT}^{4}$ Area of spherical body, $\mathrm{A}=4 \pi \mathrm{R}^{2}$ From equation (i), (ii) \& (iii), $\mathrm{P}=\sigma \times\left(4 \pi \mathrm{R}^{2}\right) \times\left(\frac{\mathrm{C}}{\mathrm{R}}\right)^{4}$ $\mathrm{P} \propto \frac{1}{\mathrm{R}^{2}}$ Hence, when radius is doubled, radiating power of sphere become one fourth.
AP EAMCET (20.04.2019) Shift-1
Heat Transfer
149464
Two bodies $A$ and $B$ are placed in an evacuated vessel maintained at a temperature of $27^{\circ} \mathrm{C}$. The temperature of $A$ is $327^{\circ} \mathrm{C}$ and that of $B$ is $227^{\circ} \mathrm{C}$. The ratio of heat loss from $A$ and $B$ is about
1 $2: 1$
2 $1: 2$
3 $4: 1$
4 $1: 4$
Explanation:
A Given that, $\mathrm{T}_{1}=327^{\circ} \mathrm{C}=327+273=600 \mathrm{~K}$ $\mathrm{T}_{2}=227^{\circ} \mathrm{C}=227+273=500 \mathrm{~K}$ $\mathrm{~T}_{0}=27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K}$ From the Stefan's law of heat loss is given by $\mathrm{E}_{1}=\sigma\left[\mathrm{T}_{1}^{4}-\mathrm{T}_{0}^{4}\right]$ $\mathrm{E}_{2}=\sigma\left[\mathrm{T}_{2}^{4}-\mathrm{T}_{0}^{4}\right]$ From dividing equation (i) \& (ii) $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\sigma\left[(600)^{4}-(300)^{4}\right]}{\sigma\left[(500)^{4}-(300)^{4}\right]}$ $\begin{aligned} \frac{\mathrm{E}_1}{\mathrm{E}_2} =\frac{\left[6^4-3^4\right]}{\left[5^4-3^4\right]} \\ \frac{\mathrm{E}_1}{\mathrm{E}_2} =\frac{1296-81}{625-81} \\ \frac{\mathrm{E}_1}{\mathrm{E}_2} =\frac{1215}{544} \\ \frac{\mathrm{E}_1}{\mathrm{E}_2} =\frac{2.23}{1} \\ \mathrm{E}_1: \mathrm{E}_2 =2.23: 1 \approx 2: 1\end{aligned}$
BITSAT-2019
Heat Transfer
149466
If the radius of a star is $R$ and it acts as a black body, what would be the temperature of the star, in which the rate of energy production is Q?
D Given that, Rate of energy production is $\mathrm{Q}$. From the Stefan's law, $\mathrm{Q}=\sigma \mathrm{AT}^{4}$ $\mathrm{~T}^{4}=\frac{\mathrm{Q}}{\sigma \mathrm{A}}$ $\mathrm{T}=\left[\frac{\mathrm{Q}}{\sigma 4 \pi \mathrm{R}^{2}}\right]^{1 / 4} \quad\left[\because=4 \pi \mathrm{R}^{2}\right]$ $\mathrm{T}=\left[\frac{\mathrm{Q}}{\sigma\left(4 \pi \mathrm{R}^{2}\right)}\right]^{1 / 4}$
BITSAT-2013
Heat Transfer
149468
If the temperature of black body increases from $300 \mathrm{~K}$ to $900 \mathrm{~K}$, then the rate of energy radiation increases by how much times?
