149565
A spherical black body of radius $12 \mathrm{~cm}$ radiates $450 \mathrm{~W}$ power at $500 \mathrm{~K}$. If the radius is one half and the temperature doubled, the power radiated in watt will be
1 225
2 450
3 900
4 1800
Explanation:
D Given, $\mathrm{r}_{1}=12 \mathrm{~cm}=0.12 \mathrm{~m}, \mathrm{r}_{2}=\frac{\mathrm{r}_{1}}{2}=0.06 \mathrm{~m}$ $\mathrm{p}_{1}=450 \mathrm{~W}, \mathrm{p}_{2}=?$ $\mathrm{~T}_{1}=500 \mathrm{~K}$ $\mathrm{~T}_{2}=500 \times 2=1000 \mathrm{~K}$ Rate of power loss from first body, $\mathrm{p}_{1}=\left(\sigma \mathrm{T}_{1}^{4}\right) 4 \pi \mathrm{r}_{1}^{2}$ Rate of power loss from second body, $\mathrm{p}_{2}=\left(\sigma \mathrm{T}_{2}^{4}\right) 4 \pi \mathrm{r}_{2}^{2}$ $\therefore \frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}= \frac{\mathrm{T}_{2}^{4} \mathrm{r}_{2}^{2}}{\mathrm{~T}_{1}^{4} \mathrm{r}_{1}^{2}}=\frac{(1000)^{4}}{(500)^{4}} \times \frac{(0.06)^{2}}{(0.12)^{2}}$ $=16 \times \frac{1}{4}=4$ $\mathrm{p}_{2} =4 \mathrm{p}_{1}=4 \times 450=1800 \mathrm{~W}$
SRMJEEE-2017
Heat Transfer
149566
Sun gives light at the rate of $1400 \mathrm{Wm}^{-2}$ of the area perpendicular to the direction of the light. Assume $\lambda$ (sunlight) $=6000 \AA$. Calculate the number of protons/sec arriving at $1 \mathrm{~m}^{2}$ area at that part of the earth
149569
At $127^{0} \mathrm{C}$, radiated energy is $2.7 \times 10^{-3} \mathrm{~J} / \mathrm{s}$. At what temperature radiated energy is $4.32 \times 10^{6} \mathrm{~J} / \mathrm{s}$.
1 $400 \mathrm{~K}$
2 $4000 \mathrm{~K}$
3 $80000 \mathrm{~K}$
4 $40000 \mathrm{~K}$
Explanation:
C Given, temperature $\left(\mathrm{T}_{1}\right)=127^{\circ} \mathrm{C}=127+$ $273=400 \mathrm{~K}$ Radiated energy at $400 \mathrm{~K}\left(\mathrm{E}_{1}\right)=2.7 \times 10^{-3}$ Radiated energy at $\mathrm{T}_{2}\left(\mathrm{E}_{2}\right)=4.32 \times 10^{6} \mathrm{~J} / \mathrm{s}$ Now, Energy radiated from a body, $\mathrm{E}=\mathrm{A} \sigma \mathrm{T}^{4} \varepsilon$ $E \propto T^{4}$ Therefore, $\frac{\mathrm{E}_{1}}{\mathrm{~T}_{1}^{4}}=\frac{\mathrm{E}_{2}}{\mathrm{~T}_{2}^{4}}$ $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{4}=\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}$ $\left(\frac{\mathrm{T}_{2}}{400}\right)^{4}=\frac{4.32 \times 10^{6}}{2.7 \times 10^{-3}}$ $\frac{\mathrm{T}_{2}}{400}=\left(\frac{4.32 \times 10^{6}}{2.7 \times 10^{-3}}\right)^{1 / 4}$ $\frac{\mathrm{T}_{2}}{400}=\left(1.6 \times 10^{9}\right)^{1 / 4}$ $\frac{\mathrm{T}_{2}}{400}=\left(16 \times 10^{8}\right)^{1 / 4}$ $\frac{\mathrm{T}_{2}}{400}=200$ $\mathrm{T}_{2}=200 \times 400$ $\mathrm{T}_{2}=80000 \mathrm{~K}$
BCECE-2004
Heat Transfer
149570
The wavelength of maximum energy, released during an atomic explosion was $2.93 \times 10^{-10} \mathrm{~m}$. Given that the Wien's constant is $2.93 \times 10^{-3} \mathrm{~m}-\mathrm{K}$ the maximum temperature attained must be of the order of :
1 $10^{-7} \mathrm{~K}$
2 $10^{7} \mathrm{~K}$
