149500
A black body is heated from $27^{\circ} \mathrm{C}$ to $127^{\circ} \mathrm{C}$. The ratio of their energies of radiation emitted will be:
1 $9: 16$
2 $27: 64$
3 $81: 256$
4 3:4
Explanation:
C We know that $\mathrm{E}=\sigma \mathrm{T}^{4}$ Where $\mathrm{E}$ is rate of emission of radiation of a body at temperature $\mathrm{T}$. $\mathrm{E}_{1}=\sigma(27+273)^{2}$ $\mathrm{E}_{2}=\sigma(127+273)^{2}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{(300)^{4}}{(400)^{4}}=\frac{81}{256}$
AIIMS-2001
Heat Transfer
149502
Calculate radiation power for sphere whose temperature is $227^{\circ} \mathrm{C}$ and radius $0.2 \mathrm{~m}$ and emissivity 0.8 .
1 $1425 \mathrm{~W}$
2 $1500 \mathrm{~W}$
3 $1255 \mathrm{~W}$
4 $1575 \mathrm{~W}$
Explanation:
A Given, Radius $=0.2 \mathrm{~m}$ And temperature $(\mathrm{T})=227+273=500 \mathrm{~K}$ Radiation power through area A is given as $\mathrm{P}=\sigma \mathrm{A} \varepsilon \mathrm{T}^{4}$ [Here $\varepsilon=$ emissivity $=0.8, \mathrm{~A}=$ surface area $=4 \pi \mathrm{r}^{2}$, $\left.\sigma=5.67 \times 10^{-8} \mathrm{w} / \mathrm{m}^{2} \mathrm{~K}^{4}\right]$ Putting these values we get $\mathrm{P}=5.67 \times 10^{-8} \times 4 \pi(0.2)^{2} \times(0.8) \times(500)^{4}$ $=1417.48 \simeq 1425 \mathrm{~W}$
AIIMS-25.05.2019(M) Shift-1
Heat Transfer
149503
Distance between sun and earth is $2 \times 10^{8} \mathrm{~km}$, temperature of sun $6000 \mathrm{~K}$, radius of sun $7 \times 10^{5}$ $\mathrm{km}$. If emissivity of earth is 0.6 , then find out temperature of earth in thermal equilibrium.
1 $400 \mathrm{~K}$
2 $300 \mathrm{~K}$
3 $500 \mathrm{~K}$
4 $600 \mathrm{~K}$
Explanation:
B Given, distance between sun and earth (d) $=2$ $\times 10^{8} \mathrm{~km}$, Temperature of sun $=6000 \mathrm{~K}$ and radius of sun $=7 \times 10^{5} \mathrm{~km}$. For thermal equilibrium, Energy received by earth = Energy emitted by earth $\frac{\mathrm{T}_{\mathrm{s}}^{4} 4 \pi \mathrm{R}_{\mathrm{s}}^{2}}{4 \pi \mathrm{d}^{2}} \times \pi \mathrm{R}_{\mathrm{e}}^{2}=\sigma \cdot \varepsilon \cdot \mathrm{T}_{2}^{4} \cdot 4 \pi \mathrm{R}_{\mathrm{e}}^{2}$ $\frac{\mathrm{T}_{\mathrm{s}}^{4} \times \mathrm{R}_{\mathrm{s}}^{2}}{4 \times \mathrm{d}^{2} \times \mathrm{e}}=\mathrm{T}_{\mathrm{e}}^{4}$ $\frac{(6000)^{4} \times\left(7 \times 10^{8}\right)^{2}}{4 \times\left(2 \times 10^{11}\right)^{2} \times 0.6}=\mathrm{T}_{\mathrm{e}}^{4}$ $\frac{36 \times 36 \times 7 \times 7}{4 \times 4 \times 0.6} \times 10^{6}=\mathrm{T}_{\mathrm{e}}^{4}$ $66.15 \times 108=\mathrm{T}_{\mathrm{e}}^{4}$ $\mathrm{T}_{\mathrm{e}}=285.19 \mathrm{~K} \simeq 300 \mathrm{~K}$