1 81
2 3
3 9
4 2
Explanation:
A Given that, $\mathrm{T}_{1}=300 \mathrm{~K}, \mathrm{~T}_{2}=900 \mathrm{~K}$ According to Stefan's law, $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{4}=\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\left(\frac{900}{300}\right)^{4}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=(3)^{4}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=81$ Hence, increase in radiation energy is 81 times.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Heat Transfer
149460
The rate of radiation of a black body at $0^{\circ} \mathrm{C}$ is $\mathrm{E} \mathrm{Js} \mathrm{s}^{-1}$. The rate of radiation of the black body at $273^{\circ} \mathrm{C}$ will be
1 $\mathrm{E} \mathrm{Js}^{-1}$
2 $4 \mathrm{E} \mathrm{Js}^{-1}$
3 $\frac{\mathrm{E}}{2} \mathrm{Js}^{-1}$
4 $16 \mathrm{E} \mathrm{Js}^{-1}$
Explanation:
D Given that, $\mathrm{E}_{1}=\mathrm{E} \mathrm{J} / \mathrm{sec}$. $\mathrm{T}_{1}=0^{\circ}+273=273 \mathrm{~K} .$ $\mathrm{T}_{2}=273+273=546 \mathrm{~K} \text { and } \mathrm{E}_{2}=?$ According to Stefan-Boltzmann's law, $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{273}{546}\right)^{4}$ $\frac{\mathrm{E}}{\mathrm{E}_{2}}=\left(\frac{1}{2}\right)^{4}$ $\mathrm{E}_{2}=16 \mathrm{E} \mathrm{J/sec}$
AP EAMCET (21.09.2020) Shift-I
Heat Transfer
149463
The temperature of spherical black body is inversely proportional to its radius. If its radius is doubled, then the power radiating from it will be
1 Doubled
2 $\frac{1}{4}$ times of initial value
3 Halved
4 four times of initial value
Explanation:
B Given that, $\mathrm{T} \propto \frac{1}{\mathrm{R}}$ Where, $\mathrm{T}=$ Temperature of the body $\mathrm{R}=$ Radius of the spherical black body $\mathrm{T}=\frac{\mathrm{C}}{\mathrm{R}}$ Power radiating from the spherical black body given by. $\rho=\sigma \mathrm{AT}^{4}$ Area of spherical body, $\mathrm{A}=4 \pi \mathrm{R}^{2}$ From equation (i), (ii) \& (iii), $\mathrm{P}=\sigma \times\left(4 \pi \mathrm{R}^{2}\right) \times\left(\frac{\mathrm{C}}{\mathrm{R}}\right)^{4}$ $\mathrm{P} \propto \frac{1}{\mathrm{R}^{2}}$ Hence, when radius is doubled, radiating power of sphere become one fourth.
AP EAMCET (20.04.2019) Shift-1
Heat Transfer
149464
Two bodies $A$ and $B$ are placed in an evacuated vessel maintained at a temperature of $27^{\circ} \mathrm{C}$. The temperature of $A$ is $327^{\circ} \mathrm{C}$ and that of $B$ is $227^{\circ} \mathrm{C}$. The ratio of heat loss from $A$ and $B$ is about
1 $2: 1$
2 $1: 2$
3 $4: 1$
4 $1: 4$
Explanation:
A Given that, $\mathrm{T}_{1}=327^{\circ} \mathrm{C}=327+273=600 \mathrm{~K}$ $\mathrm{T}_{2}=227^{\circ} \mathrm{C}=227+273=500 \mathrm{~K}$ $\mathrm{~T}_{0}=27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K}$ From the Stefan's law of heat loss is given by $\mathrm{E}_{1}=\sigma\left[\mathrm{T}_{1}^{4}-\mathrm{T}_{0}^{4}\right]$ $\mathrm{E}_{2}=\sigma\left[\mathrm{T}_{2}^{4}-\mathrm{T}_{0}^{4}\right]$ From dividing equation (i) \& (ii) $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\sigma\left[(600)^{4}-(300)^{4}\right]}{\sigma\left[(500)^{4}-(300)^{4}\right]}$ $\begin{aligned} \frac{\mathrm{E}_1}{\mathrm{E}_2} =\frac{\left[6^4-3^4\right]}{\left[5^4-3^4\right]} \\ \frac{\mathrm{E}_1}{\mathrm{E}_2} =\frac{1296-81}{625-81} \\ \frac{\mathrm{E}_1}{\mathrm{E}_2} =\frac{1215}{544} \\ \frac{\mathrm{E}_1}{\mathrm{E}_2} =\frac{2.23}{1} \\ \mathrm{E}_1: \mathrm{E}_2 =2.23: 1 \approx 2: 1\end{aligned}$
BITSAT-2019
Heat Transfer
149466
If the radius of a star is $R$ and it acts as a black body, what would be the temperature of the star, in which the rate of energy production is Q?
