D SI unit for Stefan's constant is Radiant energy $=\sigma e \mathrm{AT}^{4}$ $\sigma=\frac{\text { Radiated energy }}{\mathrm{e} \mathrm{A} \mathrm{T}^{4}}$ $=\frac{\mathrm{Watt}(\mathrm{W})}{\mathrm{m}^{2} \mathrm{~K}^{4}}$ SI unit of $\sigma=\mathrm{Wm}^{-2} \mathrm{~K}^{-4}=\mathrm{Js}^{-1} \mathrm{~m}^{-2} \mathrm{~K}^{-4}=\mathrm{J} / \mathrm{m}^{2} \mathrm{sK}^{4}$ $=\frac{\mathrm{kg} \cdot \mathrm{m}^{2}}{\mathrm{~s}^{2}} \times \frac{1}{\mathrm{~s}} \times \mathrm{m}^{-2} \mathrm{~T}^{-4}$ $=\mathrm{kg} \cdot \mathrm{s}^{-3} \mathrm{~m}^{0} \mathrm{~T}^{-4}$ Dimension of Stefan's Constant $\sigma=\mathrm{ML}^{\circ} \mathrm{T}^{-3} \mathrm{~K}^{-4}$
MHT-CET 2019
Heat Transfer
149525
Heat energy is incident on the surface at the rate of $1000 \mathrm{~J} / \mathrm{min}$. 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 $100 \mathrm{~J}$
2 $500 \mathrm{~J}$
3 $700 \mathrm{~J}$
4 $900 \mathrm{~J}$
Explanation:
B Given that, Rate of heat incident on surface $=1000 \mathrm{~J} / \mathrm{min}$ Coefficient absorption $\quad=0.8$ Coefficient of reflection $=0.1$ $\therefore$ Energy absorbed $=\text { Rate of heat incident } x$ $\text { coefficient of absorption }$ $=1000 \times 0.8$ $=800 \mathrm{~J} / \mathrm{min}$ Energy reflected $=$ Rate of heat incident $\times$ coefficient of reflection $=1000 \times 0.1=100 \mathrm{~J} / \mathrm{min}$ If energy transmitted is $\mathrm{E}$ so, energy incident $=$ energy absorbed + energy reflected + energy transmitted $1000=800+100+E$ $E=100 \mathrm{~J} / \mathrm{min}$ Thus, energy transmitted in $5 \mathrm{~min}$ $\mathrm{E}=100 \times 5$ $\mathrm{E}=500 \mathrm{~J}$
MHT-CET 2018
Heat Transfer
149526
A black body radiates heat at temperatures ' $T_{1}$ ' and ' $T_{2}, \quad\left(T_{2}>T_{1}\right)$. 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 $\lambda \mathrm{T}=\text { constant }$ $\lambda \propto \frac{1}{\mathrm{~T}}$ where $\lambda=$ wavelength $\mathrm{T}=$ Temperature $\therefore \quad \lambda \propto \frac{1}{\mathrm{n}}$ where $\mathrm{n}=$ frequency Thus $\mathrm{n} \propto \mathrm{T}$ If $\mathrm{T}_{2}>\mathrm{T}_{1}$ then $\mathrm{n}_{2}>\mathrm{n}_{1}$ Thus corresponding frequency is more at $T_{2}$.