AIIMS-25.05.2019(E) Shift-2
Heat Transfer
149504
If a body coated black at $600 \mathrm{~K}$ surrounded by atmosphere at $300 \mathrm{~K}$ has cooling rate $r_{0}$, the same body at $900 \mathrm{~K}$, surrounded by the same atmosphere will have cooling rate equal to-
1 $\frac{16}{3} r_{0}$
2 $\frac{8}{16} r_{0}$
3 $16 r_{0}$
4 $4 \mathrm{r}_{0}$
Explanation:
A Given that, $\mathrm{T}_{1}=600 \mathrm{~K}$ $\mathrm{T}_{2}=900 \mathrm{~K}, \mathrm{~T}_{\mathrm{o}}=300 \mathrm{~K}$ We know that, Cooling rate $\propto \mathrm{T}^{4}-\mathrm{T}_{0}^{4}$ $\frac{\mathrm{r}}{\mathrm{r}_{\mathrm{o}}}=\left(\frac{\mathrm{T}_{2}^{4}-\mathrm{T}_{\mathrm{o}}^{4}}{\mathrm{~T}_{1}^{4}-\mathrm{T}_{\mathrm{o}}^{4}}\right)$ $\frac{\mathrm{r}}{\mathrm{r}_{\mathrm{o}}}=\left[\frac{(900)^{4}-(300)^{4}}{(600)^{4}-(300)^{4}}\right]$ $\frac{\mathrm{r}}{\mathrm{r}_{\mathrm{o}}}=\left(\frac{16}{3}\right)$ $\mathrm{r}=\frac{16}{3} \mathrm{r}_{\mathrm{o}}$
149500
A black body is heated from $27^{\circ} \mathrm{C}$ to $127^{\circ} \mathrm{C}$. The ratio of their energies of radiation emitted will be:
1 $9: 16$
2 $27: 64$
3 $81: 256$
4 3:4
Explanation:
C We know that $\mathrm{E}=\sigma \mathrm{T}^{4}$ Where $\mathrm{E}$ is rate of emission of radiation of a body at temperature $\mathrm{T}$. $\mathrm{E}_{1}=\sigma(27+273)^{2}$ $\mathrm{E}_{2}=\sigma(127+273)^{2}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{(300)^{4}}{(400)^{4}}=\frac{81}{256}$
AIIMS-2001
Heat Transfer
149502
Calculate radiation power for sphere whose temperature is $227^{\circ} \mathrm{C}$ and radius $0.2 \mathrm{~m}$ and emissivity 0.8 .
1 $1425 \mathrm{~W}$
2 $1500 \mathrm{~W}$
3 $1255 \mathrm{~W}$
4 $1575 \mathrm{~W}$
Explanation:
A Given, Radius $=0.2 \mathrm{~m}$ And temperature $(\mathrm{T})=227+273=500 \mathrm{~K}$ Radiation power through area A is given as $\mathrm{P}=\sigma \mathrm{A} \varepsilon \mathrm{T}^{4}$ [Here $\varepsilon=$ emissivity $=0.8, \mathrm{~A}=$ surface area $=4 \pi \mathrm{r}^{2}$, $\left.\sigma=5.67 \times 10^{-8} \mathrm{w} / \mathrm{m}^{2} \mathrm{~K}^{4}\right]$ Putting these values we get $\mathrm{P}=5.67 \times 10^{-8} \times 4 \pi(0.2)^{2} \times(0.8) \times(500)^{4}$ $=1417.48 \simeq 1425 \mathrm{~W}$
AIIMS-25.05.2019(M) Shift-1
Heat Transfer
149503
Distance between sun and earth is $2 \times 10^{8} \mathrm{~km}$, temperature of sun $6000 \mathrm{~K}$, radius of sun $7 \times 10^{5}$ $\mathrm{km}$. If emissivity of earth is 0.6 , then find out temperature of earth in thermal equilibrium.
