QBManager

Heat Transfer

149500 A black body is heated from $27^{\circ} C$ to $127^{\circ} C$ . The ratio of their energies of radiation emitted will be:

1

9 : 16

2

27 : 64

3

81 : 256

4 3:4

Explanation:

C We know that
$E = σ T^{4}$
Where $E$ is rate of emission of radiation of a body at temperature $T$ .
$E_{1} = σ (27 + 273)^{2}$
$E_{2} = σ (127 + 273)^{2}$
$\frac{E_{1}}{E_{2}} = \frac{(300)^{4}}{(400)^{4}} = \frac{81}{256}$

AIIMS-2001

Heat Transfer

149502 Calculate radiation power for sphere whose temperature is $227^{\circ} C$ and radius $0.2 m$ and emissivity 0.8 .

1

1425 W

2

1500 W

3

1255 W

4

1575 W

Explanation:

A Given,
Radius $= 0.2 m$
And temperature $(T) = 227 + 273 = 500 K$
Radiation power through area A is given as
$P = σ A ε T^{4}$
[Here $ε =$ emissivity $= 0.8, A =$ surface area $= 4 π r^{2}$ , $σ = 5.67 \times 10^{- 8} w / m^{2} K^{4}]$
Putting these values we get
$P = 5.67 \times 10^{- 8} \times 4 π (0.2)^{2} \times (0.8) \times (500)^{4}$
$= 1417.48 ≃ 1425 W$

AIIMS-25.05.2019(M) Shift-1

Heat Transfer

149503 Distance between sun and earth is $2 \times 10^{8} km$ , temperature of sun $6000 K$ , radius of sun $7 \times 10^{5}$ $km$ . If emissivity of earth is 0.6 , then find out temperature of earth in thermal equilibrium.

1

400 K

2

300 K

3

500 K

4

600 K

Explanation:

B Given, distance between sun and earth (d) $= 2$ $\times 10^{8} km$ , Temperature of sun $= 6000 K$ and radius of sun $= 7 \times 10^{5} km$ .
For thermal equilibrium,
Energy received by earth = Energy emitted by earth
$\frac{T_{s}^{4} 4 π R_{s}^{2}}{4 π d^{2}} \times π R_{e}^{2} = σ \cdot ε \cdot T_{2}^{4} \cdot 4 π R_{e}^{2}$
$\frac{T_{s}^{4} \times R_{s}^{2}}{4 \times d^{2} \times e} = T_{e}^{4}$
$\frac{(6000)^{4} \times {(7 \times 10^{8})}^{2}}{4 \times {(2 \times 10^{11})}^{2} \times 0.6} = T_{e}^{4}$
$\frac{36 \times 36 \times 7 \times 7}{4 \times 4 \times 0.6} \times 10^{6} = T_{e}^{4}$
$66.15 \times 108 = T_{e}^{4}$
$T_{e} = 285.19 K ≃ 300 K$

AIIMS-25.05.2019(E) Shift-2

Heat Transfer

149504 If a body coated black at $600 K$ surrounded by atmosphere at $300 K$ has cooling rate $r_{0}$ , the same body at $900 K$ , surrounded by the same atmosphere will have cooling rate equal to-

1

\frac{16}{3} r_{0}

2

\frac{8}{16} r_{0}

3

16 r_{0}

4

4 r_{0}

Explanation:

A Given that, $T_{1} = 600 K$
$T_{2} = 900 K, T_{o} = 300 K$
We know that,
Cooling rate $\propto T^{4} - T_{0}^{4}$
$\frac{r}{r_{o}} = (\frac{T_{2}^{4} - T_{o}^{4}}{T_{1}^{4} - T_{o}^{4}})$
$\frac{r}{r_{o}} = [\frac{(900)^{4} - (300)^{4}}{(600)^{4} - (300)^{4}}]$
$\frac{r}{r_{o}} = (\frac{16}{3})$
$r = \frac{16}{3} r_{o}$

