QBManager

Heat Transfer

149524 The S.I unit and dimensions of Stefan's constant ' $σ$ ' in case of Stefan's law of radiation is

1

J / m^{2} s^{4} K, M^{1} L^{0} T^{- 3} K^{3}

2

J / m^{3} {sK}^{4}, M^{1} L^{0} T^{- 3} K^{4}

3

J / m^{3} s^{4}, M^{1} L^{0} T^{- 3} K^{- 4}

4

J / m^{2} {sK}^{4}, M^{1} L^{0} T^{- 3} K^{- 4}

Explanation:

D SI unit for Stefan's constant is
Radiant energy $= σ e {AT}^{4}$
$σ = \frac{Radiated energy}{e A T^{4}}$
$= \frac{Watt (W)}{m^{2} K^{4}}$
SI unit of $σ = {Wm}^{- 2} K^{- 4} = {Js}^{- 1} m^{- 2} K^{- 4} = J / m^{2} {sK}^{4}$
$= \frac{kg \cdot m^{2}}{s^{2}} \times \frac{1}{s} \times m^{- 2} T^{- 4}$
$= kg \cdot s^{- 3} m^{0} T^{- 4}$
Dimension of Stefan's Constant
$σ = {ML}^{\circ} T^{- 3} K^{- 4}$

MHT-CET 2019

Heat Transfer

149525 Heat energy is incident on the surface at the rate of $1000 J / \min$ . If coefficient of absorption is 0.8 and coefficient of reflection is 0.1 then heat energy transmitted by the surface in 5 minute in

1

100 J

2

500 J

3

700 J

4

900 J

Explanation:

B Given that,
Rate of heat incident on surface $= 1000 J / \min$
Coefficient absorption $= 0.8$
Coefficient of reflection $= 0.1$
$∴$ Energy absorbed
$= Rate of heat incident x$
$coefficient of absorption$
$= 1000 \times 0.8$
$= 800 J / \min$
Energy reflected $=$ Rate of heat incident $\times$ coefficient of reflection
$= 1000 \times 0.1 = 100 J / \min$
If energy transmitted is $E$
so, energy incident $=$ energy absorbed + energy reflected + energy transmitted
$1000 = 800 + 100 + E$
$E = 100 J / \min$
Thus, energy transmitted in $5 \min$
$E = 100 \times 5$
$E = 500 J$

MHT-CET 2018

Heat Transfer

149526 A black body radiates heat at temperatures ' $T_{1}$ ' and ' $T_{2}, (T_{2} > T_{1})$ . The frequency corresponding to maximum energy is

1 more at

T_{1}

2 more at

T_{2}

3 equal for

T_{1}

and

T_{2}

4 independent of

T_{1}

and

T_{2}

Explanation:

B As we know that,
Wein's displacement law is given by
$λ T = constant$
$λ \propto \frac{1}{T}$
where $λ =$ wavelength
$T =$ Temperature
$∴ λ \propto \frac{1}{n}$ where $n =$ frequency
Thus $n \propto T$
If $T_{2} > T_{1}$ then $n_{2} > n_{1}$
Thus corresponding frequency is more at $T_{2}$ .

MHT-CET 2015

Heat Transfer

149527 If $150 J$ of energy is incident on area $2 m^{2}$ . If $Q_{r}$ $= 15 J$ , coefficient of absorption is 0.6 , then amount of energy transmitted is

1

50 J

2

45 J

3

40 J

4

30 J

Explanation:

B Given that,
Incident energy $= 150 J$
Reflected energy $= 15 J$
Coefficient of absorption $= 0.6$
$∴$ Incident energy $=$ absorbed energy + reflected energy + energy transmitted
$150 = Q_{a} + Q_{r} + Q_{t}$
$∴ Q_{a} = Q \times a$
$=$ Incident energy $\times$ coefficient of absorption
So, coefficient of reflected energy $r = \frac{Q_{r}}{Q}$
and coefficient of transmitted energy $t = \frac{Q_{t}}{Q}$
$Q = Q_{a} + Q_{r} + Q_{t}$
$1 = \frac{Q_{a}}{Q} + \frac{Q_{r}}{Q} + \frac{Q_{t}}{Q}$
$1 = a + r + t$
$1 = 0.6 + \frac{15}{150} + t$
$1 = 0.6 + 0.1 + t$
$1 = 0.7 + t$
$t = 1 - 0.7 = 0.3$
$t = 0.3$
$∴ t = \frac{Q_{t}}{Q} = 0.3$
$Q_{t} = 0.3 \times Q$
$Q_{t} = 0.3 \times 150$
$Q_{t} = 45 J$

MHT-CET 2009

Heat Transfer

149529 Two stars $A$ and $B$ radiate maximum energy at the wavelength of $360 nm$ and $480 nm$ respectively. Then the ratio of the surface temperatures of $A$ and $B$ is :

1

3 : 4

2

81 : 256

3

4 : 3

4

256 : 81

Explanation:

C Given that,
Wavelength of star A, $λ_{A} = 360 nm$
Wavelength of star $B, λ_{B} = 480 nm$
Surface temperature of star $A = T_{A}$
Surface temperature of star $B = T_{B}$
According to Wein's displacement law,
$λ T =$ constant
or $λ \propto \frac{1}{T}$
$\frac{λ_{A}}{λ_{B}} = \frac{T_{B}}{T_{A}}$
$\frac{T_{A}}{T_{B}} = \frac{λ_{B}}{λ_{A}} = \frac{480}{360} = \frac{4}{3}$
$T_{A} : T_{B} = 4 : 3$

Karnataka CET-2013

Heat Transfer

149524 The S.I unit and dimensions of Stefan's constant ' $σ$ ' in case of Stefan's law of radiation is

1

J / m^{2} s^{4} K, M^{1} L^{0} T^{- 3} K^{3}

2

J / m^{3} {sK}^{4}, M^{1} L^{0} T^{- 3} K^{4}

3

J / m^{3} s^{4}, M^{1} L^{0} T^{- 3} K^{- 4}

4

J / m^{2} {sK}^{4}, M^{1} L^{0} T^{- 3} K^{- 4}

Explanation:

D SI unit for Stefan's constant is
Radiant energy $= σ e {AT}^{4}$
$σ = \frac{Radiated energy}{e A T^{4}}$
$= \frac{Watt (W)}{m^{2} K^{4}}$
SI unit of $σ = {Wm}^{- 2} K^{- 4} = {Js}^{- 1} m^{- 2} K^{- 4} = J / m^{2} {sK}^{4}$
$= \frac{kg \cdot m^{2}}{s^{2}} \times \frac{1}{s} \times m^{- 2} T^{- 4}$
$= kg \cdot s^{- 3} m^{0} T^{- 4}$
Dimension of Stefan's Constant
$σ = {ML}^{\circ} T^{- 3} K^{- 4}$

MHT-CET 2019

Heat Transfer

149525 Heat energy is incident on the surface at the rate of $1000 J / \min$ . If coefficient of absorption is 0.8 and coefficient of reflection is 0.1 then heat energy transmitted by the surface in 5 minute in

1

100 J

2

500 J

3

700 J

4

900 J

Explanation:

B Given that,
Rate of heat incident on surface $= 1000 J / \min$
Coefficient absorption $= 0.8$
Coefficient of reflection $= 0.1$
$∴$ Energy absorbed
$= Rate of heat incident x$
$coefficient of absorption$
$= 1000 \times 0.8$
$= 800 J / \min$
Energy reflected $=$ Rate of heat incident $\times$ coefficient of reflection
$= 1000 \times 0.1 = 100 J / \min$
If energy transmitted is $E$
so, energy incident $=$ energy absorbed + energy reflected + energy transmitted
$1000 = 800 + 100 + E$
$E = 100 J / \min$
Thus, energy transmitted in $5 \min$
$E = 100 \times 5$
$E = 500 J$