3 $10^{-3} \mathrm{~K}$
4 $5.86 \times 10^{7} \mathrm{~K}$
Explanation:
B Given, Wavelength of maximum energy $=2.93 \times 10^{-10} \mathrm{~m}$ Wien's Constant $=2.93 \times 10^{-3}$ From Wien's law $\lambda_{\mathrm{m}} \cdot \mathrm{T}=$ Constant $\mathrm{T}=\frac{\text { Constant }}{\lambda_{\mathrm{m}}}$ $\mathrm{T}=\frac{2.93 \times 10^{-3}}{2.93 \times 10^{-10}}$ $\mathrm{T}=10^{7} \mathrm{~K}$
BCECE-2003
Heat Transfer
149571
The surface of a black body is at a temperature $727^{\circ} \mathrm{C}$ and its cross section is $1 \mathrm{~m}^{2}$. Heat radiated from this surface in one minute in Joules is (Stefan's constant $=5.7 \times 10^{-8} \mathrm{Wm}^{-2} \mathrm{~K}^{-}$ 4)
1 $34.2 \times 10^{5}$
2 $2.5 \times 10^{5}$
3 $3.42 \times 10^{5}$
4 $2.5 \times 10^{6}$
Explanation:
A Given, Surface temperature of a black body $(\mathrm{T})=727^{\circ} \mathrm{C}$ $=727+273=1000 \mathrm{~K}$ Cross Section area $(\mathrm{A})=1 \mathrm{~m}^{2}$ Stefan's Constant $(\sigma)=5.7 \times 10^{-8} \mathrm{Wm}^{-2} \mathrm{~K}^{-4}$ We know that, $\mathrm{E}=\sigma \mathrm{T}^{4} \mathrm{~A} \varepsilon$ $\mathrm{E}=5.7 \times 10^{-8} \times(1000)^{4} \times 1 \times 1 \quad[\because \varepsilon=1]$ $\mathrm{E}=57 \times 10^{3} \mathrm{~W}$ Now, question is asked heat radiated from this surface in one minute in Joules $\mathrm{E}=57 \times 10^{3} \mathrm{~J} / \mathrm{sec} \times 60 \mathrm{~min}$ $\mathrm{E}=34.2 \times 10^{5}$ Joules
149565
A spherical black body of radius $12 \mathrm{~cm}$ radiates $450 \mathrm{~W}$ power at $500 \mathrm{~K}$. If the radius is one half and the temperature doubled, the power radiated in watt will be
1 225
2 450
3 900
4 1800
Explanation:
D Given, $\mathrm{r}_{1}=12 \mathrm{~cm}=0.12 \mathrm{~m}, \mathrm{r}_{2}=\frac{\mathrm{r}_{1}}{2}=0.06 \mathrm{~m}$ $\mathrm{p}_{1}=450 \mathrm{~W}, \mathrm{p}_{2}=?$ $\mathrm{~T}_{1}=500 \mathrm{~K}$ $\mathrm{~T}_{2}=500 \times 2=1000 \mathrm{~K}$ Rate of power loss from first body, $\mathrm{p}_{1}=\left(\sigma \mathrm{T}_{1}^{4}\right) 4 \pi \mathrm{r}_{1}^{2}$ Rate of power loss from second body, $\mathrm{p}_{2}=\left(\sigma \mathrm{T}_{2}^{4}\right) 4 \pi \mathrm{r}_{2}^{2}$ $\therefore \frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}= \frac{\mathrm{T}_{2}^{4} \mathrm{r}_{2}^{2}}{\mathrm{~T}_{1}^{4} \mathrm{r}_{1}^{2}}=\frac{(1000)^{4}}{(500)^{4}} \times \frac{(0.06)^{2}}{(0.12)^{2}}$ $=16 \times \frac{1}{4}=4$ $\mathrm{p}_{2} =4 \mathrm{p}_{1}=4 \times 450=1800 \mathrm{~W}$
SRMJEEE-2017
Heat Transfer
149566
Sun gives light at the rate of $1400 \mathrm{Wm}^{-2}$ of the area perpendicular to the direction of the light. Assume $\lambda$ (sunlight) $=6000 \AA$. Calculate the number of protons/sec arriving at $1 \mathrm{~m}^{2}$ area at that part of the earth
149569
At $127^{0} \mathrm{C}$, radiated energy is $2.7 \times 10^{-3} \mathrm{~J} / \mathrm{s}$. At what temperature radiated energy is $4.32 \times 10^{6} \mathrm{~J} / \mathrm{s}$.