BCECE-2017

Heat Transfer

149500 A black body is heated from $27^{\circ} C$ to $127^{\circ} C$ . The ratio of their energies of radiation emitted will be:

1

9 : 16

2

27 : 64

3

81 : 256

4 3:4

Explanation:

C We know that
$E = σ T^{4}$
Where $E$ is rate of emission of radiation of a body at temperature $T$ .
$E_{1} = σ (27 + 273)^{2}$
$E_{2} = σ (127 + 273)^{2}$
$\frac{E_{1}}{E_{2}} = \frac{(300)^{4}}{(400)^{4}} = \frac{81}{256}$

AIIMS-2001

Heat Transfer

149502 Calculate radiation power for sphere whose temperature is $227^{\circ} C$ and radius $0.2 m$ and emissivity 0.8 .

1

1425 W

2

1500 W

3

1255 W

4

1575 W

Explanation:

A Given,
Radius $= 0.2 m$
And temperature $(T) = 227 + 273 = 500 K$
Radiation power through area A is given as
$P = σ A ε T^{4}$
[Here $ε =$ emissivity $= 0.8, A =$ surface area $= 4 π r^{2}$ , $σ = 5.67 \times 10^{- 8} w / m^{2} K^{4}]$
Putting these values we get
$P = 5.67 \times 10^{- 8} \times 4 π (0.2)^{2} \times (0.8) \times (500)^{4}$
$= 1417.48 ≃ 1425 W$

AIIMS-25.05.2019(M) Shift-1

Heat Transfer

149503 Distance between sun and earth is $2 \times 10^{8} km$ , temperature of sun $6000 K$ , radius of sun $7 \times 10^{5}$ $km$ . If emissivity of earth is 0.6 , then find out temperature of earth in thermal equilibrium.

1

400 K

2

300 K

3

500 K

4

600 K

Explanation:

B Given, distance between sun and earth (d) $= 2$ $\times 10^{8} km$ , Temperature of sun $= 6000 K$ and radius of sun $= 7 \times 10^{5} km$ .
For thermal equilibrium,
Energy received by earth = Energy emitted by earth
$\frac{T_{s}^{4} 4 π R_{s}^{2}}{4 π d^{2}} \times π R_{e}^{2} = σ \cdot ε \cdot T_{2}^{4} \cdot 4 π R_{e}^{2}$
$\frac{T_{s}^{4} \times R_{s}^{2}}{4 \times d^{2} \times e} = T_{e}^{4}$
$\frac{(6000)^{4} \times {(7 \times 10^{8})}^{2}}{4 \times {(2 \times 10^{11})}^{2} \times 0.6} = T_{e}^{4}$
$\frac{36 \times 36 \times 7 \times 7}{4 \times 4 \times 0.6} \times 10^{6} = T_{e}^{4}$
$66.15 \times 108 = T_{e}^{4}$
$T_{e} = 285.19 K ≃ 300 K$

AIIMS-25.05.2019(E) Shift-2

Heat Transfer

149504 If a body coated black at $600 K$ surrounded by atmosphere at $300 K$ has cooling rate $r_{0}$ , the same body at $900 K$ , surrounded by the same atmosphere will have cooling rate equal to-

1

\frac{16}{3} r_{0}

2

\frac{8}{16} r_{0}

3

16 r_{0}

4

4 r_{0}

Explanation:

A Given that, $T_{1} = 600 K$
$T_{2} = 900 K, T_{o} = 300 K$
We know that,
Cooling rate $\propto T^{4} - T_{0}^{4}$
$\frac{r}{r_{o}} = (\frac{T_{2}^{4} - T_{o}^{4}}{T_{1}^{4} - T_{o}^{4}})$
$\frac{r}{r_{o}} = [\frac{(900)^{4} - (300)^{4}}{(600)^{4} - (300)^{4}}]$
$\frac{r}{r_{o}} = (\frac{16}{3})$
$r = \frac{16}{3} r_{o}$