1 $400 \mathrm{~K}$
2 $300 \mathrm{~K}$
3 $500 \mathrm{~K}$
4 $600 \mathrm{~K}$
Explanation:
B Given, distance between sun and earth (d) $=2$ $\times 10^{8} \mathrm{~km}$, Temperature of sun $=6000 \mathrm{~K}$ and radius of sun $=7 \times 10^{5} \mathrm{~km}$. For thermal equilibrium, Energy received by earth = Energy emitted by earth $\frac{\mathrm{T}_{\mathrm{s}}^{4} 4 \pi \mathrm{R}_{\mathrm{s}}^{2}}{4 \pi \mathrm{d}^{2}} \times \pi \mathrm{R}_{\mathrm{e}}^{2}=\sigma \cdot \varepsilon \cdot \mathrm{T}_{2}^{4} \cdot 4 \pi \mathrm{R}_{\mathrm{e}}^{2}$ $\frac{\mathrm{T}_{\mathrm{s}}^{4} \times \mathrm{R}_{\mathrm{s}}^{2}}{4 \times \mathrm{d}^{2} \times \mathrm{e}}=\mathrm{T}_{\mathrm{e}}^{4}$ $\frac{(6000)^{4} \times\left(7 \times 10^{8}\right)^{2}}{4 \times\left(2 \times 10^{11}\right)^{2} \times 0.6}=\mathrm{T}_{\mathrm{e}}^{4}$ $\frac{36 \times 36 \times 7 \times 7}{4 \times 4 \times 0.6} \times 10^{6}=\mathrm{T}_{\mathrm{e}}^{4}$ $66.15 \times 108=\mathrm{T}_{\mathrm{e}}^{4}$ $\mathrm{T}_{\mathrm{e}}=285.19 \mathrm{~K} \simeq 300 \mathrm{~K}$
AIIMS-25.05.2019(E) Shift-2
Heat Transfer
149504
If a body coated black at $600 \mathrm{~K}$ surrounded by atmosphere at $300 \mathrm{~K}$ has cooling rate $r_{0}$, the same body at $900 \mathrm{~K}$, surrounded by the same atmosphere will have cooling rate equal to-
1 $\frac{16}{3} r_{0}$
2 $\frac{8}{16} r_{0}$
3 $16 r_{0}$
4 $4 \mathrm{r}_{0}$
Explanation:
A Given that, $\mathrm{T}_{1}=600 \mathrm{~K}$ $\mathrm{T}_{2}=900 \mathrm{~K}, \mathrm{~T}_{\mathrm{o}}=300 \mathrm{~K}$ We know that, Cooling rate $\propto \mathrm{T}^{4}-\mathrm{T}_{0}^{4}$ $\frac{\mathrm{r}}{\mathrm{r}_{\mathrm{o}}}=\left(\frac{\mathrm{T}_{2}^{4}-\mathrm{T}_{\mathrm{o}}^{4}}{\mathrm{~T}_{1}^{4}-\mathrm{T}_{\mathrm{o}}^{4}}\right)$ $\frac{\mathrm{r}}{\mathrm{r}_{\mathrm{o}}}=\left[\frac{(900)^{4}-(300)^{4}}{(600)^{4}-(300)^{4}}\right]$ $\frac{\mathrm{r}}{\mathrm{r}_{\mathrm{o}}}=\left(\frac{16}{3}\right)$ $\mathrm{r}=\frac{16}{3} \mathrm{r}_{\mathrm{o}}$
149500
A black body is heated from $27^{\circ} \mathrm{C}$ to $127^{\circ} \mathrm{C}$. The ratio of their energies of radiation emitted will be:
1 $9: 16$
2 $27: 64$
3 $81: 256$
4 3:4
Explanation:
C We know that $\mathrm{E}=\sigma \mathrm{T}^{4}$ Where $\mathrm{E}$ is rate of emission of radiation of a body at temperature $\mathrm{T}$. $\mathrm{E}_{1}=\sigma(27+273)^{2}$ $\mathrm{E}_{2}=\sigma(127+273)^{2}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{(300)^{4}}{(400)^{4}}=\frac{81}{256}$
AIIMS-2001
Heat Transfer
149502
Calculate radiation power for sphere whose temperature is $227^{\circ} \mathrm{C}$ and radius $0.2 \mathrm{~m}$ and emissivity 0.8 .
1 $1425 \mathrm{~W}$
2 $1500 \mathrm{~W}$
3 $1255 \mathrm{~W}$
4 $1575 \mathrm{~W}$
Explanation:
A Given, Radius $=0.2 \mathrm{~m}$ And temperature $(\mathrm{T})=227+273=500 \mathrm{~K}$ Radiation power through area A is given as $\mathrm{P}=\sigma \mathrm{A} \varepsilon \mathrm{T}^{4}$ [Here $\varepsilon=$ emissivity $=0.8, \mathrm{~A}=$ surface area $=4 \pi \mathrm{r}^{2}$, $\left.\sigma=5.67 \times 10^{-8} \mathrm{w} / \mathrm{m}^{2} \mathrm{~K}^{4}\right]$ Putting these values we get $\mathrm{P}=5.67 \times 10^{-8} \times 4 \pi(0.2)^{2} \times(0.8) \times(500)^{4}$ $=1417.48 \simeq 1425 \mathrm{~W}$
AIIMS-25.05.2019(M) Shift-1
Heat Transfer
149503
Distance between sun and earth is $2 \times 10^{8} \mathrm{~km}$, temperature of sun $6000 \mathrm{~K}$, radius of sun $7 \times 10^{5}$ $\mathrm{km}$. If emissivity of earth is 0.6 , then find out temperature of earth in thermal equilibrium.