1 $400 \mathrm{~K}$
2 $4000 \mathrm{~K}$
3 $80000 \mathrm{~K}$
4 $40000 \mathrm{~K}$
Explanation:
C Given, temperature $\left(\mathrm{T}_{1}\right)=127^{\circ} \mathrm{C}=127+$ $273=400 \mathrm{~K}$ Radiated energy at $400 \mathrm{~K}\left(\mathrm{E}_{1}\right)=2.7 \times 10^{-3}$ Radiated energy at $\mathrm{T}_{2}\left(\mathrm{E}_{2}\right)=4.32 \times 10^{6} \mathrm{~J} / \mathrm{s}$ Now, Energy radiated from a body, $\mathrm{E}=\mathrm{A} \sigma \mathrm{T}^{4} \varepsilon$ $E \propto T^{4}$ Therefore, $\frac{\mathrm{E}_{1}}{\mathrm{~T}_{1}^{4}}=\frac{\mathrm{E}_{2}}{\mathrm{~T}_{2}^{4}}$ $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{4}=\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}$ $\left(\frac{\mathrm{T}_{2}}{400}\right)^{4}=\frac{4.32 \times 10^{6}}{2.7 \times 10^{-3}}$ $\frac{\mathrm{T}_{2}}{400}=\left(\frac{4.32 \times 10^{6}}{2.7 \times 10^{-3}}\right)^{1 / 4}$ $\frac{\mathrm{T}_{2}}{400}=\left(1.6 \times 10^{9}\right)^{1 / 4}$ $\frac{\mathrm{T}_{2}}{400}=\left(16 \times 10^{8}\right)^{1 / 4}$ $\frac{\mathrm{T}_{2}}{400}=200$ $\mathrm{T}_{2}=200 \times 400$ $\mathrm{T}_{2}=80000 \mathrm{~K}$
BCECE-2004
Heat Transfer
149570
The wavelength of maximum energy, released during an atomic explosion was $2.93 \times 10^{-10} \mathrm{~m}$. Given that the Wien's constant is $2.93 \times 10^{-3} \mathrm{~m}-\mathrm{K}$ the maximum temperature attained must be of the order of :
1 $10^{-7} \mathrm{~K}$
2 $10^{7} \mathrm{~K}$
3 $10^{-3} \mathrm{~K}$
4 $5.86 \times 10^{7} \mathrm{~K}$
Explanation:
B Given, Wavelength of maximum energy $=2.93 \times 10^{-10} \mathrm{~m}$ Wien's Constant $=2.93 \times 10^{-3}$ From Wien's law $\lambda_{\mathrm{m}} \cdot \mathrm{T}=$ Constant $\mathrm{T}=\frac{\text { Constant }}{\lambda_{\mathrm{m}}}$ $\mathrm{T}=\frac{2.93 \times 10^{-3}}{2.93 \times 10^{-10}}$ $\mathrm{T}=10^{7} \mathrm{~K}$
BCECE-2003
Heat Transfer
149571
The surface of a black body is at a temperature $727^{\circ} \mathrm{C}$ and its cross section is $1 \mathrm{~m}^{2}$. Heat radiated from this surface in one minute in Joules is (Stefan's constant $=5.7 \times 10^{-8} \mathrm{Wm}^{-2} \mathrm{~K}^{-}$ 4)
1 $34.2 \times 10^{5}$
2 $2.5 \times 10^{5}$
3 $3.42 \times 10^{5}$