MHT-CET 2015
Heat Transfer
149527
If $150 \mathrm{~J}$ of energy is incident on area $2 \mathrm{~m}^{2}$. If $Q_{\mathrm{r}}$ $=15 \mathrm{~J}$, coefficient of absorption is 0.6 , then amount of energy transmitted is
1 $50 \mathrm{~J}$
2 $45 \mathrm{~J}$
3 $40 \mathrm{~J}$
4 $30 \mathrm{~J}$
Explanation:
B Given that, Incident energy $=150 \mathrm{~J}$ Reflected energy $=15 \mathrm{~J}$ Coefficient of absorption $=0.6$ $\therefore$ Incident energy $=$ absorbed energy + reflected energy + energy transmitted $150=Q_{a}+Q_{r}+Q_{t}$ $\therefore \quad \mathrm{Q}_{\mathrm{a}}=\mathrm{Q} \times \mathrm{a}$ $=$ Incident energy $\times$ coefficient of absorption So, coefficient of reflected energy $r=\frac{Q_{r}}{Q}$ and coefficient of transmitted energy $t=\frac{Q_{t}}{Q}$ $\mathrm{Q}=\mathrm{Q}_{\mathrm{a}}+\mathrm{Q}_{\mathrm{r}}+\mathrm{Q}_{\mathrm{t}}$ $1=\frac{Q_{a}}{Q}+\frac{Q_{r}}{Q}+\frac{Q_{t}}{Q}$ $1=\mathrm{a}+\mathrm{r}+\mathrm{t}$ $1=0.6+\frac{15}{150}+\mathrm{t}$ $1=0.6+0.1+\mathrm{t}$ $1=0.7+\mathrm{t}$ $\mathrm{t}=1-0.7=0.3$ $\mathrm{t}=0.3$ $\therefore \quad \mathrm{t}=\frac{\mathrm{Q}_{\mathrm{t}}}{\mathrm{Q}}=0.3$ $\mathrm{Q}_{\mathrm{t}}=0.3 \times \mathrm{Q}$ $\mathrm{Q}_{\mathrm{t}}=0.3 \times 150$ $\mathrm{Q}_{\mathrm{t}}=45 \mathrm{~J}$
MHT-CET 2009
Heat Transfer
149529
Two stars $A$ and $B$ radiate maximum energy at the wavelength of $360 \mathrm{~nm}$ and $480 \mathrm{~nm}$ 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, $\lambda_{\mathrm{A}}=360 \mathrm{~nm}$ Wavelength of star $\mathrm{B}, \lambda_{\mathrm{B}}=480 \mathrm{~nm}$ Surface temperature of star $A=T_{A}$ Surface temperature of star $B=T_{B}$ According to Wein's displacement law, $\lambda \mathrm{T}=$ constant or $\lambda \propto \frac{1}{\mathrm{~T}}$ $\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{\mathrm{T}_{\mathrm{B}}}{\mathrm{T}_{\mathrm{A}}}$ $\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\frac{\lambda_{\mathrm{B}}}{\lambda_{\mathrm{A}}}=\frac{480}{360}=\frac{4}{3}$ $\mathrm{T}_{\mathrm{A}}: \mathrm{T}_{\mathrm{B}}=4: 3$
D SI unit for Stefan's constant is Radiant energy $=\sigma e \mathrm{AT}^{4}$ $\sigma=\frac{\text { Radiated energy }}{\mathrm{e} \mathrm{A} \mathrm{T}^{4}}$ $=\frac{\mathrm{Watt}(\mathrm{W})}{\mathrm{m}^{2} \mathrm{~K}^{4}}$ SI unit of $\sigma=\mathrm{Wm}^{-2} \mathrm{~K}^{-4}=\mathrm{Js}^{-1} \mathrm{~m}^{-2} \mathrm{~K}^{-4}=\mathrm{J} / \mathrm{m}^{2} \mathrm{sK}^{4}$ $=\frac{\mathrm{kg} \cdot \mathrm{m}^{2}}{\mathrm{~s}^{2}} \times \frac{1}{\mathrm{~s}} \times \mathrm{m}^{-2} \mathrm{~T}^{-4}$ $=\mathrm{kg} \cdot \mathrm{s}^{-3} \mathrm{~m}^{0} \mathrm{~T}^{-4}$ Dimension of Stefan's Constant $\sigma=\mathrm{ML}^{\circ} \mathrm{T}^{-3} \mathrm{~K}^{-4}$
MHT-CET 2019
Heat Transfer
149525
Heat energy is incident on the surface at the rate of $1000 \mathrm{~J} / \mathrm{min}$. 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 $100 \mathrm{~J}$
