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Heat Transfer

149600 The surface temperature is the sun which has maximum energy emission at $500 nm$ is $6000 K$ . The temperature of a star which has maximum energy emission at $400 nm$ will be

1

8500 K

2

4500 K

3

7500 K

4

6500 K

Explanation:

C Given data,
$λ_{1} = 500 nm$
$λ_{2} = 400 nm$
$T_{1} = 6000 K$
$T_{2} =$ ?
According to Wein's Displacement law-
$λ_{1} T_{1} = λ_{2} T_{2}$
Where, $λ_{1} =$ wavelength at $T_{1}$
and $T_{2}$
$500 \times 6000 = 400 \times T_{2}$
$T_{2} = \frac{500 \times 6000}{400}$
$T_{2} = \frac{30, 00000}{400}$
$T_{2} = 7500 K$

JIPMER-2012

Heat Transfer

149601 The temperature of the black body increases from $T$ to $2 T$ . the factor by which the rate of emission will increase, is ?

1 4

2 2

3 16

4 8

Explanation:

C Given, $T_{1} = T, T_{2} = 2 T$
According to Stefan's law-
$∴ E = σ T^{4}$
Where $σ$ is Stefan's constant
$∴ \frac{E_{1}}{E_{2}} = \frac{T^{4}}{(2 T)^{4}}$
$\frac{E_{1}}{E_{2}} = \frac{1}{2^{4}} = \frac{1}{16}$
$E_{2} = 16 E_{1}$
Hence, rate of emission increases sixteen time.

JIPMER-2005

Heat Transfer

149602 A metal rod at a temperature of $145^{\circ} C$ , radiates energy at a rate of $17 W$ . If its temperature is increased to $273^{\circ} C$ , then it will radiate at the rate of

1

49.6 W

2

17.5 W

3

5.3 W

4

67.5 W

Explanation:

A Given that,
$T_{1} = 145^{\circ} C = 273 + 145 = 418 K$
$E_{1} = 17 W$
$T_{2} = 273^{\circ} C = 273 + 273 = 546 K$
According to Stefan's law,
$E \propto T^{4}$
$\frac{E_{1}}{E_{2}} = {(\frac{T_{1}}{T_{2}})}^{4}$
$\frac{E_{1}}{E_{2}} = {(\frac{418}{546})}^{4}$
$\frac{E_{1}}{E_{2}} = 0.343$
Therefore final radiated energy,
$E_{2} = \frac{E_{1}}{0.343} = \frac{17}{0.343}$
$E_{2} = 49.6 W$

JIPMER-2016

Heat Transfer

149605 When the temperature of a body is increased from $27^{\circ} C$ to $127^{\circ} C$ . the radiation emitted by it increases by a factor of

1

\frac{256}{81}

2

\frac{15}{19}

3

\frac{4}{5}

4

\frac{12}{27}

Explanation:

A Given that,
$T_{1} = 27^{\circ} C = 27 + 273 = 300 K$
$T_{2} = 127^{\circ} C = 127 + 273 = 400 K$
According to stefan's law,
$E \propto T^{4}$
$E = σ T^{4}$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}^{4}}{T_{2}^{4}}$
$\frac{E_{2}}{E_{1}} = \frac{T_{2}^{4}}{T_{1}^{4}}$
$\frac{E_{2}}{E_{1}} = {(\frac{400}{300})}^{4}$
$\frac{E_{2}}{E_{1}} = \frac{256}{81}$

AP EAMCET-03.09.2021

Heat Transfer

149600 The surface temperature is the sun which has maximum energy emission at $500 nm$ is $6000 K$ . The temperature of a star which has maximum energy emission at $400 nm$ will be

1

8500 K

2

4500 K

3

7500 K

4

6500 K

Explanation:

C Given data,
$λ_{1} = 500 nm$
$λ_{2} = 400 nm$
$T_{1} = 6000 K$
$T_{2} =$ ?
According to Wein's Displacement law-
$λ_{1} T_{1} = λ_{2} T_{2}$
Where, $λ_{1} =$ wavelength at $T_{1}$
and $T_{2}$
$500 \times 6000 = 400 \times T_{2}$
$T_{2} = \frac{500 \times 6000}{400}$
$T_{2} = \frac{30, 00000}{400}$
$T_{2} = 7500 K$

JIPMER-2012

Heat Transfer

149601 The temperature of the black body increases from $T$ to $2 T$ . the factor by which the rate of emission will increase, is ?

