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Heat Transfer

149455 The radiation energy density per unit wavelength at a temperature $T$ has a maximum at a wavelength $λ_{0}$ . At temperature $2 T$ , it will have a maximum at a wavelength:

1

4 λ_{0}

2

2 λ_{0}

3

λ_{0} / 2

4

λ_{0} / 4

Explanation:

C Given that,
$λ_{0}$ is the maximum wavelength corresponding to temperature $T$ .
If the temperature $2 T$ then corresponding wavelength is $λ$
According to Wien's law,
$λ_{m} T =$ Constant
$\frac{{(λ_{m})}_{1}}{{(λ_{m})}_{2}} = \frac{T_{2}}{T_{1}}$
$\frac{λ_{0}}{λ} = \frac{2 T}{T}$
$λ = \frac{λ_{0}}{2}$

UPSEE - 2004

Heat Transfer

149456 The total energy radiated from a black body source is collected for $1 \min$ and is used to heat a quantity of water. The temperature of water is found to increase from $20^{\circ} C$ to ${20.5}^{\circ} C$ . If the absolute temperature of the black body is doubled and the experiment is repeated with the same quantity of water at $20^{\circ} C$ , the temperature of water will be :

1

21^{\circ} C

2

22^{\circ} C

3

24^{\circ} C

4

28^{\circ} C

Explanation:

D From Stefan's law of radiation, energy radiated by a black body (E) is proportional to fourth power of absolute temperature $(T)$ .
$E \propto T^{4}$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}^{4}}{T_{2}^{4}}$
Given, $T_{1} = T, T_{2} = 2 T$
$∴ \frac{E_{1}}{E_{2}} = \frac{(T)^{4}}{(2 T)^{4}} = \frac{1}{2^{4}} = \frac{1}{16}$
Heat taken by water from radiation
$E = ms Δ T$
Where, $s$ is specific heat, $Δ T$ the change in temperature and $m$ is the mass.
$∴ E = m \times 1 \times (20.5 - 20)$
$E = m \times 0.5$
When energy supplied is 16 times the previous one, then let temperature rise to $T^{'}$ .
$∴ 16 E = m \times 1 \times (T^{'} - 20)$
Dividing Eq. (i) by (ii), we get
$\frac{1}{16} = \frac{0.5}{T^{'} - 20}$
$T^{'} - 20 = 16 \times 0.5 = 8$
$T^{'} = 20 + 8 = 28^{\circ} C$

UPSEE - 2004

Heat Transfer

149457 The wavelength of maximum emitted energy $(λ_{m})$ of a body at $700 K$ is $4.08 μ m$ . If the temperature of the body is raised to $1400 K$ , then the value of $λ_{m}$ will be

1

1.02 μ m

2

16.32 μ m

3

8.16 μ m

4

2.04 μ m

Explanation:

D Given that,
$λ_{m_{1}} = 4.08 μ m, T_{1} = 700 K .$
$T_{2} = 1400 K, λ_{m_{2}} = ?$
From Wien's displacement law,
$λ_{m} \propto \frac{1}{T}$
$\frac{λ_{m_{2}}}{λ_{m_{1}}} = \frac{T_{1}}{T_{2}}$
$λ_{m_{2}} = \frac{T_{1}}{T_{2}} \times λ_{m_{1}}$
$= \frac{700}{1400} \times 4.08$
$λ_{m_{2}} = 2.04 μ m$

AP EAMCET (22.09.2020) Shift-II

Heat Transfer

149458 A rectangular metal plate $8 cm \times 4 cm$ at $127^{\circ} C$ emits $E {Js}^{- 1}$ . If both length and breadth are halved and the temperature is raised to $327^{\circ} C$ , the rate of emission is

1

(\frac{9}{4}) E {Js}^{- 1}

2

(\frac{81}{64}) E {Js}^{- 1}

3

(\frac{27}{8}) E {Js}^{- 1}

4

(\frac{10}{7}) E {Js}^{- 1}

Explanation:

B Given that,
Area of rectangular box $= 8 cm \times 4 cm$
$(T_{1}) = 127^{\circ} C + 273 = 400 K$ and $E_{1} = E J / \sec$
If length $= \frac{8}{2} = 4 cm$
Breadth $= \frac{4}{2} = 2 cm$ .
And Temperature $(T_{2}) = 327 + 273 = 600 K$
We know, Stefan-Boltzmann's law,
$E = σ {AT}^{4}$
$\frac{E_{1}}{E_{2}} = (\frac{A_{1}}{A_{2}}) {(\frac{T_{1}}{T_{2}})}^{4}$
$\frac{E}{E_{2}} = (\frac{8 \times 4}{4 \times 2}) {(\frac{400}{600})}^{4}$
$\frac{E}{E_{2}} = \frac{4 \times 16}{81}$
$E_{2} = \frac{81}{64} E {Js}^{- 1}$

AP EAMCET (21.09.2020) Shift-II

Heat Transfer

149455 The radiation energy density per unit wavelength at a temperature $T$ has a maximum at a wavelength $λ_{0}$ . At temperature $2 T$ , it will have a maximum at a wavelength:

1

4 λ_{0}

2

2 λ_{0}

3

λ_{0} / 2

4

λ_{0} / 4

Explanation:

C Given that,
$λ_{0}$ is the maximum wavelength corresponding to temperature $T$ .
If the temperature $2 T$ then corresponding wavelength is $λ$
According to Wien's law,
$λ_{m} T =$ Constant
$\frac{{(λ_{m})}_{1}}{{(λ_{m})}_{2}} = \frac{T_{2}}{T_{1}}$
$\frac{λ_{0}}{λ} = \frac{2 T}{T}$
$λ = \frac{λ_{0}}{2}$

UPSEE - 2004

Heat Transfer

149456 The total energy radiated from a black body source is collected for $1 \min$ and is used to heat a quantity of water. The temperature of water is found to increase from $20^{\circ} C$ to ${20.5}^{\circ} C$ . If the absolute temperature of the black body is doubled and the experiment is repeated with the same quantity of water at $20^{\circ} C$ , the temperature of water will be :

1

21^{\circ} C

2

22^{\circ} C

3

24^{\circ} C

4

28^{\circ} C

Explanation:

D From Stefan's law of radiation, energy radiated by a black body (E) is proportional to fourth power of absolute temperature $(T)$ .
$E \propto T^{4}$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}^{4}}{T_{2}^{4}}$
Given, $T_{1} = T, T_{2} = 2 T$
$∴ \frac{E_{1}}{E_{2}} = \frac{(T)^{4}}{(2 T)^{4}} = \frac{1}{2^{4}} = \frac{1}{16}$
Heat taken by water from radiation
$E = ms Δ T$
Where, $s$ is specific heat, $Δ T$ the change in temperature and $m$ is the mass.
$∴ E = m \times 1 \times (20.5 - 20)$
$E = m \times 0.5$
When energy supplied is 16 times the previous one, then let temperature rise to $T^{'}$ .
$∴ 16 E = m \times 1 \times (T^{'} - 20)$
Dividing Eq. (i) by (ii), we get
$\frac{1}{16} = \frac{0.5}{T^{'} - 20}$
$T^{'} - 20 = 16 \times 0.5 = 8$
$T^{'} = 20 + 8 = 28^{\circ} C$

UPSEE - 2004

Heat Transfer

149457 The wavelength of maximum emitted energy $(λ_{m})$ of a body at $700 K$ is $4.08 μ m$ . If the temperature of the body is raised to $1400 K$ , then the value of $λ_{m}$ will be

1

1.02 μ m

2

16.32 μ m

3

8.16 μ m

4

2.04 μ m

Explanation:

D Given that,
$λ_{m_{1}} = 4.08 μ m, T_{1} = 700 K .$
$T_{2} = 1400 K, λ_{m_{2}} = ?$
From Wien's displacement law,
$λ_{m} \propto \frac{1}{T}$
$\frac{λ_{m_{2}}}{λ_{m_{1}}} = \frac{T_{1}}{T_{2}}$
$λ_{m_{2}} = \frac{T_{1}}{T_{2}} \times λ_{m_{1}}$
$= \frac{700}{1400} \times 4.08$
$λ_{m_{2}} = 2.04 μ m$

AP EAMCET (22.09.2020) Shift-II

Heat Transfer

149458 A rectangular metal plate $8 cm \times 4 cm$ at $127^{\circ} C$ emits $E {Js}^{- 1}$ . If both length and breadth are halved and the temperature is raised to $327^{\circ} C$ , the rate of emission is

1

(\frac{9}{4}) E {Js}^{- 1}

2

(\frac{81}{64}) E {Js}^{- 1}

3

(\frac{27}{8}) E {Js}^{- 1}

4

(\frac{10}{7}) E {Js}^{- 1}

Explanation:

B Given that,
Area of rectangular box $= 8 cm \times 4 cm$
$(T_{1}) = 127^{\circ} C + 273 = 400 K$ and $E_{1} = E J / \sec$
If length $= \frac{8}{2} = 4 cm$
Breadth $= \frac{4}{2} = 2 cm$ .
And Temperature $(T_{2}) = 327 + 273 = 600 K$
We know, Stefan-Boltzmann's law,
$E = σ {AT}^{4}$
$\frac{E_{1}}{E_{2}} = (\frac{A_{1}}{A_{2}}) {(\frac{T_{1}}{T_{2}})}^{4}$
$\frac{E}{E_{2}} = (\frac{8 \times 4}{4 \times 2}) {(\frac{400}{600})}^{4}$
$\frac{E}{E_{2}} = \frac{4 \times 16}{81}$
$E_{2} = \frac{81}{64} E {Js}^{- 1}$

AP EAMCET (21.09.2020) Shift-II

Heat Transfer

149455 The radiation energy density per unit wavelength at a temperature $T$ has a maximum at a wavelength $λ_{0}$ . At temperature $2 T$ , it will have a maximum at a wavelength:

1

4 λ_{0}

2

2 λ_{0}

3

λ_{0} / 2

4

λ_{0} / 4

Explanation:

C Given that,
$λ_{0}$ is the maximum wavelength corresponding to temperature $T$ .
If the temperature $2 T$ then corresponding wavelength is $λ$
According to Wien's law,
$λ_{m} T =$ Constant
$\frac{{(λ_{m})}_{1}}{{(λ_{m})}_{2}} = \frac{T_{2}}{T_{1}}$
$\frac{λ_{0}}{λ} = \frac{2 T}{T}$
$λ = \frac{λ_{0}}{2}$

UPSEE - 2004

Heat Transfer

149456 The total energy radiated from a black body source is collected for $1 \min$ and is used to heat a quantity of water. The temperature of water is found to increase from $20^{\circ} C$ to ${20.5}^{\circ} C$ . If the absolute temperature of the black body is doubled and the experiment is repeated with the same quantity of water at $20^{\circ} C$ , the temperature of water will be :

1

21^{\circ} C

2

22^{\circ} C

3

24^{\circ} C

4

28^{\circ} C

Explanation:

D From Stefan's law of radiation, energy radiated by a black body (E) is proportional to fourth power of absolute temperature $(T)$ .
$E \propto T^{4}$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}^{4}}{T_{2}^{4}}$
Given, $T_{1} = T, T_{2} = 2 T$
$∴ \frac{E_{1}}{E_{2}} = \frac{(T)^{4}}{(2 T)^{4}} = \frac{1}{2^{4}} = \frac{1}{16}$
Heat taken by water from radiation
$E = ms Δ T$
Where, $s$ is specific heat, $Δ T$ the change in temperature and $m$ is the mass.
$∴ E = m \times 1 \times (20.5 - 20)$
$E = m \times 0.5$
When energy supplied is 16 times the previous one, then let temperature rise to $T^{'}$ .
$∴ 16 E = m \times 1 \times (T^{'} - 20)$
Dividing Eq. (i) by (ii), we get
$\frac{1}{16} = \frac{0.5}{T^{'} - 20}$
$T^{'} - 20 = 16 \times 0.5 = 8$
$T^{'} = 20 + 8 = 28^{\circ} C$