D Given that, Rate of energy production is $\mathrm{Q}$. From the Stefan's law, $\mathrm{Q}=\sigma \mathrm{AT}^{4}$ $\mathrm{~T}^{4}=\frac{\mathrm{Q}}{\sigma \mathrm{A}}$ $\mathrm{T}=\left[\frac{\mathrm{Q}}{\sigma 4 \pi \mathrm{R}^{2}}\right]^{1 / 4} \quad\left[\because=4 \pi \mathrm{R}^{2}\right]$ $\mathrm{T}=\left[\frac{\mathrm{Q}}{\sigma\left(4 \pi \mathrm{R}^{2}\right)}\right]^{1 / 4}$
BITSAT-2013
Heat Transfer
149468
If the temperature of black body increases from $300 \mathrm{~K}$ to $900 \mathrm{~K}$, then the rate of energy radiation increases by how much times?
1 81
2 3
3 9
4 2
Explanation:
A Given that, $\mathrm{T}_{1}=300 \mathrm{~K}, \mathrm{~T}_{2}=900 \mathrm{~K}$ According to Stefan's law, $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{4}=\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\left(\frac{900}{300}\right)^{4}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=(3)^{4}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=81$ Hence, increase in radiation energy is 81 times.
149460
The rate of radiation of a black body at $0^{\circ} \mathrm{C}$ is $\mathrm{E} \mathrm{Js} \mathrm{s}^{-1}$. The rate of radiation of the black body at $273^{\circ} \mathrm{C}$ will be
1 $\mathrm{E} \mathrm{Js}^{-1}$
2 $4 \mathrm{E} \mathrm{Js}^{-1}$
3 $\frac{\mathrm{E}}{2} \mathrm{Js}^{-1}$
4 $16 \mathrm{E} \mathrm{Js}^{-1}$
Explanation:
D Given that, $\mathrm{E}_{1}=\mathrm{E} \mathrm{J} / \mathrm{sec}$. $\mathrm{T}_{1}=0^{\circ}+273=273 \mathrm{~K} .$ $\mathrm{T}_{2}=273+273=546 \mathrm{~K} \text { and } \mathrm{E}_{2}=?$ According to Stefan-Boltzmann's law, $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{273}{546}\right)^{4}$ $\frac{\mathrm{E}}{\mathrm{E}_{2}}=\left(\frac{1}{2}\right)^{4}$ $\mathrm{E}_{2}=16 \mathrm{E} \mathrm{J/sec}$
AP EAMCET (21.09.2020) Shift-I
Heat Transfer
149463
The temperature of spherical black body is inversely proportional to its radius. If its radius is doubled, then the power radiating from it will be
1 Doubled
2 $\frac{1}{4}$ times of initial value
3 Halved
4 four times of initial value
Explanation:
B Given that, $\mathrm{T} \propto \frac{1}{\mathrm{R}}$ Where, $\mathrm{T}=$ Temperature of the body $\mathrm{R}=$ Radius of the spherical black body $\mathrm{T}=\frac{\mathrm{C}}{\mathrm{R}}$ Power radiating from the spherical black body given by. $\rho=\sigma \mathrm{AT}^{4}$ Area of spherical body, $\mathrm{A}=4 \pi \mathrm{R}^{2}$ From equation (i), (ii) \& (iii), $\mathrm{P}=\sigma \times\left(4 \pi \mathrm{R}^{2}\right) \times\left(\frac{\mathrm{C}}{\mathrm{R}}\right)^{4}$ $\mathrm{P} \propto \frac{1}{\mathrm{R}^{2}}$ Hence, when radius is doubled, radiating power of sphere become one fourth.
