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A black body radiates energy at the rate of E $W / m^{2}$ at a high temperature TK. When the temperature is reduced to $\frac{T}{2} K$ , the radiant energy will be

$\frac{E}{16}$

$\frac{E}{4}$

$4 E$

$16 E$

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Explanation

(a) $E \propto T^{4} \Rightarrow \frac{E_{1}}{E_{2}} = \frac{T^{4}}{T^{4}} \times 2^{4} \Rightarrow E_{2} = \frac{E}{16}$

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An object is at a temperature of $400 ° C$ . At what temperature would it radiate energy twice as fast? The temperature of the surroundings may be assumed to be negligible .

$200 ° C$

200 K

$800 ° C$

800 K

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Explanation

(d) $\frac{E_{2}}{E_{1}} = {(\frac{T_{2}}{T_{1}})}^{4} \Rightarrow \frac{2}{1} = {(\frac{T}{400 + 273})}^{4} = {(\frac{T}{673})}^{4} \Rightarrow T = 2^{1 / 4} \times 673 = 800 K$

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A black body at a temperature of 227 $° C$ radiates heat energy at the rate of 5 $cal / {cm}^{2} - \sec$ . At a temperature of $727 ° C$ , the rate of heat radiated per unit area in $cal / {cm}^{2}$ will be

160

250

500

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Explanation

(a) $\frac{E_{2}}{E_{1}} = {(\frac{T_{2}}{T_{1}})}^{4} = (\frac{273 + 727}{237 + 227}) = \frac{{(1000)}^{4}}{{(500)}^{4}} = 16 \Rightarrow E_{2} = 80$

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Energy is being emitted from the surface of a black body at 127 $° C$ temperature at the rate of $1.0 \times 10^{6} J / \sec - m^{2}$ . Temperature of the black body at which the rate of energy emission is $16.0 \times 10^{6} J / \sec - m^{2}$ will be -

254 $° C$

508 $° C$

527 $° C$

727 $° C$

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Explanation

(c) $\frac{E_{2}}{E_{1}} = {(\frac{T_{2}}{T_{1}})}^{4} \Rightarrow T_{2} = {(\frac{E_{2}}{E_{1}})}^{1 / 4} \times T_{1} = {(16)}^{1 / 4} \times (273 + 127) \Rightarrow T_{2} = 800 k = 527 ° C$

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If temperature of a black body increases from $7 ° C$ to $287 ° C$ , then the rate of energy radiation increases by

${(\frac{287}{7})}^{4}$

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Explanation

(b) For a block body rate of energy $\frac{Q}{t} = P = {AσT}^{4}$

$P \propto T^{4} \Rightarrow \frac{P_{1}}{P_{2}} = {(\frac{T_{1}}{T_{2}})}^{4} = {\{\frac{(273 + 7)}{(273 + 287)}\}}^{4} = \frac{1}{16}$

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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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