4 $2.5 \times 10^{6}$
Explanation:
A Given, Surface temperature of a black body $(\mathrm{T})=727^{\circ} \mathrm{C}$ $=727+273=1000 \mathrm{~K}$ Cross Section area $(\mathrm{A})=1 \mathrm{~m}^{2}$ Stefan's Constant $(\sigma)=5.7 \times 10^{-8} \mathrm{Wm}^{-2} \mathrm{~K}^{-4}$ We know that, $\mathrm{E}=\sigma \mathrm{T}^{4} \mathrm{~A} \varepsilon$ $\mathrm{E}=5.7 \times 10^{-8} \times(1000)^{4} \times 1 \times 1 \quad[\because \varepsilon=1]$ $\mathrm{E}=57 \times 10^{3} \mathrm{~W}$ Now, question is asked heat radiated from this surface in one minute in Joules $\mathrm{E}=57 \times 10^{3} \mathrm{~J} / \mathrm{sec} \times 60 \mathrm{~min}$ $\mathrm{E}=34.2 \times 10^{5}$ Joules
149565
A spherical black body of radius $12 \mathrm{~cm}$ radiates $450 \mathrm{~W}$ power at $500 \mathrm{~K}$. If the radius is one half and the temperature doubled, the power radiated in watt will be
1 225
2 450
3 900
4 1800
Explanation:
D Given, $\mathrm{r}_{1}=12 \mathrm{~cm}=0.12 \mathrm{~m}, \mathrm{r}_{2}=\frac{\mathrm{r}_{1}}{2}=0.06 \mathrm{~m}$ $\mathrm{p}_{1}=450 \mathrm{~W}, \mathrm{p}_{2}=?$ $\mathrm{~T}_{1}=500 \mathrm{~K}$ $\mathrm{~T}_{2}=500 \times 2=1000 \mathrm{~K}$ Rate of power loss from first body, $\mathrm{p}_{1}=\left(\sigma \mathrm{T}_{1}^{4}\right) 4 \pi \mathrm{r}_{1}^{2}$ Rate of power loss from second body, $\mathrm{p}_{2}=\left(\sigma \mathrm{T}_{2}^{4}\right) 4 \pi \mathrm{r}_{2}^{2}$ $\therefore \frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}= \frac{\mathrm{T}_{2}^{4} \mathrm{r}_{2}^{2}}{\mathrm{~T}_{1}^{4} \mathrm{r}_{1}^{2}}=\frac{(1000)^{4}}{(500)^{4}} \times \frac{(0.06)^{2}}{(0.12)^{2}}$ $=16 \times \frac{1}{4}=4$ $\mathrm{p}_{2} =4 \mathrm{p}_{1}=4 \times 450=1800 \mathrm{~W}$
SRMJEEE-2017
Heat Transfer
149566
Sun gives light at the rate of $1400 \mathrm{Wm}^{-2}$ of the area perpendicular to the direction of the light. Assume $\lambda$ (sunlight) $=6000 \AA$. Calculate the number of protons/sec arriving at $1 \mathrm{~m}^{2}$ area at that part of the earth
149569
At $127^{0} \mathrm{C}$, radiated energy is $2.7 \times 10^{-3} \mathrm{~J} / \mathrm{s}$. At what temperature radiated energy is $4.32 \times 10^{6} \mathrm{~J} / \mathrm{s}$.
