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Two identical metal balls at temperature 200 $° C$ and 400 $° C$ kept in air at 27 $° C$ . The ratio of net heat loss by these bodies is

1/4

1/2

1/16

$\frac{473^{4} - 300^{4}}{673^{4} - 300^{4}}$

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Explanation

(d) If temperature of surrounding is considered then
net loss of energy of a body by radiation

$Q = Aεσ (T^{4} - T_{0}^{4}) t \Rightarrow Q \propto (T^{4} - T_{0}^{4}) \Rightarrow \frac{Q_{1}}{Q_{2}} = \frac{T_{1}^{4} - T_{0}^{4}}{T_{2}^{4} - T_{0}^{4}} = \frac{{(273 + 200)}^{4} - {(273 + 27)}^{4}}{{(273 + 400)}^{4} - {(273 + 27)}^{4}} = \frac{{(473)}^{2} - {(300)}^{4}}{{(673)}^{4} - {(300)}^{4}}$

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A black body radiates 20 W at temperature 227 $° C$ . If temperature of the black body is changed to 727 $° C$ then its radiating power will be

120W

240 W

320 W

360 W

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Explanation

$\Rightarrow \frac{P_{2}}{P_{1}} = {(\frac{T_{2}}{T_{1}})}^{4} \Rightarrow \frac{P_{2}}{20} = {(\frac{273 + 727}{273 + 227})}^{4} \Rightarrow \frac{P_{2}}{20} = 2^{4} \Rightarrow P_{2} = 320 W$

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The radiation emitted by a star A is 10,000 times that of the sun. If the surface temperatures of the sun and the star A are 6000 K and 2000 K respectively, the ratio of the radii of the star A and the sun is

300 : 1

600 : 1

900 : 1

1200 : 1

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Explanation

(c) $Q \propto {AT}^{4} \propto r^{2} T^{4} \frac{Q_{star}}{Q_{sun}} = \frac{r_{star}^{2} . T_{star}^{4}}{r_{sun}^{2} . T_{sun}^{4}} \Rightarrow \frac{10000}{1} = \frac{r_{star}^{2}}{r_{sun}^{2}} \times {(\frac{6000}{2000})}^{4} \Rightarrow \frac{r_{star}}{r_{sun}} = \frac{100 \times 9}{1} = \frac{900}{1}$

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A black body radiates at the rate of W watts at a temperature T. If the temperature of the body is reduced to T/3, it will radiate at the rate of (in Watts)

$\frac{W}{81}$

$\frac{W}{27}$

$\frac{W}{9}$

$\frac{W}{3}$

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Explanation

(a) $P = (\frac{Q}{t}) \propto T^{4} \Rightarrow \frac{W}{P_{2}} = {(\frac{T}{T / 3})}^{4} \Rightarrow P_{2} = \frac{W}{81}$

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Star A has radius r surface temperature T while star B has radius 4r and surface temperature T/2. The ratio of the power of two stars, $P_{A}$ : $P_{B}$ is

16 : 1

1 : 16

1 : 1

1 : 4

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Explanation

$\Rightarrow \frac{P_{2}}{P_{1}} = {(\frac{r_{2}}{r_{1}})}^{2} \times {(\frac{T_{2}}{T_{1}})}^{4} = {(\frac{4 r}{r})}^{2} \times {(\frac{T / 2}{\frac{T}{}})}^{4} = 1$

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Suppose the sun expands so that its radius becomes 100 times its present radius and its surface temperature becomes half of its present value. The total energy emitted by it then will increase by a factor of

$10^{4}$

625

256

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Explanation

(b) $\frac{Q_{2}}{Q_{1}} = {(\frac{r_{2}^{2}}{r_{1}^{2}})}^{2} \times {(\frac{T_{2}}{T_{1}})}^{4} = {(\frac{100}{1})}^{2} \times {(\frac{1}{2})}^{2} = 625$

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If the sun’s surface radiates heat at $6.3 \times 10^{7} {Wm}^{- 2}$ . Calculate the temperature of the sun assuming it to be a black body $(σ = 5.7 \times 10^{- 8} {Wm}^{- 2} K^{- 4})$

$5.8 \times 10^{3} K$

$8.5 \times 10^{3} K$

$3.5 \times 10^{8} K$

$5.3 \times 10^{8} K$

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Explanation

(a) From Stefan’s law $E = {σT}^{4}$

$T^{4} = \frac{E}{σ} = \frac{6.3 \times 10^{7}}{5.7 \times 10^{- 8}} = 1.105 \times 10^{15} = 0.1105 \times 10^{16} T = 0.58 \times 10^{4} K = 5.8 \times 10^{3} K$

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The value of Stefan’s constant is

$5.67 \times 10^{- 8} W / m^{2} - K^{4}$

$5.67 \times 10^{- 5} W / m^{2} - K^{4}$

$5.67 \times 10^{- 11} W / m^{2} - K^{4}$

None of these

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Explanation

(a) The value of stefan's constant is 5.67 $\times 10^{- 8} W / m^{2} - K^{4}$

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Rate of cooling at 600K, if surrounding temperature is 300K is R. The rate of cooling at 900K is

$\frac{16}{3} R$

$\frac{2}{3} R$

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Explanation

(a) Rate of cooling ∝ $(T^{4} - T_{0}^{4})$

$\Rightarrow \frac{R_{1}}{R_{2}} = (\frac{T_{1}^{4} - T_{0}^{4}}{T_{2}^{4} - T_{0}^{4}}) \Rightarrow \frac{R}{R_{2}} = \frac{{(600)}^{4} - {(300)}^{4}}{{(900)}^{4} - {(300)}^{4}} or R_{2} = \frac{16}{3} R$

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A black body of surface area 10 ${cm}^{2}$ is heated to 127°C and is suspended in a room at temperature 27°C. The initial rate of loss of heat from the body at the room temperature will be

2.99 W

1.89 W

1.18 W

0.99 W

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Explanation

(d) Loss of heat $∆ Q = Aεσ (T^{4} - T_{0}^{4}) t$

$\Rightarrow$ Rate of loss of heat $\frac{∆ Q}{t} = Aεσ (T^{4} - T_{0}^{4})$

$= 10 \times 10^{- 4} \times 1 \times 5.67 \times 10^{- 8} \{{(273 + 127)}^{4} - {(273 + 27)}^{4}\} = 0.99 W$

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