MHT-CET 2018

Heat Transfer

149526 A black body radiates heat at temperatures ' $T_{1}$ ' and ' $T_{2}, (T_{2} > T_{1})$ . The frequency corresponding to maximum energy is

1 more at

T_{1}

2 more at

T_{2}

3 equal for

T_{1}

and

T_{2}

4 independent of

T_{1}

and

T_{2}

Explanation:

B As we know that,
Wein's displacement law is given by
$λ T = constant$
$λ \propto \frac{1}{T}$
where $λ =$ wavelength
$T =$ Temperature
$∴ λ \propto \frac{1}{n}$ where $n =$ frequency
Thus $n \propto T$
If $T_{2} > T_{1}$ then $n_{2} > n_{1}$
Thus corresponding frequency is more at $T_{2}$ .

MHT-CET 2015

Heat Transfer

149527 If $150 J$ of energy is incident on area $2 m^{2}$ . If $Q_{r}$ $= 15 J$ , coefficient of absorption is 0.6 , then amount of energy transmitted is

1

50 J

2

45 J

3

40 J

4

30 J

Explanation:

B Given that,
Incident energy $= 150 J$
Reflected energy $= 15 J$
Coefficient of absorption $= 0.6$
$∴$ Incident energy $=$ absorbed energy + reflected energy + energy transmitted
$150 = Q_{a} + Q_{r} + Q_{t}$
$∴ Q_{a} = Q \times a$
$=$ Incident energy $\times$ coefficient of absorption
So, coefficient of reflected energy $r = \frac{Q_{r}}{Q}$
and coefficient of transmitted energy $t = \frac{Q_{t}}{Q}$
$Q = Q_{a} + Q_{r} + Q_{t}$
$1 = \frac{Q_{a}}{Q} + \frac{Q_{r}}{Q} + \frac{Q_{t}}{Q}$
$1 = a + r + t$
$1 = 0.6 + \frac{15}{150} + t$
$1 = 0.6 + 0.1 + t$
$1 = 0.7 + t$
$t = 1 - 0.7 = 0.3$
$t = 0.3$
$∴ t = \frac{Q_{t}}{Q} = 0.3$
$Q_{t} = 0.3 \times Q$
$Q_{t} = 0.3 \times 150$
$Q_{t} = 45 J$

MHT-CET 2009

Heat Transfer

149529 Two stars $A$ and $B$ radiate maximum energy at the wavelength of $360 nm$ and $480 nm$ respectively. Then the ratio of the surface temperatures of $A$ and $B$ is :

1

3 : 4

2

81 : 256

3

4 : 3

4

256 : 81

Explanation:

C Given that,
Wavelength of star A, $λ_{A} = 360 nm$
Wavelength of star $B, λ_{B} = 480 nm$
Surface temperature of star $A = T_{A}$
Surface temperature of star $B = T_{B}$
According to Wein's displacement law,
$λ T =$ constant
or $λ \propto \frac{1}{T}$
$\frac{λ_{A}}{λ_{B}} = \frac{T_{B}}{T_{A}}$
$\frac{T_{A}}{T_{B}} = \frac{λ_{B}}{λ_{A}} = \frac{480}{360} = \frac{4}{3}$
$T_{A} : T_{B} = 4 : 3$

Karnataka CET-2013

Heat Transfer

149524 The S.I unit and dimensions of Stefan's constant ' $σ$ ' in case of Stefan's law of radiation is

1

J / m^{2} s^{4} K, M^{1} L^{0} T^{- 3} K^{3}

2

J / m^{3} {sK}^{4}, M^{1} L^{0} T^{- 3} K^{4}

3

J / m^{3} s^{4}, M^{1} L^{0} T^{- 3} K^{- 4}

4

J / m^{2} {sK}^{4}, M^{1} L^{0} T^{- 3} K^{- 4}

Explanation:

D SI unit for Stefan's constant is
Radiant energy $= σ e {AT}^{4}$
$σ = \frac{Radiated energy}{e A T^{4}}$
$= \frac{Watt (W)}{m^{2} K^{4}}$
SI unit of $σ = {Wm}^{- 2} K^{- 4} = {Js}^{- 1} m^{- 2} K^{- 4} = J / m^{2} {sK}^{4}$
$= \frac{kg \cdot m^{2}}{s^{2}} \times \frac{1}{s} \times m^{- 2} T^{- 4}$
$= kg \cdot s^{- 3} m^{0} T^{- 4}$
Dimension of Stefan's Constant
$σ = {ML}^{\circ} T^{- 3} K^{- 4}$

MHT-CET 2019

Heat Transfer

149525 Heat energy is incident on the surface at the rate of $1000 J / \min$ . If coefficient of absorption is 0.8 and coefficient of reflection is 0.1 then heat energy transmitted by the surface in 5 minute in

1

100 J

2

500 J

3

700 J

4

900 J

Explanation:

B Given that,
Rate of heat incident on surface $= 1000 J / \min$
Coefficient absorption $= 0.8$
Coefficient of reflection $= 0.1$
$∴$ Energy absorbed
$= Rate of heat incident x$
$coefficient of absorption$
$= 1000 \times 0.8$
$= 800 J / \min$
Energy reflected $=$ Rate of heat incident $\times$ coefficient of reflection
$= 1000 \times 0.1 = 100 J / \min$
If energy transmitted is $E$
so, energy incident $=$ energy absorbed + energy reflected + energy transmitted
$1000 = 800 + 100 + E$
$E = 100 J / \min$
Thus, energy transmitted in $5 \min$
$E = 100 \times 5$
$E = 500 J$

MHT-CET 2018

Heat Transfer

149526 A black body radiates heat at temperatures ' $T_{1}$ ' and ' $T_{2}, (T_{2} > T_{1})$ . The frequency corresponding to maximum energy is

1 more at

T_{1}

2 more at

T_{2}

3 equal for

T_{1}

and

T_{2}

4 independent of

T_{1}

and

T_{2}

Explanation:

B As we know that,
Wein's displacement law is given by
$λ T = constant$
$λ \propto \frac{1}{T}$
where $λ =$ wavelength
$T =$ Temperature
$∴ λ \propto \frac{1}{n}$ where $n =$ frequency
Thus $n \propto T$
If $T_{2} > T_{1}$ then $n_{2} > n_{1}$
Thus corresponding frequency is more at $T_{2}$ .

MHT-CET 2015

Heat Transfer

149527 If $150 J$ of energy is incident on area $2 m^{2}$ . If $Q_{r}$ $= 15 J$ , coefficient of absorption is 0.6 , then amount of energy transmitted is

1

50 J

2

45 J

3

40 J

4

30 J

Explanation:

B Given that,
Incident energy $= 150 J$
Reflected energy $= 15 J$
Coefficient of absorption $= 0.6$
$∴$ Incident energy $=$ absorbed energy + reflected energy + energy transmitted
$150 = Q_{a} + Q_{r} + Q_{t}$
$∴ Q_{a} = Q \times a$
$=$ Incident energy $\times$ coefficient of absorption
So, coefficient of reflected energy $r = \frac{Q_{r}}{Q}$
and coefficient of transmitted energy $t = \frac{Q_{t}}{Q}$
$Q = Q_{a} + Q_{r} + Q_{t}$
$1 = \frac{Q_{a}}{Q} + \frac{Q_{r}}{Q} + \frac{Q_{t}}{Q}$
$1 = a + r + t$
$1 = 0.6 + \frac{15}{150} + t$
$1 = 0.6 + 0.1 + t$
$1 = 0.7 + t$
$t = 1 - 0.7 = 0.3$
$t = 0.3$
$∴ t = \frac{Q_{t}}{Q} = 0.3$
$Q_{t} = 0.3 \times Q$
$Q_{t} = 0.3 \times 150$
$Q_{t} = 45 J$

MHT-CET 2009

Heat Transfer

149529 Two stars $A$ and $B$ radiate maximum energy at the wavelength of $360 nm$ and $480 nm$ respectively. Then the ratio of the surface temperatures of $A$ and $B$ is :