1 $400 \mathrm{~K}$
2 $4000 \mathrm{~K}$
3 $80000 \mathrm{~K}$
4 $40000 \mathrm{~K}$
Explanation:
C Given, temperature $\left(\mathrm{T}_{1}\right)=127^{\circ} \mathrm{C}=127+$ $273=400 \mathrm{~K}$ Radiated energy at $400 \mathrm{~K}\left(\mathrm{E}_{1}\right)=2.7 \times 10^{-3}$ Radiated energy at $\mathrm{T}_{2}\left(\mathrm{E}_{2}\right)=4.32 \times 10^{6} \mathrm{~J} / \mathrm{s}$ Now, Energy radiated from a body, $\mathrm{E}=\mathrm{A} \sigma \mathrm{T}^{4} \varepsilon$ $E \propto T^{4}$ Therefore, $\frac{\mathrm{E}_{1}}{\mathrm{~T}_{1}^{4}}=\frac{\mathrm{E}_{2}}{\mathrm{~T}_{2}^{4}}$ $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{4}=\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}$ $\left(\frac{\mathrm{T}_{2}}{400}\right)^{4}=\frac{4.32 \times 10^{6}}{2.7 \times 10^{-3}}$ $\frac{\mathrm{T}_{2}}{400}=\left(\frac{4.32 \times 10^{6}}{2.7 \times 10^{-3}}\right)^{1 / 4}$ $\frac{\mathrm{T}_{2}}{400}=\left(1.6 \times 10^{9}\right)^{1 / 4}$ $\frac{\mathrm{T}_{2}}{400}=\left(16 \times 10^{8}\right)^{1 / 4}$ $\frac{\mathrm{T}_{2}}{400}=200$ $\mathrm{T}_{2}=200 \times 400$ $\mathrm{T}_{2}=80000 \mathrm{~K}$
BCECE-2004
Heat Transfer
149570
The wavelength of maximum energy, released during an atomic explosion was $2.93 \times 10^{-10} \mathrm{~m}$. Given that the Wien's constant is $2.93 \times 10^{-3} \mathrm{~m}-\mathrm{K}$ the maximum temperature attained must be of the order of :
1 $10^{-7} \mathrm{~K}$
2 $10^{7} \mathrm{~K}$
3 $10^{-3} \mathrm{~K}$
4 $5.86 \times 10^{7} \mathrm{~K}$
Explanation:
B Given, Wavelength of maximum energy $=2.93 \times 10^{-10} \mathrm{~m}$ Wien's Constant $=2.93 \times 10^{-3}$ From Wien's law $\lambda_{\mathrm{m}} \cdot \mathrm{T}=$ Constant $\mathrm{T}=\frac{\text { Constant }}{\lambda_{\mathrm{m}}}$ $\mathrm{T}=\frac{2.93 \times 10^{-3}}{2.93 \times 10^{-10}}$ $\mathrm{T}=10^{7} \mathrm{~K}$
BCECE-2003
Heat Transfer
149571
The surface of a black body is at a temperature $727^{\circ} \mathrm{C}$ and its cross section is $1 \mathrm{~m}^{2}$. Heat radiated from this surface in one minute in Joules is (Stefan's constant $=5.7 \times 10^{-8} \mathrm{Wm}^{-2} \mathrm{~K}^{-}$ 4)
1 $34.2 \times 10^{5}$
2 $2.5 \times 10^{5}$
3 $3.42 \times 10^{5}$
4 $2.5 \times 10^{6}$
Explanation:
A Given, Surface temperature of a black body $(\mathrm{T})=727^{\circ} \mathrm{C}$ $=727+273=1000 \mathrm{~K}$ Cross Section area $(\mathrm{A})=1 \mathrm{~m}^{2}$ Stefan's Constant $(\sigma)=5.7 \times 10^{-8} \mathrm{Wm}^{-2} \mathrm{~K}^{-4}$ We know that, $\mathrm{E}=\sigma \mathrm{T}^{4} \mathrm{~A} \varepsilon$ $\mathrm{E}=5.7 \times 10^{-8} \times(1000)^{4} \times 1 \times 1 \quad[\because \varepsilon=1]$ $\mathrm{E}=57 \times 10^{3} \mathrm{~W}$ Now, question is asked heat radiated from this surface in one minute in Joules $\mathrm{E}=57 \times 10^{3} \mathrm{~J} / \mathrm{sec} \times 60 \mathrm{~min}$ $\mathrm{E}=34.2 \times 10^{5}$ Joules
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Heat Transfer