UPSEE - 2004

Heat Transfer

149457 The wavelength of maximum emitted energy $(λ_{m})$ of a body at $700 K$ is $4.08 μ m$ . If the temperature of the body is raised to $1400 K$ , then the value of $λ_{m}$ will be

1

1.02 μ m

2

16.32 μ m

3

8.16 μ m

4

2.04 μ m

Explanation:

D Given that,
$λ_{m_{1}} = 4.08 μ m, T_{1} = 700 K .$
$T_{2} = 1400 K, λ_{m_{2}} = ?$
From Wien's displacement law,
$λ_{m} \propto \frac{1}{T}$
$\frac{λ_{m_{2}}}{λ_{m_{1}}} = \frac{T_{1}}{T_{2}}$
$λ_{m_{2}} = \frac{T_{1}}{T_{2}} \times λ_{m_{1}}$
$= \frac{700}{1400} \times 4.08$
$λ_{m_{2}} = 2.04 μ m$

AP EAMCET (22.09.2020) Shift-II

Heat Transfer

149458 A rectangular metal plate $8 cm \times 4 cm$ at $127^{\circ} C$ emits $E {Js}^{- 1}$ . If both length and breadth are halved and the temperature is raised to $327^{\circ} C$ , the rate of emission is

1

(\frac{9}{4}) E {Js}^{- 1}

2

(\frac{81}{64}) E {Js}^{- 1}

3

(\frac{27}{8}) E {Js}^{- 1}

4

(\frac{10}{7}) E {Js}^{- 1}

Explanation:

B Given that,
Area of rectangular box $= 8 cm \times 4 cm$
$(T_{1}) = 127^{\circ} C + 273 = 400 K$ and $E_{1} = E J / \sec$
If length $= \frac{8}{2} = 4 cm$
Breadth $= \frac{4}{2} = 2 cm$ .
And Temperature $(T_{2}) = 327 + 273 = 600 K$
We know, Stefan-Boltzmann's law,
$E = σ {AT}^{4}$
$\frac{E_{1}}{E_{2}} = (\frac{A_{1}}{A_{2}}) {(\frac{T_{1}}{T_{2}})}^{4}$
$\frac{E}{E_{2}} = (\frac{8 \times 4}{4 \times 2}) {(\frac{400}{600})}^{4}$
$\frac{E}{E_{2}} = \frac{4 \times 16}{81}$
$E_{2} = \frac{81}{64} E {Js}^{- 1}$

AP EAMCET (21.09.2020) Shift-II

Heat Transfer

149455 The radiation energy density per unit wavelength at a temperature $T$ has a maximum at a wavelength $λ_{0}$ . At temperature $2 T$ , it will have a maximum at a wavelength:

1

4 λ_{0}

2

2 λ_{0}

3

λ_{0} / 2

4

λ_{0} / 4

Explanation:

C Given that,
$λ_{0}$ is the maximum wavelength corresponding to temperature $T$ .
If the temperature $2 T$ then corresponding wavelength is $λ$
According to Wien's law,
$λ_{m} T =$ Constant
$\frac{{(λ_{m})}_{1}}{{(λ_{m})}_{2}} = \frac{T_{2}}{T_{1}}$
$\frac{λ_{0}}{λ} = \frac{2 T}{T}$
$λ = \frac{λ_{0}}{2}$