1

3 : 4

2

81 : 256

3

4 : 3

4

256 : 81

Explanation:

C Given that,
Wavelength of star A, $λ_{A} = 360 nm$
Wavelength of star $B, λ_{B} = 480 nm$
Surface temperature of star $A = T_{A}$
Surface temperature of star $B = T_{B}$
According to Wein's displacement law,
$λ T =$ constant
or $λ \propto \frac{1}{T}$
$\frac{λ_{A}}{λ_{B}} = \frac{T_{B}}{T_{A}}$
$\frac{T_{A}}{T_{B}} = \frac{λ_{B}}{λ_{A}} = \frac{480}{360} = \frac{4}{3}$
$T_{A} : T_{B} = 4 : 3$

Karnataka CET-2013

Heat Transfer

149524 The S.I unit and dimensions of Stefan's constant ' $σ$ ' in case of Stefan's law of radiation is

1

J / m^{2} s^{4} K, M^{1} L^{0} T^{- 3} K^{3}

2

J / m^{3} {sK}^{4}, M^{1} L^{0} T^{- 3} K^{4}

3

J / m^{3} s^{4}, M^{1} L^{0} T^{- 3} K^{- 4}

4

J / m^{2} {sK}^{4}, M^{1} L^{0} T^{- 3} K^{- 4}

Explanation:

D SI unit for Stefan's constant is
Radiant energy $= σ e {AT}^{4}$
$σ = \frac{Radiated energy}{e A T^{4}}$
$= \frac{Watt (W)}{m^{2} K^{4}}$
SI unit of $σ = {Wm}^{- 2} K^{- 4} = {Js}^{- 1} m^{- 2} K^{- 4} = J / m^{2} {sK}^{4}$
$= \frac{kg \cdot m^{2}}{s^{2}} \times \frac{1}{s} \times m^{- 2} T^{- 4}$
$= kg \cdot s^{- 3} m^{0} T^{- 4}$
Dimension of Stefan's Constant
$σ = {ML}^{\circ} T^{- 3} K^{- 4}$

MHT-CET 2019

Heat Transfer

149525 Heat energy is incident on the surface at the rate of $1000 J / \min$ . If coefficient of absorption is 0.8 and coefficient of reflection is 0.1 then heat energy transmitted by the surface in 5 minute in

1

100 J

2

500 J

3

700 J

4

900 J

Explanation:

B Given that,
Rate of heat incident on surface $= 1000 J / \min$
Coefficient absorption $= 0.8$
Coefficient of reflection $= 0.1$
$∴$ Energy absorbed
$= Rate of heat incident x$
$coefficient of absorption$
$= 1000 \times 0.8$
$= 800 J / \min$
Energy reflected $=$ Rate of heat incident $\times$ coefficient of reflection
$= 1000 \times 0.1 = 100 J / \min$
If energy transmitted is $E$
so, energy incident $=$ energy absorbed + energy reflected + energy transmitted
$1000 = 800 + 100 + E$
$E = 100 J / \min$
Thus, energy transmitted in $5 \min$
$E = 100 \times 5$
$E = 500 J$

MHT-CET 2018

Heat Transfer

149526 A black body radiates heat at temperatures ' $T_{1}$ ' and ' $T_{2}, (T_{2} > T_{1})$ . The frequency corresponding to maximum energy is

1 more at

T_{1}

2 more at

T_{2}

3 equal for

T_{1}

and

T_{2}

4 independent of

T_{1}

and

T_{2}

Explanation:

B As we know that,
Wein's displacement law is given by
$λ T = constant$
$λ \propto \frac{1}{T}$
where $λ =$ wavelength
$T =$ Temperature
$∴ λ \propto \frac{1}{n}$ where $n =$ frequency
Thus $n \propto T$
If $T_{2} > T_{1}$ then $n_{2} > n_{1}$
Thus corresponding frequency is more at $T_{2}$ .