2 $500 \mathrm{~J}$
3 $700 \mathrm{~J}$
4 $900 \mathrm{~J}$
Explanation:
B Given that, Rate of heat incident on surface $=1000 \mathrm{~J} / \mathrm{min}$ Coefficient absorption $\quad=0.8$ Coefficient of reflection $=0.1$ $\therefore$ Energy absorbed $=\text { Rate of heat incident } x$ $\text { coefficient of absorption }$ $=1000 \times 0.8$ $=800 \mathrm{~J} / \mathrm{min}$ Energy reflected $=$ Rate of heat incident $\times$ coefficient of reflection $=1000 \times 0.1=100 \mathrm{~J} / \mathrm{min}$ If energy transmitted is $\mathrm{E}$ so, energy incident $=$ energy absorbed + energy reflected + energy transmitted $1000=800+100+E$ $E=100 \mathrm{~J} / \mathrm{min}$ Thus, energy transmitted in $5 \mathrm{~min}$ $\mathrm{E}=100 \times 5$ $\mathrm{E}=500 \mathrm{~J}$
MHT-CET 2018
Heat Transfer
149526
A black body radiates heat at temperatures ' $T_{1}$ ' and ' $T_{2}, \quad\left(T_{2}>T_{1}\right)$. 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 $\lambda \mathrm{T}=\text { constant }$ $\lambda \propto \frac{1}{\mathrm{~T}}$ where $\lambda=$ wavelength $\mathrm{T}=$ Temperature $\therefore \quad \lambda \propto \frac{1}{\mathrm{n}}$ where $\mathrm{n}=$ frequency Thus $\mathrm{n} \propto \mathrm{T}$ If $\mathrm{T}_{2}>\mathrm{T}_{1}$ then $\mathrm{n}_{2}>\mathrm{n}_{1}$ Thus corresponding frequency is more at $T_{2}$.
MHT-CET 2015
Heat Transfer
149527
If $150 \mathrm{~J}$ of energy is incident on area $2 \mathrm{~m}^{2}$. If $Q_{\mathrm{r}}$ $=15 \mathrm{~J}$, coefficient of absorption is 0.6 , then amount of energy transmitted is
1 $50 \mathrm{~J}$
2 $45 \mathrm{~J}$
3 $40 \mathrm{~J}$
4 $30 \mathrm{~J}$
Explanation:
B Given that, Incident energy $=150 \mathrm{~J}$ Reflected energy $=15 \mathrm{~J}$ Coefficient of absorption $=0.6$ $\therefore$ Incident energy $=$ absorbed energy + reflected energy + energy transmitted $150=Q_{a}+Q_{r}+Q_{t}$ $\therefore \quad \mathrm{Q}_{\mathrm{a}}=\mathrm{Q} \times \mathrm{a}$ $=$ Incident energy $\times$ coefficient of absorption So, coefficient of reflected energy $r=\frac{Q_{r}}{Q}$ and coefficient of transmitted energy $t=\frac{Q_{t}}{Q}$ $\mathrm{Q}=\mathrm{Q}_{\mathrm{a}}+\mathrm{Q}_{\mathrm{r}}+\mathrm{Q}_{\mathrm{t}}$ $1=\frac{Q_{a}}{Q}+\frac{Q_{r}}{Q}+\frac{Q_{t}}{Q}$ $1=\mathrm{a}+\mathrm{r}+\mathrm{t}$ $1=0.6+\frac{15}{150}+\mathrm{t}$ $1=0.6+0.1+\mathrm{t}$ $1=0.7+\mathrm{t}$ $\mathrm{t}=1-0.7=0.3$ $\mathrm{t}=0.3$ $\therefore \quad \mathrm{t}=\frac{\mathrm{Q}_{\mathrm{t}}}{\mathrm{Q}}=0.3$ $\mathrm{Q}_{\mathrm{t}}=0.3 \times \mathrm{Q}$ $\mathrm{Q}_{\mathrm{t}}=0.3 \times 150$ $\mathrm{Q}_{\mathrm{t}}=45 \mathrm{~J}$
MHT-CET 2009
Heat Transfer
149529