BCECE-2017

Heat Transfer

149500 A black body is heated from $27^{\circ} C$ to $127^{\circ} C$ . The ratio of their energies of radiation emitted will be:

1

9 : 16

2

27 : 64

3

81 : 256

4 3:4

Explanation:

C We know that
$E = σ T^{4}$
Where $E$ is rate of emission of radiation of a body at temperature $T$ .
$E_{1} = σ (27 + 273)^{2}$
$E_{2} = σ (127 + 273)^{2}$
$\frac{E_{1}}{E_{2}} = \frac{(300)^{4}}{(400)^{4}} = \frac{81}{256}$

AIIMS-2001

Heat Transfer

149502 Calculate radiation power for sphere whose temperature is $227^{\circ} C$ and radius $0.2 m$ and emissivity 0.8 .

1

1425 W

2

1500 W

3

1255 W

4

1575 W

Explanation:

A Given,
Radius $= 0.2 m$
And temperature $(T) = 227 + 273 = 500 K$
Radiation power through area A is given as
$P = σ A ε T^{4}$
[Here $ε =$ emissivity $= 0.8, A =$ surface area $= 4 π r^{2}$ , $σ = 5.67 \times 10^{- 8} w / m^{2} K^{4}]$
Putting these values we get
$P = 5.67 \times 10^{- 8} \times 4 π (0.2)^{2} \times (0.8) \times (500)^{4}$
$= 1417.48 ≃ 1425 W$

AIIMS-25.05.2019(M) Shift-1

Heat Transfer

149503 Distance between sun and earth is $2 \times 10^{8} km$ , temperature of sun $6000 K$ , radius of sun $7 \times 10^{5}$ $km$ . If emissivity of earth is 0.6 , then find out temperature of earth in thermal equilibrium.

1

400 K

2

300 K

3

500 K

4

600 K

Explanation:

B Given, distance between sun and earth (d) $= 2$ $\times 10^{8} km$ , Temperature of sun $= 6000 K$ and radius of sun $= 7 \times 10^{5} km$ .
For thermal equilibrium,
Energy received by earth = Energy emitted by earth
$\frac{T_{s}^{4} 4 π R_{s}^{2}}{4 π d^{2}} \times π R_{e}^{2} = σ \cdot ε \cdot T_{2}^{4} \cdot 4 π R_{e}^{2}$
$\frac{T_{s}^{4} \times R_{s}^{2}}{4 \times d^{2} \times e} = T_{e}^{4}$
$\frac{(6000)^{4} \times {(7 \times 10^{8})}^{2}}{4 \times {(2 \times 10^{11})}^{2} \times 0.6} = T_{e}^{4}$
$\frac{36 \times 36 \times 7 \times 7}{4 \times 4 \times 0.6} \times 10^{6} = T_{e}^{4}$
$66.15 \times 108 = T_{e}^{4}$
$T_{e} = 285.19 K ≃ 300 K$

AIIMS-25.05.2019(E) Shift-2

Heat Transfer

149504 If a body coated black at $600 K$ surrounded by atmosphere at $300 K$ has cooling rate $r_{0}$ , the same body at $900 K$ , surrounded by the same atmosphere will have cooling rate equal to-

1

\frac{16}{3} r_{0}

2

\frac{8}{16} r_{0}

3

16 r_{0}

4

4 r_{0}

Explanation:

A Given that, $T_{1} = 600 K$
$T_{2} = 900 K, T_{o} = 300 K$
We know that,
Cooling rate $\propto T^{4} - T_{0}^{4}$
$\frac{r}{r_{o}} = (\frac{T_{2}^{4} - T_{o}^{4}}{T_{1}^{4} - T_{o}^{4}})$
$\frac{r}{r_{o}} = [\frac{(900)^{4} - (300)^{4}}{(600)^{4} - (300)^{4}}]$
$\frac{r}{r_{o}} = (\frac{16}{3})$
$r = \frac{16}{3} r_{o}$