1 4

2 2

3 16

4 8

Explanation:

C Given, $T_{1} = T, T_{2} = 2 T$
According to Stefan's law-
$∴ E = σ T^{4}$
Where $σ$ is Stefan's constant
$∴ \frac{E_{1}}{E_{2}} = \frac{T^{4}}{(2 T)^{4}}$
$\frac{E_{1}}{E_{2}} = \frac{1}{2^{4}} = \frac{1}{16}$
$E_{2} = 16 E_{1}$
Hence, rate of emission increases sixteen time.

JIPMER-2005

Heat Transfer

149602 A metal rod at a temperature of $145^{\circ} C$ , radiates energy at a rate of $17 W$ . If its temperature is increased to $273^{\circ} C$ , then it will radiate at the rate of

1

49.6 W

2

17.5 W

3

5.3 W

4

67.5 W

Explanation:

A Given that,
$T_{1} = 145^{\circ} C = 273 + 145 = 418 K$
$E_{1} = 17 W$
$T_{2} = 273^{\circ} C = 273 + 273 = 546 K$
According to Stefan's law,
$E \propto T^{4}$
$\frac{E_{1}}{E_{2}} = {(\frac{T_{1}}{T_{2}})}^{4}$
$\frac{E_{1}}{E_{2}} = {(\frac{418}{546})}^{4}$
$\frac{E_{1}}{E_{2}} = 0.343$
Therefore final radiated energy,
$E_{2} = \frac{E_{1}}{0.343} = \frac{17}{0.343}$
$E_{2} = 49.6 W$

JIPMER-2016

Heat Transfer

149605 When the temperature of a body is increased from $27^{\circ} C$ to $127^{\circ} C$ . the radiation emitted by it increases by a factor of

1

\frac{256}{81}

2

\frac{15}{19}

3

\frac{4}{5}

4

\frac{12}{27}

Explanation:

A Given that,
$T_{1} = 27^{\circ} C = 27 + 273 = 300 K$
$T_{2} = 127^{\circ} C = 127 + 273 = 400 K$
According to stefan's law,
$E \propto T^{4}$
$E = σ T^{4}$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}^{4}}{T_{2}^{4}}$
$\frac{E_{2}}{E_{1}} = \frac{T_{2}^{4}}{T_{1}^{4}}$
$\frac{E_{2}}{E_{1}} = {(\frac{400}{300})}^{4}$
$\frac{E_{2}}{E_{1}} = \frac{256}{81}$

AP EAMCET-03.09.2021

Heat Transfer

149600 The surface temperature is the sun which has maximum energy emission at $500 nm$ is $6000 K$ . The temperature of a star which has maximum energy emission at $400 nm$ will be

1

8500 K

2

4500 K

3

7500 K

4

6500 K

Explanation:

C Given data,
$λ_{1} = 500 nm$
$λ_{2} = 400 nm$
$T_{1} = 6000 K$
$T_{2} =$ ?
According to Wein's Displacement law-
$λ_{1} T_{1} = λ_{2} T_{2}$
Where, $λ_{1} =$ wavelength at $T_{1}$
and $T_{2}$
$500 \times 6000 = 400 \times T_{2}$
$T_{2} = \frac{500 \times 6000}{400}$
$T_{2} = \frac{30, 00000}{400}$
$T_{2} = 7500 K$

JIPMER-2012

Heat Transfer

149601 The temperature of the black body increases from $T$ to $2 T$ . the factor by which the rate of emission will increase, is ?

1 4

2 2

3 16

4 8

Explanation:

C Given, $T_{1} = T, T_{2} = 2 T$
According to Stefan's law-
$∴ E = σ T^{4}$
Where $σ$ is Stefan's constant
$∴ \frac{E_{1}}{E_{2}} = \frac{T^{4}}{(2 T)^{4}}$
$\frac{E_{1}}{E_{2}} = \frac{1}{2^{4}} = \frac{1}{16}$
$E_{2} = 16 E_{1}$
Hence, rate of emission increases sixteen time.

JIPMER-2005

Heat Transfer

149602 A metal rod at a temperature of $145^{\circ} C$ , radiates energy at a rate of $17 W$ . If its temperature is increased to $273^{\circ} C$ , then it will radiate at the rate of

1

49.6 W

2

17.5 W

3

5.3 W

4

67.5 W

Explanation:

A Given that,
$T_{1} = 145^{\circ} C = 273 + 145 = 418 K$
$E_{1} = 17 W$
$T_{2} = 273^{\circ} C = 273 + 273 = 546 K$
According to Stefan's law,
$E \propto T^{4}$
$\frac{E_{1}}{E_{2}} = {(\frac{T_{1}}{T_{2}})}^{4}$
$\frac{E_{1}}{E_{2}} = {(\frac{418}{546})}^{4}$
$\frac{E_{1}}{E_{2}} = 0.343$
Therefore final radiated energy,
$E_{2} = \frac{E_{1}}{0.343} = \frac{17}{0.343}$
$E_{2} = 49.6 W$