Two stars $A$ and $B$ radiate maximum energy at the wavelength of $360 \mathrm{~nm}$ and $480 \mathrm{~nm}$ 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, $\lambda_{\mathrm{A}}=360 \mathrm{~nm}$ Wavelength of star $\mathrm{B}, \lambda_{\mathrm{B}}=480 \mathrm{~nm}$ Surface temperature of star $A=T_{A}$ Surface temperature of star $B=T_{B}$ According to Wein's displacement law, $\lambda \mathrm{T}=$ constant or $\lambda \propto \frac{1}{\mathrm{~T}}$ $\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{\mathrm{T}_{\mathrm{B}}}{\mathrm{T}_{\mathrm{A}}}$ $\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\frac{\lambda_{\mathrm{B}}}{\lambda_{\mathrm{A}}}=\frac{480}{360}=\frac{4}{3}$ $\mathrm{T}_{\mathrm{A}}: \mathrm{T}_{\mathrm{B}}=4: 3$
D SI unit for Stefan's constant is Radiant energy $=\sigma e \mathrm{AT}^{4}$ $\sigma=\frac{\text { Radiated energy }}{\mathrm{e} \mathrm{A} \mathrm{T}^{4}}$ $=\frac{\mathrm{Watt}(\mathrm{W})}{\mathrm{m}^{2} \mathrm{~K}^{4}}$ SI unit of $\sigma=\mathrm{Wm}^{-2} \mathrm{~K}^{-4}=\mathrm{Js}^{-1} \mathrm{~m}^{-2} \mathrm{~K}^{-4}=\mathrm{J} / \mathrm{m}^{2} \mathrm{sK}^{4}$ $=\frac{\mathrm{kg} \cdot \mathrm{m}^{2}}{\mathrm{~s}^{2}} \times \frac{1}{\mathrm{~s}} \times \mathrm{m}^{-2} \mathrm{~T}^{-4}$ $=\mathrm{kg} \cdot \mathrm{s}^{-3} \mathrm{~m}^{0} \mathrm{~T}^{-4}$ Dimension of Stefan's Constant $\sigma=\mathrm{ML}^{\circ} \mathrm{T}^{-3} \mathrm{~K}^{-4}$
MHT-CET 2019
Heat Transfer
149525
Heat energy is incident on the surface at the rate of $1000 \mathrm{~J} / \mathrm{min}$. 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 $100 \mathrm{~J}$
2 $500 \mathrm{~J}$
3 $700 \mathrm{~J}$
4 $900 \mathrm{~J}$
Explanation:
B Given that, Rate of heat incident on surface $=1000 \mathrm{~J} / \mathrm{min}$ Coefficient absorption $\quad=0.8$ Coefficient of reflection $=0.1$ $\therefore$ Energy absorbed $=\text { Rate of heat incident } x$ $\text { coefficient of absorption }$ $=1000 \times 0.8$ $=800 \mathrm{~J} / \mathrm{min}$ Energy reflected $=$ Rate of heat incident $\times$ coefficient of reflection $=1000 \times 0.1=100 \mathrm{~J} / \mathrm{min}$ If energy transmitted is $\mathrm{E}$ so, energy incident $=$ energy absorbed + energy reflected + energy transmitted $1000=800+100+E$ $E=100 \mathrm{~J} / \mathrm{min}$ Thus, energy transmitted in $5 \mathrm{~min}$ $\mathrm{E}=100 \times 5$ $\mathrm{E}=500 \mathrm{~J}$
MHT-CET 2018
Heat Transfer
149526
A black body radiates heat at temperatures ' $T_{1}$ ' and ' $T_{2}, \quad\left(T_{2}>T_{1}\right)$. 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 $\lambda \mathrm{T}=\text { constant }$ $\lambda \propto \frac{1}{\mathrm{~T}}$ where $\lambda=$ wavelength $\mathrm{T}=$ Temperature $\therefore \quad \lambda \propto \frac{1}{\mathrm{n}}$ where $\mathrm{n}=$ frequency Thus $\mathrm{n} \propto \mathrm{T}$ If $\mathrm{T}_{2}>\mathrm{T}_{1}$ then $\mathrm{n}_{2}>\mathrm{n}_{1}$ Thus corresponding frequency is more at $T_{2}$.