UPSEE - 2004

Heat Transfer

149456 The total energy radiated from a black body source is collected for $1 \min$ and is used to heat a quantity of water. The temperature of water is found to increase from $20^{\circ} C$ to ${20.5}^{\circ} C$ . If the absolute temperature of the black body is doubled and the experiment is repeated with the same quantity of water at $20^{\circ} C$ , the temperature of water will be :

1

21^{\circ} C

2

22^{\circ} C

3

24^{\circ} C

4

28^{\circ} C

Explanation:

D From Stefan's law of radiation, energy radiated by a black body (E) is proportional to fourth power of absolute temperature $(T)$ .
$E \propto T^{4}$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}^{4}}{T_{2}^{4}}$
Given, $T_{1} = T, T_{2} = 2 T$
$∴ \frac{E_{1}}{E_{2}} = \frac{(T)^{4}}{(2 T)^{4}} = \frac{1}{2^{4}} = \frac{1}{16}$
Heat taken by water from radiation
$E = ms Δ T$
Where, $s$ is specific heat, $Δ T$ the change in temperature and $m$ is the mass.
$∴ E = m \times 1 \times (20.5 - 20)$
$E = m \times 0.5$
When energy supplied is 16 times the previous one, then let temperature rise to $T^{'}$ .
$∴ 16 E = m \times 1 \times (T^{'} - 20)$
Dividing Eq. (i) by (ii), we get
$\frac{1}{16} = \frac{0.5}{T^{'} - 20}$
$T^{'} - 20 = 16 \times 0.5 = 8$
$T^{'} = 20 + 8 = 28^{\circ} C$

UPSEE - 2004

Heat Transfer

149457 The wavelength of maximum emitted energy $(λ_{m})$ of a body at $700 K$ is $4.08 μ m$ . If the temperature of the body is raised to $1400 K$ , then the value of $λ_{m}$ will be

1

1.02 μ m

2

16.32 μ m

3

8.16 μ m

4

2.04 μ m

Explanation:

D Given that,
$λ_{m_{1}} = 4.08 μ m, T_{1} = 700 K .$
$T_{2} = 1400 K, λ_{m_{2}} = ?$
From Wien's displacement law,
$λ_{m} \propto \frac{1}{T}$
$\frac{λ_{m_{2}}}{λ_{m_{1}}} = \frac{T_{1}}{T_{2}}$
$λ_{m_{2}} = \frac{T_{1}}{T_{2}} \times λ_{m_{1}}$
$= \frac{700}{1400} \times 4.08$
$λ_{m_{2}} = 2.04 μ m$

AP EAMCET (22.09.2020) Shift-II

Heat Transfer

149458 A rectangular metal plate $8 cm \times 4 cm$ at $127^{\circ} C$ emits $E {Js}^{- 1}$ . If both length and breadth are halved and the temperature is raised to $327^{\circ} C$ , the rate of emission is

1

(\frac{9}{4}) E {Js}^{- 1}

2

(\frac{81}{64}) E {Js}^{- 1}

3

(\frac{27}{8}) E {Js}^{- 1}

4

(\frac{10}{7}) E {Js}^{- 1}

Explanation:

B Given that,
Area of rectangular box $= 8 cm \times 4 cm$
$(T_{1}) = 127^{\circ} C + 273 = 400 K$ and $E_{1} = E J / \sec$
If length $= \frac{8}{2} = 4 cm$
Breadth $= \frac{4}{2} = 2 cm$ .
And Temperature $(T_{2}) = 327 + 273 = 600 K$
We know, Stefan-Boltzmann's law,
$E = σ {AT}^{4}$
$\frac{E_{1}}{E_{2}} = (\frac{A_{1}}{A_{2}}) {(\frac{T_{1}}{T_{2}})}^{4}$
$\frac{E}{E_{2}} = (\frac{8 \times 4}{4 \times 2}) {(\frac{400}{600})}^{4}$
$\frac{E}{E_{2}} = \frac{4 \times 16}{81}$
$E_{2} = \frac{81}{64} E {Js}^{- 1}$

AP EAMCET (21.09.2020) Shift-II

Question Bank Series

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