149565
A spherical black body of radius $12 \mathrm{~cm}$ radiates $450 \mathrm{~W}$ power at $500 \mathrm{~K}$. If the radius is one half and the temperature doubled, the power radiated in watt will be
1 225
2 450
3 900
4 1800
Explanation:
D Given, $\mathrm{r}_{1}=12 \mathrm{~cm}=0.12 \mathrm{~m}, \mathrm{r}_{2}=\frac{\mathrm{r}_{1}}{2}=0.06 \mathrm{~m}$ $\mathrm{p}_{1}=450 \mathrm{~W}, \mathrm{p}_{2}=?$ $\mathrm{~T}_{1}=500 \mathrm{~K}$ $\mathrm{~T}_{2}=500 \times 2=1000 \mathrm{~K}$ Rate of power loss from first body, $\mathrm{p}_{1}=\left(\sigma \mathrm{T}_{1}^{4}\right) 4 \pi \mathrm{r}_{1}^{2}$ Rate of power loss from second body, $\mathrm{p}_{2}=\left(\sigma \mathrm{T}_{2}^{4}\right) 4 \pi \mathrm{r}_{2}^{2}$ $\therefore \frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}= \frac{\mathrm{T}_{2}^{4} \mathrm{r}_{2}^{2}}{\mathrm{~T}_{1}^{4} \mathrm{r}_{1}^{2}}=\frac{(1000)^{4}}{(500)^{4}} \times \frac{(0.06)^{2}}{(0.12)^{2}}$ $=16 \times \frac{1}{4}=4$ $\mathrm{p}_{2} =4 \mathrm{p}_{1}=4 \times 450=1800 \mathrm{~W}$
SRMJEEE-2017
Heat Transfer
149566
Sun gives light at the rate of $1400 \mathrm{Wm}^{-2}$ of the area perpendicular to the direction of the light. Assume $\lambda$ (sunlight) $=6000 \AA$. Calculate the number of protons/sec arriving at $1 \mathrm{~m}^{2}$ area at that part of the earth
149569
At $127^{0} \mathrm{C}$, radiated energy is $2.7 \times 10^{-3} \mathrm{~J} / \mathrm{s}$. At what temperature radiated energy is $4.32 \times 10^{6} \mathrm{~J} / \mathrm{s}$.
1 $400 \mathrm{~K}$
2 $4000 \mathrm{~K}$
3 $80000 \mathrm{~K}$
4 $40000 \mathrm{~K}$
Explanation:
C Given, temperature $\left(\mathrm{T}_{1}\right)=127^{\circ} \mathrm{C}=127+$ $273=400 \mathrm{~K}$ Radiated energy at $400 \mathrm{~K}\left(\mathrm{E}_{1}\right)=2.7 \times 10^{-3}$ Radiated energy at $\mathrm{T}_{2}\left(\mathrm{E}_{2}\right)=4.32 \times 10^{6} \mathrm{~J} / \mathrm{s}$ Now, Energy radiated from a body, $\mathrm{E}=\mathrm{A} \sigma \mathrm{T}^{4} \varepsilon$ $E \propto T^{4}$ Therefore, $\frac{\mathrm{E}_{1}}{\mathrm{~T}_{1}^{4}}=\frac{\mathrm{E}_{2}}{\mathrm{~T}_{2}^{4}}$ $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{4}=\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}$ $\left(\frac{\mathrm{T}_{2}}{400}\right)^{4}=\frac{4.32 \times 10^{6}}{2.7 \times 10^{-3}}$ $\frac{\mathrm{T}_{2}}{400}=\left(\frac{4.32 \times 10^{6}}{2.7 \times 10^{-3}}\right)^{1 / 4}$ $\frac{\mathrm{T}_{2}}{400}=\left(1.6 \times 10^{9}\right)^{1 / 4}$ $\frac{\mathrm{T}_{2}}{400}=\left(16 \times 10^{8}\right)^{1 / 4}$ $\frac{\mathrm{T}_{2}}{400}=200$ $\mathrm{T}_{2}=200 \times 400$ $\mathrm{T}_{2}=80000 \mathrm{~K}$
BCECE-2004
Heat Transfer
149570
The wavelength of maximum energy, released during an atomic explosion was $2.93 \times 10^{-10} \mathrm{~m}$. Given that the Wien's constant is $2.93 \times 10^{-3} \mathrm{~m}-\mathrm{K}$ the maximum temperature attained must be of the order of :
1 $10^{-7} \mathrm{~K}$
2 $10^{7} \mathrm{~K}$