MHT-CET 2015
Heat Transfer
149527
If $150 \mathrm{~J}$ of energy is incident on area $2 \mathrm{~m}^{2}$. If $Q_{\mathrm{r}}$ $=15 \mathrm{~J}$, coefficient of absorption is 0.6 , then amount of energy transmitted is
1 $50 \mathrm{~J}$
2 $45 \mathrm{~J}$
3 $40 \mathrm{~J}$
4 $30 \mathrm{~J}$
Explanation:
B Given that, Incident energy $=150 \mathrm{~J}$ Reflected energy $=15 \mathrm{~J}$ Coefficient of absorption $=0.6$ $\therefore$ Incident energy $=$ absorbed energy + reflected energy + energy transmitted $150=Q_{a}+Q_{r}+Q_{t}$ $\therefore \quad \mathrm{Q}_{\mathrm{a}}=\mathrm{Q} \times \mathrm{a}$ $=$ Incident energy $\times$ coefficient of absorption So, coefficient of reflected energy $r=\frac{Q_{r}}{Q}$ and coefficient of transmitted energy $t=\frac{Q_{t}}{Q}$ $\mathrm{Q}=\mathrm{Q}_{\mathrm{a}}+\mathrm{Q}_{\mathrm{r}}+\mathrm{Q}_{\mathrm{t}}$ $1=\frac{Q_{a}}{Q}+\frac{Q_{r}}{Q}+\frac{Q_{t}}{Q}$ $1=\mathrm{a}+\mathrm{r}+\mathrm{t}$ $1=0.6+\frac{15}{150}+\mathrm{t}$ $1=0.6+0.1+\mathrm{t}$ $1=0.7+\mathrm{t}$ $\mathrm{t}=1-0.7=0.3$ $\mathrm{t}=0.3$ $\therefore \quad \mathrm{t}=\frac{\mathrm{Q}_{\mathrm{t}}}{\mathrm{Q}}=0.3$ $\mathrm{Q}_{\mathrm{t}}=0.3 \times \mathrm{Q}$ $\mathrm{Q}_{\mathrm{t}}=0.3 \times 150$ $\mathrm{Q}_{\mathrm{t}}=45 \mathrm{~J}$
MHT-CET 2009
Heat Transfer
149529
Two stars $A$ and $B$ radiate maximum energy at the wavelength of $360 \mathrm{~nm}$ and $480 \mathrm{~nm}$ 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, $\lambda_{\mathrm{A}}=360 \mathrm{~nm}$ Wavelength of star $\mathrm{B}, \lambda_{\mathrm{B}}=480 \mathrm{~nm}$ Surface temperature of star $A=T_{A}$ Surface temperature of star $B=T_{B}$ According to Wein's displacement law, $\lambda \mathrm{T}=$ constant or $\lambda \propto \frac{1}{\mathrm{~T}}$ $\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{\mathrm{T}_{\mathrm{B}}}{\mathrm{T}_{\mathrm{A}}}$ $\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\frac{\lambda_{\mathrm{B}}}{\lambda_{\mathrm{A}}}=\frac{480}{360}=\frac{4}{3}$ $\mathrm{T}_{\mathrm{A}}: \mathrm{T}_{\mathrm{B}}=4: 3$
D SI unit for Stefan's constant is Radiant energy $=\sigma e \mathrm{AT}^{4}$ $\sigma=\frac{\text { Radiated energy }}{\mathrm{e} \mathrm{A} \mathrm{T}^{4}}$ $=\frac{\mathrm{Watt}(\mathrm{W})}{\mathrm{m}^{2} \mathrm{~K}^{4}}$ SI unit of $\sigma=\mathrm{Wm}^{-2} \mathrm{~K}^{-4}=\mathrm{Js}^{-1} \mathrm{~m}^{-2} \mathrm{~K}^{-4}=\mathrm{J} / \mathrm{m}^{2} \mathrm{sK}^{4}$ $=\frac{\mathrm{kg} \cdot \mathrm{m}^{2}}{\mathrm{~s}^{2}} \times \frac{1}{\mathrm{~s}} \times \mathrm{m}^{-2} \mathrm{~T}^{-4}$ $=\mathrm{kg} \cdot \mathrm{s}^{-3} \mathrm{~m}^{0} \mathrm{~T}^{-4}$ Dimension of Stefan's Constant $\sigma=\mathrm{ML}^{\circ} \mathrm{T}^{-3} \mathrm{~K}^{-4}$
MHT-CET 2019
Heat Transfer
149525
Heat energy is incident on the surface at the rate of $1000 \mathrm{~J} / \mathrm{min}$. 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 $100 \mathrm{~J}$