JIPMER-2016

Heat Transfer

149605 When the temperature of a body is increased from $27^{\circ} C$ to $127^{\circ} C$ . the radiation emitted by it increases by a factor of

1

\frac{256}{81}

2

\frac{15}{19}

3

\frac{4}{5}

4

\frac{12}{27}

Explanation:

A Given that,
$T_{1} = 27^{\circ} C = 27 + 273 = 300 K$
$T_{2} = 127^{\circ} C = 127 + 273 = 400 K$
According to stefan's law,
$E \propto T^{4}$
$E = σ T^{4}$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}^{4}}{T_{2}^{4}}$
$\frac{E_{2}}{E_{1}} = \frac{T_{2}^{4}}{T_{1}^{4}}$
$\frac{E_{2}}{E_{1}} = {(\frac{400}{300})}^{4}$
$\frac{E_{2}}{E_{1}} = \frac{256}{81}$

AP EAMCET-03.09.2021

Heat Transfer

149600 The surface temperature is the sun which has maximum energy emission at $500 nm$ is $6000 K$ . The temperature of a star which has maximum energy emission at $400 nm$ will be

1

8500 K

2

4500 K

3

7500 K

4

6500 K

Explanation:

C Given data,
$λ_{1} = 500 nm$
$λ_{2} = 400 nm$
$T_{1} = 6000 K$
$T_{2} =$ ?
According to Wein's Displacement law-
$λ_{1} T_{1} = λ_{2} T_{2}$
Where, $λ_{1} =$ wavelength at $T_{1}$
and $T_{2}$
$500 \times 6000 = 400 \times T_{2}$
$T_{2} = \frac{500 \times 6000}{400}$
$T_{2} = \frac{30, 00000}{400}$
$T_{2} = 7500 K$

JIPMER-2012

Heat Transfer

149601 The temperature of the black body increases from $T$ to $2 T$ . the factor by which the rate of emission will increase, is ?

1 4

2 2

3 16

4 8

Explanation:

C Given, $T_{1} = T, T_{2} = 2 T$
According to Stefan's law-
$∴ E = σ T^{4}$
Where $σ$ is Stefan's constant
$∴ \frac{E_{1}}{E_{2}} = \frac{T^{4}}{(2 T)^{4}}$
$\frac{E_{1}}{E_{2}} = \frac{1}{2^{4}} = \frac{1}{16}$
$E_{2} = 16 E_{1}$
Hence, rate of emission increases sixteen time.

JIPMER-2005

Heat Transfer

149602 A metal rod at a temperature of $145^{\circ} C$ , radiates energy at a rate of $17 W$ . If its temperature is increased to $273^{\circ} C$ , then it will radiate at the rate of

1

49.6 W

2

17.5 W

3

5.3 W

4

67.5 W

Explanation:

A Given that,
$T_{1} = 145^{\circ} C = 273 + 145 = 418 K$
$E_{1} = 17 W$
$T_{2} = 273^{\circ} C = 273 + 273 = 546 K$
According to Stefan's law,
$E \propto T^{4}$
$\frac{E_{1}}{E_{2}} = {(\frac{T_{1}}{T_{2}})}^{4}$
$\frac{E_{1}}{E_{2}} = {(\frac{418}{546})}^{4}$
$\frac{E_{1}}{E_{2}} = 0.343$
Therefore final radiated energy,
$E_{2} = \frac{E_{1}}{0.343} = \frac{17}{0.343}$
$E_{2} = 49.6 W$

JIPMER-2016

Heat Transfer

149605 When the temperature of a body is increased from $27^{\circ} C$ to $127^{\circ} C$ . the radiation emitted by it increases by a factor of

1

\frac{256}{81}

2

\frac{15}{19}

3

\frac{4}{5}

4

\frac{12}{27}

Explanation:

A Given that,
$T_{1} = 27^{\circ} C = 27 + 273 = 300 K$
$T_{2} = 127^{\circ} C = 127 + 273 = 400 K$
According to stefan's law,
$E \propto T^{4}$
$E = σ T^{4}$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}^{4}}{T_{2}^{4}}$
$\frac{E_{2}}{E_{1}} = \frac{T_{2}^{4}}{T_{1}^{4}}$
$\frac{E_{2}}{E_{1}} = {(\frac{400}{300})}^{4}$
$\frac{E_{2}}{E_{1}} = \frac{256}{81}$

AP EAMCET-03.09.2021

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