MHT-CET 2015

Heat Transfer

149527 If $150 J$ of energy is incident on area $2 m^{2}$ . If $Q_{r}$ $= 15 J$ , coefficient of absorption is 0.6 , then amount of energy transmitted is

1

50 J

2

45 J

3

40 J

4

30 J

Explanation:

B Given that,
Incident energy $= 150 J$
Reflected energy $= 15 J$
Coefficient of absorption $= 0.6$
$∴$ Incident energy $=$ absorbed energy + reflected energy + energy transmitted
$150 = Q_{a} + Q_{r} + Q_{t}$
$∴ Q_{a} = Q \times a$
$=$ Incident energy $\times$ coefficient of absorption
So, coefficient of reflected energy $r = \frac{Q_{r}}{Q}$
and coefficient of transmitted energy $t = \frac{Q_{t}}{Q}$
$Q = Q_{a} + Q_{r} + Q_{t}$
$1 = \frac{Q_{a}}{Q} + \frac{Q_{r}}{Q} + \frac{Q_{t}}{Q}$
$1 = a + r + t$
$1 = 0.6 + \frac{15}{150} + t$
$1 = 0.6 + 0.1 + t$
$1 = 0.7 + t$
$t = 1 - 0.7 = 0.3$
$t = 0.3$
$∴ t = \frac{Q_{t}}{Q} = 0.3$
$Q_{t} = 0.3 \times Q$
$Q_{t} = 0.3 \times 150$
$Q_{t} = 45 J$

MHT-CET 2009

Heat Transfer

149529 Two stars $A$ and $B$ radiate maximum energy at the wavelength of $360 nm$ and $480 nm$ respectively. Then the ratio of the surface temperatures of $A$ and $B$ is :

1

3 : 4

2

81 : 256

3

4 : 3

4

256 : 81

Explanation:

C Given that,
Wavelength of star A, $λ_{A} = 360 nm$
Wavelength of star $B, λ_{B} = 480 nm$
Surface temperature of star $A = T_{A}$
Surface temperature of star $B = T_{B}$
According to Wein's displacement law,
$λ T =$ constant
or $λ \propto \frac{1}{T}$
$\frac{λ_{A}}{λ_{B}} = \frac{T_{B}}{T_{A}}$
$\frac{T_{A}}{T_{B}} = \frac{λ_{B}}{λ_{A}} = \frac{480}{360} = \frac{4}{3}$
$T_{A} : T_{B} = 4 : 3$

Karnataka CET-2013

Heat Transfer

149524 The S.I unit and dimensions of Stefan's constant ' $σ$ ' in case of Stefan's law of radiation is

1

J / m^{2} s^{4} K, M^{1} L^{0} T^{- 3} K^{3}

2

J / m^{3} {sK}^{4}, M^{1} L^{0} T^{- 3} K^{4}

3

J / m^{3} s^{4}, M^{1} L^{0} T^{- 3} K^{- 4}

4

J / m^{2} {sK}^{4}, M^{1} L^{0} T^{- 3} K^{- 4}

Explanation:

D SI unit for Stefan's constant is
Radiant energy $= σ e {AT}^{4}$
$σ = \frac{Radiated energy}{e A T^{4}}$
$= \frac{Watt (W)}{m^{2} K^{4}}$
SI unit of $σ = {Wm}^{- 2} K^{- 4} = {Js}^{- 1} m^{- 2} K^{- 4} = J / m^{2} {sK}^{4}$
$= \frac{kg \cdot m^{2}}{s^{2}} \times \frac{1}{s} \times m^{- 2} T^{- 4}$
$= kg \cdot s^{- 3} m^{0} T^{- 4}$
Dimension of Stefan's Constant
$σ = {ML}^{\circ} T^{- 3} K^{- 4}$

MHT-CET 2019

Heat Transfer

149525 Heat energy is incident on the surface at the rate of $1000 J / \min$ . If coefficient of absorption is 0.8 and coefficient of reflection is 0.1 then heat energy transmitted by the surface in 5 minute in

1

100 J

2

500 J

3

700 J

4

900 J

Explanation:

B Given that,
Rate of heat incident on surface $= 1000 J / \min$
Coefficient absorption $= 0.8$
Coefficient of reflection $= 0.1$
$∴$ Energy absorbed
$= Rate of heat incident x$
$coefficient of absorption$
$= 1000 \times 0.8$
$= 800 J / \min$
Energy reflected $=$ Rate of heat incident $\times$ coefficient of reflection
$= 1000 \times 0.1 = 100 J / \min$
If energy transmitted is $E$
so, energy incident $=$ energy absorbed + energy reflected + energy transmitted
$1000 = 800 + 100 + E$
$E = 100 J / \min$
Thus, energy transmitted in $5 \min$
$E = 100 \times 5$
$E = 500 J$

MHT-CET 2018

Heat Transfer

149526 A black body radiates heat at temperatures ' $T_{1}$ ' and ' $T_{2}, (T_{2} > T_{1})$ . The frequency corresponding to maximum energy is

1 more at

T_{1}

2 more at

T_{2}

3 equal for

T_{1}

and

T_{2}

4 independent of

T_{1}

and

T_{2}

Explanation:

B As we know that,
Wein's displacement law is given by
$λ T = constant$
$λ \propto \frac{1}{T}$
where $λ =$ wavelength
$T =$ Temperature
$∴ λ \propto \frac{1}{n}$ where $n =$ frequency
Thus $n \propto T$
If $T_{2} > T_{1}$ then $n_{2} > n_{1}$
Thus corresponding frequency is more at $T_{2}$ .

MHT-CET 2015

Heat Transfer

149527 If $150 J$ of energy is incident on area $2 m^{2}$ . If $Q_{r}$ $= 15 J$ , coefficient of absorption is 0.6 , then amount of energy transmitted is

1

50 J

2

45 J

3

40 J

4

30 J

Explanation:

B Given that,
Incident energy $= 150 J$
Reflected energy $= 15 J$
Coefficient of absorption $= 0.6$
$∴$ Incident energy $=$ absorbed energy + reflected energy + energy transmitted
$150 = Q_{a} + Q_{r} + Q_{t}$
$∴ Q_{a} = Q \times a$
$=$ Incident energy $\times$ coefficient of absorption
So, coefficient of reflected energy $r = \frac{Q_{r}}{Q}$
and coefficient of transmitted energy $t = \frac{Q_{t}}{Q}$
$Q = Q_{a} + Q_{r} + Q_{t}$
$1 = \frac{Q_{a}}{Q} + \frac{Q_{r}}{Q} + \frac{Q_{t}}{Q}$
$1 = a + r + t$
$1 = 0.6 + \frac{15}{150} + t$
$1 = 0.6 + 0.1 + t$
$1 = 0.7 + t$
$t = 1 - 0.7 = 0.3$
$t = 0.3$
$∴ t = \frac{Q_{t}}{Q} = 0.3$
$Q_{t} = 0.3 \times Q$
$Q_{t} = 0.3 \times 150$
$Q_{t} = 45 J$

MHT-CET 2009

Heat Transfer

149529 Two stars $A$ and $B$ radiate maximum energy at the wavelength of $360 nm$ and $480 nm$ respectively. Then the ratio of the surface temperatures of $A$ and $B$ is :

1

3 : 4

2

81 : 256

3

4 : 3

4

256 : 81

Explanation:

C Given that,
Wavelength of star A, $λ_{A} = 360 nm$
Wavelength of star $B, λ_{B} = 480 nm$
Surface temperature of star $A = T_{A}$
Surface temperature of star $B = T_{B}$
According to Wein's displacement law,
$λ T =$ constant
or $λ \propto \frac{1}{T}$
$\frac{λ_{A}}{λ_{B}} = \frac{T_{B}}{T_{A}}$
$\frac{T_{A}}{T_{B}} = \frac{λ_{B}}{λ_{A}} = \frac{480}{360} = \frac{4}{3}$
$T_{A} : T_{B} = 4 : 3$

Karnataka CET-2013