2 $500 \mathrm{~J}$
3 $700 \mathrm{~J}$
4 $900 \mathrm{~J}$
Explanation:
B Given that, Rate of heat incident on surface $=1000 \mathrm{~J} / \mathrm{min}$ Coefficient absorption $\quad=0.8$ Coefficient of reflection $=0.1$ $\therefore$ Energy absorbed $=\text { Rate of heat incident } x$ $\text { coefficient of absorption }$ $=1000 \times 0.8$ $=800 \mathrm{~J} / \mathrm{min}$ Energy reflected $=$ Rate of heat incident $\times$ coefficient of reflection $=1000 \times 0.1=100 \mathrm{~J} / \mathrm{min}$ If energy transmitted is $\mathrm{E}$ so, energy incident $=$ energy absorbed + energy reflected + energy transmitted $1000=800+100+E$ $E=100 \mathrm{~J} / \mathrm{min}$ Thus, energy transmitted in $5 \mathrm{~min}$ $\mathrm{E}=100 \times 5$ $\mathrm{E}=500 \mathrm{~J}$
MHT-CET 2018
Heat Transfer
149526
A black body radiates heat at temperatures ' $T_{1}$ ' and ' $T_{2}, \quad\left(T_{2}>T_{1}\right)$. 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 $\lambda \mathrm{T}=\text { constant }$ $\lambda \propto \frac{1}{\mathrm{~T}}$ where $\lambda=$ wavelength $\mathrm{T}=$ Temperature $\therefore \quad \lambda \propto \frac{1}{\mathrm{n}}$ where $\mathrm{n}=$ frequency Thus $\mathrm{n} \propto \mathrm{T}$ If $\mathrm{T}_{2}>\mathrm{T}_{1}$ then $\mathrm{n}_{2}>\mathrm{n}_{1}$ Thus corresponding frequency is more at $T_{2}$.
MHT-CET 2015
Heat Transfer
149527
If $150 \mathrm{~J}$ of energy is incident on area $2 \mathrm{~m}^{2}$. If $Q_{\mathrm{r}}$ $=15 \mathrm{~J}$, coefficient of absorption is 0.6 , then amount of energy transmitted is
1 $50 \mathrm{~J}$
2 $45 \mathrm{~J}$
3 $40 \mathrm{~J}$
4 $30 \mathrm{~J}$
Explanation:
B Given that, Incident energy $=150 \mathrm{~J}$ Reflected energy $=15 \mathrm{~J}$ Coefficient of absorption $=0.6$ $\therefore$ Incident energy $=$ absorbed energy + reflected energy + energy transmitted $150=Q_{a}+Q_{r}+Q_{t}$ $\therefore \quad \mathrm{Q}_{\mathrm{a}}=\mathrm{Q} \times \mathrm{a}$ $=$ Incident energy $\times$ coefficient of absorption So, coefficient of reflected energy $r=\frac{Q_{r}}{Q}$ and coefficient of transmitted energy $t=\frac{Q_{t}}{Q}$ $\mathrm{Q}=\mathrm{Q}_{\mathrm{a}}+\mathrm{Q}_{\mathrm{r}}+\mathrm{Q}_{\mathrm{t}}$ $1=\frac{Q_{a}}{Q}+\frac{Q_{r}}{Q}+\frac{Q_{t}}{Q}$ $1=\mathrm{a}+\mathrm{r}+\mathrm{t}$ $1=0.6+\frac{15}{150}+\mathrm{t}$ $1=0.6+0.1+\mathrm{t}$ $1=0.7+\mathrm{t}$ $\mathrm{t}=1-0.7=0.3$ $\mathrm{t}=0.3$ $\therefore \quad \mathrm{t}=\frac{\mathrm{Q}_{\mathrm{t}}}{\mathrm{Q}}=0.3$ $\mathrm{Q}_{\mathrm{t}}=0.3 \times \mathrm{Q}$ $\mathrm{Q}_{\mathrm{t}}=0.3 \times 150$ $\mathrm{Q}_{\mathrm{t}}=45 \mathrm{~J}$
MHT-CET 2009
Heat Transfer
149529