1 $400 \mathrm{~K}$
2 $300 \mathrm{~K}$
3 $500 \mathrm{~K}$
4 $600 \mathrm{~K}$
Explanation:
B Given, distance between sun and earth (d) $=2$ $\times 10^{8} \mathrm{~km}$, Temperature of sun $=6000 \mathrm{~K}$ and radius of sun $=7 \times 10^{5} \mathrm{~km}$. For thermal equilibrium, Energy received by earth = Energy emitted by earth $\frac{\mathrm{T}_{\mathrm{s}}^{4} 4 \pi \mathrm{R}_{\mathrm{s}}^{2}}{4 \pi \mathrm{d}^{2}} \times \pi \mathrm{R}_{\mathrm{e}}^{2}=\sigma \cdot \varepsilon \cdot \mathrm{T}_{2}^{4} \cdot 4 \pi \mathrm{R}_{\mathrm{e}}^{2}$ $\frac{\mathrm{T}_{\mathrm{s}}^{4} \times \mathrm{R}_{\mathrm{s}}^{2}}{4 \times \mathrm{d}^{2} \times \mathrm{e}}=\mathrm{T}_{\mathrm{e}}^{4}$ $\frac{(6000)^{4} \times\left(7 \times 10^{8}\right)^{2}}{4 \times\left(2 \times 10^{11}\right)^{2} \times 0.6}=\mathrm{T}_{\mathrm{e}}^{4}$ $\frac{36 \times 36 \times 7 \times 7}{4 \times 4 \times 0.6} \times 10^{6}=\mathrm{T}_{\mathrm{e}}^{4}$ $66.15 \times 108=\mathrm{T}_{\mathrm{e}}^{4}$ $\mathrm{T}_{\mathrm{e}}=285.19 \mathrm{~K} \simeq 300 \mathrm{~K}$
AIIMS-25.05.2019(E) Shift-2
Heat Transfer
149504
If a body coated black at $600 \mathrm{~K}$ surrounded by atmosphere at $300 \mathrm{~K}$ has cooling rate $r_{0}$, the same body at $900 \mathrm{~K}$, surrounded by the same atmosphere will have cooling rate equal to-
1 $\frac{16}{3} r_{0}$
2 $\frac{8}{16} r_{0}$
3 $16 r_{0}$
4 $4 \mathrm{r}_{0}$
Explanation:
A Given that, $\mathrm{T}_{1}=600 \mathrm{~K}$ $\mathrm{T}_{2}=900 \mathrm{~K}, \mathrm{~T}_{\mathrm{o}}=300 \mathrm{~K}$ We know that, Cooling rate $\propto \mathrm{T}^{4}-\mathrm{T}_{0}^{4}$ $\frac{\mathrm{r}}{\mathrm{r}_{\mathrm{o}}}=\left(\frac{\mathrm{T}_{2}^{4}-\mathrm{T}_{\mathrm{o}}^{4}}{\mathrm{~T}_{1}^{4}-\mathrm{T}_{\mathrm{o}}^{4}}\right)$ $\frac{\mathrm{r}}{\mathrm{r}_{\mathrm{o}}}=\left[\frac{(900)^{4}-(300)^{4}}{(600)^{4}-(300)^{4}}\right]$ $\frac{\mathrm{r}}{\mathrm{r}_{\mathrm{o}}}=\left(\frac{16}{3}\right)$ $\mathrm{r}=\frac{16}{3} \mathrm{r}_{\mathrm{o}}$
149500
A black body is heated from $27^{\circ} \mathrm{C}$ to $127^{\circ} \mathrm{C}$. The ratio of their energies of radiation emitted will be:
1 $9: 16$
2 $27: 64$
3 $81: 256$
4 3:4
Explanation:
C We know that $\mathrm{E}=\sigma \mathrm{T}^{4}$ Where $\mathrm{E}$ is rate of emission of radiation of a body at temperature $\mathrm{T}$. $\mathrm{E}_{1}=\sigma(27+273)^{2}$ $\mathrm{E}_{2}=\sigma(127+273)^{2}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{(300)^{4}}{(400)^{4}}=\frac{81}{256}$
AIIMS-2001
Heat Transfer
149502
Calculate radiation power for sphere whose temperature is $227^{\circ} \mathrm{C}$ and radius $0.2 \mathrm{~m}$ and emissivity 0.8 .