3 $10^{-3} \mathrm{~K}$
4 $5.86 \times 10^{7} \mathrm{~K}$
Explanation:
B Given, Wavelength of maximum energy $=2.93 \times 10^{-10} \mathrm{~m}$ Wien's Constant $=2.93 \times 10^{-3}$ From Wien's law $\lambda_{\mathrm{m}} \cdot \mathrm{T}=$ Constant $\mathrm{T}=\frac{\text { Constant }}{\lambda_{\mathrm{m}}}$ $\mathrm{T}=\frac{2.93 \times 10^{-3}}{2.93 \times 10^{-10}}$ $\mathrm{T}=10^{7} \mathrm{~K}$
BCECE-2003
Heat Transfer
149571
The surface of a black body is at a temperature $727^{\circ} \mathrm{C}$ and its cross section is $1 \mathrm{~m}^{2}$. Heat radiated from this surface in one minute in Joules is (Stefan's constant $=5.7 \times 10^{-8} \mathrm{Wm}^{-2} \mathrm{~K}^{-}$ 4)
1 $34.2 \times 10^{5}$
2 $2.5 \times 10^{5}$
3 $3.42 \times 10^{5}$
4 $2.5 \times 10^{6}$
Explanation:
A Given, Surface temperature of a black body $(\mathrm{T})=727^{\circ} \mathrm{C}$ $=727+273=1000 \mathrm{~K}$ Cross Section area $(\mathrm{A})=1 \mathrm{~m}^{2}$ Stefan's Constant $(\sigma)=5.7 \times 10^{-8} \mathrm{Wm}^{-2} \mathrm{~K}^{-4}$ We know that, $\mathrm{E}=\sigma \mathrm{T}^{4} \mathrm{~A} \varepsilon$ $\mathrm{E}=5.7 \times 10^{-8} \times(1000)^{4} \times 1 \times 1 \quad[\because \varepsilon=1]$ $\mathrm{E}=57 \times 10^{3} \mathrm{~W}$ Now, question is asked heat radiated from this surface in one minute in Joules $\mathrm{E}=57 \times 10^{3} \mathrm{~J} / \mathrm{sec} \times 60 \mathrm{~min}$ $\mathrm{E}=34.2 \times 10^{5}$ Joules
149565
A spherical black body of radius $12 \mathrm{~cm}$ radiates $450 \mathrm{~W}$ power at $500 \mathrm{~K}$. If the radius is one half and the temperature doubled, the power radiated in watt will be
1 225
2 450
3 900
4 1800
Explanation:
D Given, $\mathrm{r}_{1}=12 \mathrm{~cm}=0.12 \mathrm{~m}, \mathrm{r}_{2}=\frac{\mathrm{r}_{1}}{2}=0.06 \mathrm{~m}$ $\mathrm{p}_{1}=450 \mathrm{~W}, \mathrm{p}_{2}=?$ $\mathrm{~T}_{1}=500 \mathrm{~K}$ $\mathrm{~T}_{2}=500 \times 2=1000 \mathrm{~K}$ Rate of power loss from first body, $\mathrm{p}_{1}=\left(\sigma \mathrm{T}_{1}^{4}\right) 4 \pi \mathrm{r}_{1}^{2}$ Rate of power loss from second body, $\mathrm{p}_{2}=\left(\sigma \mathrm{T}_{2}^{4}\right) 4 \pi \mathrm{r}_{2}^{2}$ $\therefore \frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}= \frac{\mathrm{T}_{2}^{4} \mathrm{r}_{2}^{2}}{\mathrm{~T}_{1}^{4} \mathrm{r}_{1}^{2}}=\frac{(1000)^{4}}{(500)^{4}} \times \frac{(0.06)^{2}}{(0.12)^{2}}$ $=16 \times \frac{1}{4}=4$ $\mathrm{p}_{2} =4 \mathrm{p}_{1}=4 \times 450=1800 \mathrm{~W}$
SRMJEEE-2017
Heat Transfer
149566
Sun gives light at the rate of $1400 \mathrm{Wm}^{-2}$ of the area perpendicular to the direction of the light. Assume $\lambda$ (sunlight) $=6000 \AA$. Calculate the number of protons/sec arriving at $1 \mathrm{~m}^{2}$ area at that part of the earth
149569
At $127^{0} \mathrm{C}$, radiated energy is $2.7 \times 10^{-3} \mathrm{~J} / \mathrm{s}$. At what temperature radiated energy is $4.32 \times 10^{6} \mathrm{~J} / \mathrm{s}$.