Two stars $A$ and $B$ radiate maximum energy at the wavelength of $360 \mathrm{~nm}$ and $480 \mathrm{~nm}$ 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, $\lambda_{\mathrm{A}}=360 \mathrm{~nm}$ Wavelength of star $\mathrm{B}, \lambda_{\mathrm{B}}=480 \mathrm{~nm}$ Surface temperature of star $A=T_{A}$ Surface temperature of star $B=T_{B}$ According to Wein's displacement law, $\lambda \mathrm{T}=$ constant or $\lambda \propto \frac{1}{\mathrm{~T}}$ $\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{\mathrm{T}_{\mathrm{B}}}{\mathrm{T}_{\mathrm{A}}}$ $\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\frac{\lambda_{\mathrm{B}}}{\lambda_{\mathrm{A}}}=\frac{480}{360}=\frac{4}{3}$ $\mathrm{T}_{\mathrm{A}}: \mathrm{T}_{\mathrm{B}}=4: 3$
D SI unit for Stefan's constant is Radiant energy $=\sigma e \mathrm{AT}^{4}$ $\sigma=\frac{\text { Radiated energy }}{\mathrm{e} \mathrm{A} \mathrm{T}^{4}}$ $=\frac{\mathrm{Watt}(\mathrm{W})}{\mathrm{m}^{2} \mathrm{~K}^{4}}$ SI unit of $\sigma=\mathrm{Wm}^{-2} \mathrm{~K}^{-4}=\mathrm{Js}^{-1} \mathrm{~m}^{-2} \mathrm{~K}^{-4}=\mathrm{J} / \mathrm{m}^{2} \mathrm{sK}^{4}$ $=\frac{\mathrm{kg} \cdot \mathrm{m}^{2}}{\mathrm{~s}^{2}} \times \frac{1}{\mathrm{~s}} \times \mathrm{m}^{-2} \mathrm{~T}^{-4}$ $=\mathrm{kg} \cdot \mathrm{s}^{-3} \mathrm{~m}^{0} \mathrm{~T}^{-4}$ Dimension of Stefan's Constant $\sigma=\mathrm{ML}^{\circ} \mathrm{T}^{-3} \mathrm{~K}^{-4}$
MHT-CET 2019
Heat Transfer
149525
Heat energy is incident on the surface at the rate of $1000 \mathrm{~J} / \mathrm{min}$. 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 $100 \mathrm{~J}$
2 $500 \mathrm{~J}$
3 $700 \mathrm{~J}$
4 $900 \mathrm{~J}$
Explanation:
B Given that, Rate of heat incident on surface $=1000 \mathrm{~J} / \mathrm{min}$ Coefficient absorption $\quad=0.8$ Coefficient of reflection $=0.1$ $\therefore$ Energy absorbed $=\text { Rate of heat incident } x$ $\text { coefficient of absorption }$ $=1000 \times 0.8$ $=800 \mathrm{~J} / \mathrm{min}$ Energy reflected $=$ Rate of heat incident $\times$ coefficient of reflection $=1000 \times 0.1=100 \mathrm{~J} / \mathrm{min}$ If energy transmitted is $\mathrm{E}$ so, energy incident $=$ energy absorbed + energy reflected + energy transmitted $1000=800+100+E$ $E=100 \mathrm{~J} / \mathrm{min}$ Thus, energy transmitted in $5 \mathrm{~min}$ $\mathrm{E}=100 \times 5$ $\mathrm{E}=500 \mathrm{~J}$
MHT-CET 2018
Heat Transfer
149526
A black body radiates heat at temperatures ' $T_{1}$ ' and ' $T_{2}, \quad\left(T_{2}>T_{1}\right)$. 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 $\lambda \mathrm{T}=\text { constant }$ $\lambda \propto \frac{1}{\mathrm{~T}}$ where $\lambda=$ wavelength $\mathrm{T}=$ Temperature $\therefore \quad \lambda \propto \frac{1}{\mathrm{n}}$ where $\mathrm{n}=$ frequency Thus $\mathrm{n} \propto \mathrm{T}$ If $\mathrm{T}_{2}>\mathrm{T}_{1}$ then $\mathrm{n}_{2}>\mathrm{n}_{1}$ Thus corresponding frequency is more at $T_{2}$.