AP EAMCET (20.04.2019) Shift-1
Heat Transfer
149464
Two bodies $A$ and $B$ are placed in an evacuated vessel maintained at a temperature of $27^{\circ} \mathrm{C}$. The temperature of $A$ is $327^{\circ} \mathrm{C}$ and that of $B$ is $227^{\circ} \mathrm{C}$. The ratio of heat loss from $A$ and $B$ is about
1 $2: 1$
2 $1: 2$
3 $4: 1$
4 $1: 4$
Explanation:
A Given that, $\mathrm{T}_{1}=327^{\circ} \mathrm{C}=327+273=600 \mathrm{~K}$ $\mathrm{T}_{2}=227^{\circ} \mathrm{C}=227+273=500 \mathrm{~K}$ $\mathrm{~T}_{0}=27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K}$ From the Stefan's law of heat loss is given by $\mathrm{E}_{1}=\sigma\left[\mathrm{T}_{1}^{4}-\mathrm{T}_{0}^{4}\right]$ $\mathrm{E}_{2}=\sigma\left[\mathrm{T}_{2}^{4}-\mathrm{T}_{0}^{4}\right]$ From dividing equation (i) \& (ii) $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\sigma\left[(600)^{4}-(300)^{4}\right]}{\sigma\left[(500)^{4}-(300)^{4}\right]}$ $\begin{aligned} \frac{\mathrm{E}_1}{\mathrm{E}_2} =\frac{\left[6^4-3^4\right]}{\left[5^4-3^4\right]} \\ \frac{\mathrm{E}_1}{\mathrm{E}_2} =\frac{1296-81}{625-81} \\ \frac{\mathrm{E}_1}{\mathrm{E}_2} =\frac{1215}{544} \\ \frac{\mathrm{E}_1}{\mathrm{E}_2} =\frac{2.23}{1} \\ \mathrm{E}_1: \mathrm{E}_2 =2.23: 1 \approx 2: 1\end{aligned}$
BITSAT-2019
Heat Transfer
149466
If the radius of a star is $R$ and it acts as a black body, what would be the temperature of the star, in which the rate of energy production is Q?
D Given that, Rate of energy production is $\mathrm{Q}$. From the Stefan's law, $\mathrm{Q}=\sigma \mathrm{AT}^{4}$ $\mathrm{~T}^{4}=\frac{\mathrm{Q}}{\sigma \mathrm{A}}$ $\mathrm{T}=\left[\frac{\mathrm{Q}}{\sigma 4 \pi \mathrm{R}^{2}}\right]^{1 / 4} \quad\left[\because=4 \pi \mathrm{R}^{2}\right]$ $\mathrm{T}=\left[\frac{\mathrm{Q}}{\sigma\left(4 \pi \mathrm{R}^{2}\right)}\right]^{1 / 4}$
BITSAT-2013
Heat Transfer
149468
If the temperature of black body increases from $300 \mathrm{~K}$ to $900 \mathrm{~K}$, then the rate of energy radiation increases by how much times?
1 81
2 3
3 9
4 2
Explanation:
A Given that, $\mathrm{T}_{1}=300 \mathrm{~K}, \mathrm{~T}_{2}=900 \mathrm{~K}$ According to Stefan's law, $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{4}=\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\left(\frac{900}{300}\right)^{4}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=(3)^{4}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=81$ Hence, increase in radiation energy is 81 times.
149460
The rate of radiation of a black body at $0^{\circ} \mathrm{C}$ is $\mathrm{E} \mathrm{Js} \mathrm{s}^{-1}$. The rate of radiation of the black body at $273^{\circ} \mathrm{C}$ will be
1 $\mathrm{E} \mathrm{Js}^{-1}$
2 $4 \mathrm{E} \mathrm{Js}^{-1}$
3 $\frac{\mathrm{E}}{2} \mathrm{Js}^{-1}$
4 $16 \mathrm{E} \mathrm{Js}^{-1}$
Explanation:
D Given that, $\mathrm{E}_{1}=\mathrm{E} \mathrm{J} / \mathrm{sec}$. $\mathrm{T}_{1}=0^{\circ}+273=273 \mathrm{~K} .$ $\mathrm{T}_{2}=273+273=546 \mathrm{~K} \text { and } \mathrm{E}_{2}=?$ According to Stefan-Boltzmann's law, $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{273}{546}\right)^{4}$ $\frac{\mathrm{E}}{\mathrm{E}_{2}}=\left(\frac{1}{2}\right)^{4}$ $\mathrm{E}_{2}=16 \mathrm{E} \mathrm{J/sec}$
AP EAMCET (21.09.2020) Shift-I
Heat Transfer
149463
The temperature of spherical black body is inversely proportional to its radius. If its radius is doubled, then the power radiating from it will be
1 Doubled
2 $\frac{1}{4}$ times of initial value
3 Halved
4 four times of initial value
Explanation:
B Given that, $\mathrm{T} \propto \frac{1}{\mathrm{R}}$ Where, $\mathrm{T}=$ Temperature of the body $\mathrm{R}=$ Radius of the spherical black body $\mathrm{T}=\frac{\mathrm{C}}{\mathrm{R}}$ Power radiating from the spherical black body given by. $\rho=\sigma \mathrm{AT}^{4}$ Area of spherical body, $\mathrm{A}=4 \pi \mathrm{R}^{2}$ From equation (i), (ii) \& (iii), $\mathrm{P}=\sigma \times\left(4 \pi \mathrm{R}^{2}\right) \times\left(\frac{\mathrm{C}}{\mathrm{R}}\right)^{4}$ $\mathrm{P} \propto \frac{1}{\mathrm{R}^{2}}$ Hence, when radius is doubled, radiating power of sphere become one fourth.
