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The area of a hole of heat furnace is $10^{- 4} m^{2}$ . It radiates $1.58 \times 10^{5}$ calories of heat per hour. If the emissivity of the furnace is 0.80, then its temperature is

1500 K

2000 K

2500 K

3000 K

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Explanation

$\Rightarrow \frac{1.58 \times 10^{5} \times \times 4.2}{60 \times 60} = 5.6 \times 10^{- 8} \times 10^{- 4} \times 0.8 \times T^{4} T \approx 2500 K$

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Two spheres P and Q, of same colour having radii 8 cm and 2 cm are maintained at temperatures 127 $° C$ and 527 $° C$ respectively. The ratio of energy radiated by P and Q is

0.054

0.0034

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Explanation

$\Rightarrow Q \propto {AT}^{4} \propto r^{2} T^{4} (∵ A = 4 {πr}^{2}) \Rightarrow \frac{Q_{P}}{Q_{Q}} = {(\frac{r_{P}}{r_{Q}})}^{2} {(\frac{T_{P}}{T_{Q}})}^{4} = {(\frac{8}{2})}^{2} {\{\frac{(273 + 127)}{(273 + 527)}\}}^{4} = 1$

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A body radiates energy 5W at a temperature of 127 $° C$ . If the temperature is increased to 927 $° C$ , then it radiates energy at the rate of

410 W

81 W

405 W

200 W

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Explanation

$\Rightarrow \frac{P_{1}}{P_{2}} = {(\frac{T_{1}}{T_{2}})}^{4} = {(\frac{927 + 273}{127 + 273})}^{4} \Rightarrow P_{1} = 405 W$

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The temperatures of two bodies A and B are respectively 727 $° C$ and 327 $° C$ . The ratio of the rates of heat radiated by them is

727:327

5 : 3

25 : 9

625 : 81

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Explanation

(d) $Q \propto T^{4} \Rightarrow \frac{H_{A}}{H_{B}} = {(\frac{273 + 727}{273 + 327})}^{4} = {(\frac{10}{6})}^{4} = {(\frac{5}{3})}^{4} = \frac{625}{81}$

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The energy emitted per second by a black body at 27 $° C$ is 10 J. If the temperature of the black body is increased to 327 $° C$ , the energy emitted per second will be

20 J

40 J

80 J

160 J

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Explanation

(d) $Q_{Black body} = {AσT}^{4} t \Rightarrow Q \propto T^{4} \Rightarrow Q_{2} = Q_{1} {(\frac{T_{2}}{T_{1}})}^{4} = 10 {(\frac{273 + 327}{273 + 27})}^{4} = 10 {(\frac{600}{300})}^{4} = 160 J$

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The radiant energy from the sun incident normally at the surface of earth is $20 kcal / m^{2} \min$ . What would have been the radiant energy incident normally on the earth, if the sun had a temperature twice of the present one ?

$160 kcal / m^{2} \min$

$40 kcal / m^{2} \min$

$320 kcal / m^{2} \min$

$80 kcal / m^{2} \min$

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Explanation

(c) $\frac{E_{2}}{E_{1}} = {(\frac{T_{2}}{T_{1}})}^{4} \Rightarrow \frac{E_{2}}{20} = {(\frac{2 T}{T})}^{4} = 16 \Rightarrow E_{2} = 320 kcal / m^{2} \min$

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If the temperature of the sun (black body) is doubled, the rate of energy received on earth will be increased by a factor of

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Explanation

(d) Amount of energy radiated ∝ $(Temperature)^{4}$

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The ratio of energy of emitted radiation of a black body at 27 $° C$ and 927 $° C$ is

1 : 4

1 : 16

1 : 64

1 : 256

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Explanation

(d) $\frac{Q_{1}}{Q_{2}} = {(\frac{T_{1}}{T_{2}})}^{4} = {(\frac{273 + 27}{273 + 927})}^{4} = {(\frac{1}{4})}^{4} = \frac{1}{256}$

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Two spherical black bodies of radii $r_{1}$ and $r_{2}$ and with surface temperature $T_{1}$ and $T_{2}$ respectively radiate the same power. Then the ratio of $r_{1}$ and $r_{2}$ will be

${(\frac{T_{2}}{T_{1}})}^{2}$

${(\frac{T_{2}}{T_{1}})}^{4}$

${(\frac{T_{1}}{T_{2}})}^{2}$

${(\frac{T_{1}}{T_{2}})}^{4}$

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Explanation

(a) For black body, $P = {AεσT}^{4}$ . For same power $A \propto \frac{1}{T^{4}}$

$\Rightarrow {(\frac{r_{1}}{r_{2}})}^{2} = {(\frac{T_{2}}{T_{1}})}^{4} \Rightarrow \frac{r_{1}}{r_{2}} = {(\frac{T_{2}}{T_{1}})}^{2}$

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A black body is at a temperature 300 K. It emits energy at a rate, which is proportional to

300

${(300)}^{2}$

${(300)}^{3}$

${(300)}^{4}$

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Explanation

(d) $E \propto T^{4}$

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