MHT-CET 2015
Heat Transfer
149527
If $150 \mathrm{~J}$ of energy is incident on area $2 \mathrm{~m}^{2}$. If $Q_{\mathrm{r}}$ $=15 \mathrm{~J}$, coefficient of absorption is 0.6 , then amount of energy transmitted is
1 $50 \mathrm{~J}$
2 $45 \mathrm{~J}$
3 $40 \mathrm{~J}$
4 $30 \mathrm{~J}$
Explanation:
B Given that, Incident energy $=150 \mathrm{~J}$ Reflected energy $=15 \mathrm{~J}$ Coefficient of absorption $=0.6$ $\therefore$ Incident energy $=$ absorbed energy + reflected energy + energy transmitted $150=Q_{a}+Q_{r}+Q_{t}$ $\therefore \quad \mathrm{Q}_{\mathrm{a}}=\mathrm{Q} \times \mathrm{a}$ $=$ Incident energy $\times$ coefficient of absorption So, coefficient of reflected energy $r=\frac{Q_{r}}{Q}$ and coefficient of transmitted energy $t=\frac{Q_{t}}{Q}$ $\mathrm{Q}=\mathrm{Q}_{\mathrm{a}}+\mathrm{Q}_{\mathrm{r}}+\mathrm{Q}_{\mathrm{t}}$ $1=\frac{Q_{a}}{Q}+\frac{Q_{r}}{Q}+\frac{Q_{t}}{Q}$ $1=\mathrm{a}+\mathrm{r}+\mathrm{t}$ $1=0.6+\frac{15}{150}+\mathrm{t}$ $1=0.6+0.1+\mathrm{t}$ $1=0.7+\mathrm{t}$ $\mathrm{t}=1-0.7=0.3$ $\mathrm{t}=0.3$ $\therefore \quad \mathrm{t}=\frac{\mathrm{Q}_{\mathrm{t}}}{\mathrm{Q}}=0.3$ $\mathrm{Q}_{\mathrm{t}}=0.3 \times \mathrm{Q}$ $\mathrm{Q}_{\mathrm{t}}=0.3 \times 150$ $\mathrm{Q}_{\mathrm{t}}=45 \mathrm{~J}$
MHT-CET 2009
Heat Transfer
149529
Two stars $A$ and $B$ radiate maximum energy at the wavelength of $360 \mathrm{~nm}$ and $480 \mathrm{~nm}$ 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, $\lambda_{\mathrm{A}}=360 \mathrm{~nm}$ Wavelength of star $\mathrm{B}, \lambda_{\mathrm{B}}=480 \mathrm{~nm}$ Surface temperature of star $A=T_{A}$ Surface temperature of star $B=T_{B}$ According to Wein's displacement law, $\lambda \mathrm{T}=$ constant or $\lambda \propto \frac{1}{\mathrm{~T}}$ $\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{\mathrm{T}_{\mathrm{B}}}{\mathrm{T}_{\mathrm{A}}}$ $\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\frac{\lambda_{\mathrm{B}}}{\lambda_{\mathrm{A}}}=\frac{480}{360}=\frac{4}{3}$ $\mathrm{T}_{\mathrm{A}}: \mathrm{T}_{\mathrm{B}}=4: 3$