1 $1425 \mathrm{~W}$
2 $1500 \mathrm{~W}$
3 $1255 \mathrm{~W}$
4 $1575 \mathrm{~W}$
Explanation:
A Given, Radius $=0.2 \mathrm{~m}$ And temperature $(\mathrm{T})=227+273=500 \mathrm{~K}$ Radiation power through area A is given as $\mathrm{P}=\sigma \mathrm{A} \varepsilon \mathrm{T}^{4}$ [Here $\varepsilon=$ emissivity $=0.8, \mathrm{~A}=$ surface area $=4 \pi \mathrm{r}^{2}$, $\left.\sigma=5.67 \times 10^{-8} \mathrm{w} / \mathrm{m}^{2} \mathrm{~K}^{4}\right]$ Putting these values we get $\mathrm{P}=5.67 \times 10^{-8} \times 4 \pi(0.2)^{2} \times(0.8) \times(500)^{4}$ $=1417.48 \simeq 1425 \mathrm{~W}$
AIIMS-25.05.2019(M) Shift-1
Heat Transfer
149503
Distance between sun and earth is $2 \times 10^{8} \mathrm{~km}$, temperature of sun $6000 \mathrm{~K}$, radius of sun $7 \times 10^{5}$ $\mathrm{km}$. If emissivity of earth is 0.6 , then find out temperature of earth in thermal equilibrium.
1 $400 \mathrm{~K}$
2 $300 \mathrm{~K}$
3 $500 \mathrm{~K}$
4 $600 \mathrm{~K}$
Explanation:
B Given, distance between sun and earth (d) $=2$ $\times 10^{8} \mathrm{~km}$, Temperature of sun $=6000 \mathrm{~K}$ and radius of sun $=7 \times 10^{5} \mathrm{~km}$. For thermal equilibrium, Energy received by earth = Energy emitted by earth $\frac{\mathrm{T}_{\mathrm{s}}^{4} 4 \pi \mathrm{R}_{\mathrm{s}}^{2}}{4 \pi \mathrm{d}^{2}} \times \pi \mathrm{R}_{\mathrm{e}}^{2}=\sigma \cdot \varepsilon \cdot \mathrm{T}_{2}^{4} \cdot 4 \pi \mathrm{R}_{\mathrm{e}}^{2}$ $\frac{\mathrm{T}_{\mathrm{s}}^{4} \times \mathrm{R}_{\mathrm{s}}^{2}}{4 \times \mathrm{d}^{2} \times \mathrm{e}}=\mathrm{T}_{\mathrm{e}}^{4}$ $\frac{(6000)^{4} \times\left(7 \times 10^{8}\right)^{2}}{4 \times\left(2 \times 10^{11}\right)^{2} \times 0.6}=\mathrm{T}_{\mathrm{e}}^{4}$ $\frac{36 \times 36 \times 7 \times 7}{4 \times 4 \times 0.6} \times 10^{6}=\mathrm{T}_{\mathrm{e}}^{4}$ $66.15 \times 108=\mathrm{T}_{\mathrm{e}}^{4}$ $\mathrm{T}_{\mathrm{e}}=285.19 \mathrm{~K} \simeq 300 \mathrm{~K}$
AIIMS-25.05.2019(E) Shift-2
Heat Transfer
149504
If a body coated black at $600 \mathrm{~K}$ surrounded by atmosphere at $300 \mathrm{~K}$ has cooling rate $r_{0}$, the same body at $900 \mathrm{~K}$, surrounded by the same atmosphere will have cooling rate equal to-
1 $\frac{16}{3} r_{0}$
2 $\frac{8}{16} r_{0}$
3 $16 r_{0}$
4 $4 \mathrm{r}_{0}$
Explanation:
A Given that, $\mathrm{T}_{1}=600 \mathrm{~K}$ $\mathrm{T}_{2}=900 \mathrm{~K}, \mathrm{~T}_{\mathrm{o}}=300 \mathrm{~K}$ We know that, Cooling rate $\propto \mathrm{T}^{4}-\mathrm{T}_{0}^{4}$ $\frac{\mathrm{r}}{\mathrm{r}_{\mathrm{o}}}=\left(\frac{\mathrm{T}_{2}^{4}-\mathrm{T}_{\mathrm{o}}^{4}}{\mathrm{~T}_{1}^{4}-\mathrm{T}_{\mathrm{o}}^{4}}\right)$ $\frac{\mathrm{r}}{\mathrm{r}_{\mathrm{o}}}=\left[\frac{(900)^{4}-(300)^{4}}{(600)^{4}-(300)^{4}}\right]$ $\frac{\mathrm{r}}{\mathrm{r}_{\mathrm{o}}}=\left(\frac{16}{3}\right)$ $\mathrm{r}=\frac{16}{3} \mathrm{r}_{\mathrm{o}}$