BCECE-2017

Heat Transfer

149500 A black body is heated from $27^{\circ} C$ to $127^{\circ} C$ . The ratio of their energies of radiation emitted will be:

1

9 : 16

2

27 : 64

3

81 : 256

4 3:4

Explanation:

C We know that
$E = σ T^{4}$
Where $E$ is rate of emission of radiation of a body at temperature $T$ .
$E_{1} = σ (27 + 273)^{2}$
$E_{2} = σ (127 + 273)^{2}$
$\frac{E_{1}}{E_{2}} = \frac{(300)^{4}}{(400)^{4}} = \frac{81}{256}$

AIIMS-2001

Heat Transfer

149502 Calculate radiation power for sphere whose temperature is $227^{\circ} C$ and radius $0.2 m$ and emissivity 0.8 .

1

1425 W

2

1500 W

3

1255 W

4

1575 W

Explanation:

A Given,
Radius $= 0.2 m$
And temperature $(T) = 227 + 273 = 500 K$
Radiation power through area A is given as
$P = σ A ε T^{4}$
[Here $ε =$ emissivity $= 0.8, A =$ surface area $= 4 π r^{2}$ , $σ = 5.67 \times 10^{- 8} w / m^{2} K^{4}]$
Putting these values we get
$P = 5.67 \times 10^{- 8} \times 4 π (0.2)^{2} \times (0.8) \times (500)^{4}$
$= 1417.48 ≃ 1425 W$

AIIMS-25.05.2019(M) Shift-1

Heat Transfer

149503 Distance between sun and earth is $2 \times 10^{8} km$ , temperature of sun $6000 K$ , radius of sun $7 \times 10^{5}$ $km$ . If emissivity of earth is 0.6 , then find out temperature of earth in thermal equilibrium.

1

400 K

2

300 K

3

500 K

4

600 K

Explanation:

B Given, distance between sun and earth (d) $= 2$ $\times 10^{8} km$ , Temperature of sun $= 6000 K$ and radius of sun $= 7 \times 10^{5} km$ .
For thermal equilibrium,
Energy received by earth = Energy emitted by earth
$\frac{T_{s}^{4} 4 π R_{s}^{2}}{4 π d^{2}} \times π R_{e}^{2} = σ \cdot ε \cdot T_{2}^{4} \cdot 4 π R_{e}^{2}$
$\frac{T_{s}^{4} \times R_{s}^{2}}{4 \times d^{2} \times e} = T_{e}^{4}$
$\frac{(6000)^{4} \times {(7 \times 10^{8})}^{2}}{4 \times {(2 \times 10^{11})}^{2} \times 0.6} = T_{e}^{4}$
$\frac{36 \times 36 \times 7 \times 7}{4 \times 4 \times 0.6} \times 10^{6} = T_{e}^{4}$
$66.15 \times 108 = T_{e}^{4}$
$T_{e} = 285.19 K ≃ 300 K$

AIIMS-25.05.2019(E) Shift-2

Heat Transfer

149504 If a body coated black at $600 K$ surrounded by atmosphere at $300 K$ has cooling rate $r_{0}$ , the same body at $900 K$ , surrounded by the same atmosphere will have cooling rate equal to-

1

\frac{16}{3} r_{0}

2

\frac{8}{16} r_{0}

3

16 r_{0}

4

4 r_{0}

Explanation:

A Given that, $T_{1} = 600 K$
$T_{2} = 900 K, T_{o} = 300 K$
We know that,
Cooling rate $\propto T^{4} - T_{0}^{4}$
$\frac{r}{r_{o}} = (\frac{T_{2}^{4} - T_{o}^{4}}{T_{1}^{4} - T_{o}^{4}})$
$\frac{r}{r_{o}} = [\frac{(900)^{4} - (300)^{4}}{(600)^{4} - (300)^{4}}]$
$\frac{r}{r_{o}} = (\frac{16}{3})$
$r = \frac{16}{3} r_{o}$

BCECE-2017

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