1 $400 \mathrm{~K}$
2 $4000 \mathrm{~K}$
3 $80000 \mathrm{~K}$
4 $40000 \mathrm{~K}$
Explanation:
C Given, temperature $\left(\mathrm{T}_{1}\right)=127^{\circ} \mathrm{C}=127+$ $273=400 \mathrm{~K}$ Radiated energy at $400 \mathrm{~K}\left(\mathrm{E}_{1}\right)=2.7 \times 10^{-3}$ Radiated energy at $\mathrm{T}_{2}\left(\mathrm{E}_{2}\right)=4.32 \times 10^{6} \mathrm{~J} / \mathrm{s}$ Now, Energy radiated from a body, $\mathrm{E}=\mathrm{A} \sigma \mathrm{T}^{4} \varepsilon$ $E \propto T^{4}$ Therefore, $\frac{\mathrm{E}_{1}}{\mathrm{~T}_{1}^{4}}=\frac{\mathrm{E}_{2}}{\mathrm{~T}_{2}^{4}}$ $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{4}=\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}$ $\left(\frac{\mathrm{T}_{2}}{400}\right)^{4}=\frac{4.32 \times 10^{6}}{2.7 \times 10^{-3}}$ $\frac{\mathrm{T}_{2}}{400}=\left(\frac{4.32 \times 10^{6}}{2.7 \times 10^{-3}}\right)^{1 / 4}$ $\frac{\mathrm{T}_{2}}{400}=\left(1.6 \times 10^{9}\right)^{1 / 4}$ $\frac{\mathrm{T}_{2}}{400}=\left(16 \times 10^{8}\right)^{1 / 4}$ $\frac{\mathrm{T}_{2}}{400}=200$ $\mathrm{T}_{2}=200 \times 400$ $\mathrm{T}_{2}=80000 \mathrm{~K}$
BCECE-2004
Heat Transfer
149570
The wavelength of maximum energy, released during an atomic explosion was $2.93 \times 10^{-10} \mathrm{~m}$. Given that the Wien's constant is $2.93 \times 10^{-3} \mathrm{~m}-\mathrm{K}$ the maximum temperature attained must be of the order of :
1 $10^{-7} \mathrm{~K}$
2 $10^{7} \mathrm{~K}$
3 $10^{-3} \mathrm{~K}$
4 $5.86 \times 10^{7} \mathrm{~K}$
Explanation:
B Given, Wavelength of maximum energy $=2.93 \times 10^{-10} \mathrm{~m}$ Wien's Constant $=2.93 \times 10^{-3}$ From Wien's law $\lambda_{\mathrm{m}} \cdot \mathrm{T}=$ Constant $\mathrm{T}=\frac{\text { Constant }}{\lambda_{\mathrm{m}}}$ $\mathrm{T}=\frac{2.93 \times 10^{-3}}{2.93 \times 10^{-10}}$ $\mathrm{T}=10^{7} \mathrm{~K}$
BCECE-2003
Heat Transfer
149571
The surface of a black body is at a temperature $727^{\circ} \mathrm{C}$ and its cross section is $1 \mathrm{~m}^{2}$. Heat radiated from this surface in one minute in Joules is (Stefan's constant $=5.7 \times 10^{-8} \mathrm{Wm}^{-2} \mathrm{~K}^{-}$ 4)
1 $34.2 \times 10^{5}$
2 $2.5 \times 10^{5}$
3 $3.42 \times 10^{5}$
4 $2.5 \times 10^{6}$
Explanation:
A Given, Surface temperature of a black body $(\mathrm{T})=727^{\circ} \mathrm{C}$ $=727+273=1000 \mathrm{~K}$ Cross Section area $(\mathrm{A})=1 \mathrm{~m}^{2}$ Stefan's Constant $(\sigma)=5.7 \times 10^{-8} \mathrm{Wm}^{-2} \mathrm{~K}^{-4}$ We know that, $\mathrm{E}=\sigma \mathrm{T}^{4} \mathrm{~A} \varepsilon$ $\mathrm{E}=5.7 \times 10^{-8} \times(1000)^{4} \times 1 \times 1 \quad[\because \varepsilon=1]$ $\mathrm{E}=57 \times 10^{3} \mathrm{~W}$ Now, question is asked heat radiated from this surface in one minute in Joules $\mathrm{E}=57 \times 10^{3} \mathrm{~J} / \mathrm{sec} \times 60 \mathrm{~min}$ $\mathrm{E}=34.2 \times 10^{5}$ Joules