AP EAMCET (20.04.2019) Shift-1
Heat Transfer
149464
Two bodies $A$ and $B$ are placed in an evacuated vessel maintained at a temperature of $27^{\circ} \mathrm{C}$. The temperature of $A$ is $327^{\circ} \mathrm{C}$ and that of $B$ is $227^{\circ} \mathrm{C}$. The ratio of heat loss from $A$ and $B$ is about
1 $2: 1$
2 $1: 2$
3 $4: 1$
4 $1: 4$
Explanation:
A Given that, $\mathrm{T}_{1}=327^{\circ} \mathrm{C}=327+273=600 \mathrm{~K}$ $\mathrm{T}_{2}=227^{\circ} \mathrm{C}=227+273=500 \mathrm{~K}$ $\mathrm{~T}_{0}=27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K}$ From the Stefan's law of heat loss is given by $\mathrm{E}_{1}=\sigma\left[\mathrm{T}_{1}^{4}-\mathrm{T}_{0}^{4}\right]$ $\mathrm{E}_{2}=\sigma\left[\mathrm{T}_{2}^{4}-\mathrm{T}_{0}^{4}\right]$ From dividing equation (i) \& (ii) $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\sigma\left[(600)^{4}-(300)^{4}\right]}{\sigma\left[(500)^{4}-(300)^{4}\right]}$ $\begin{aligned} \frac{\mathrm{E}_1}{\mathrm{E}_2} =\frac{\left[6^4-3^4\right]}{\left[5^4-3^4\right]} \\ \frac{\mathrm{E}_1}{\mathrm{E}_2} =\frac{1296-81}{625-81} \\ \frac{\mathrm{E}_1}{\mathrm{E}_2} =\frac{1215}{544} \\ \frac{\mathrm{E}_1}{\mathrm{E}_2} =\frac{2.23}{1} \\ \mathrm{E}_1: \mathrm{E}_2 =2.23: 1 \approx 2: 1\end{aligned}$
BITSAT-2019
Heat Transfer
149466
If the radius of a star is $R$ and it acts as a black body, what would be the temperature of the star, in which the rate of energy production is Q?
D Given that, Rate of energy production is $\mathrm{Q}$. From the Stefan's law, $\mathrm{Q}=\sigma \mathrm{AT}^{4}$ $\mathrm{~T}^{4}=\frac{\mathrm{Q}}{\sigma \mathrm{A}}$ $\mathrm{T}=\left[\frac{\mathrm{Q}}{\sigma 4 \pi \mathrm{R}^{2}}\right]^{1 / 4} \quad\left[\because=4 \pi \mathrm{R}^{2}\right]$ $\mathrm{T}=\left[\frac{\mathrm{Q}}{\sigma\left(4 \pi \mathrm{R}^{2}\right)}\right]^{1 / 4}$
BITSAT-2013
Heat Transfer
149468
If the temperature of black body increases from $300 \mathrm{~K}$ to $900 \mathrm{~K}$, then the rate of energy radiation increases by how much times?
1 81
2 3
3 9
4 2
Explanation:
A Given that, $\mathrm{T}_{1}=300 \mathrm{~K}, \mathrm{~T}_{2}=900 \mathrm{~K}$ According to Stefan's law, $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{4}=\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\left(\frac{900}{300}\right)^{4}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=(3)^{4}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=81$ Hence, increase in radiation energy is 81 times.