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 beA- 250B- 500C- 80D- 160

The correct option is C 80 E_2/E_1=(T_2/T_1)^4=(273+727/237+227)=(1000)^4/(500)^4=16⇒ E_2=80 ...

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Byju's Answer

Standard XII

Chemistry

Normality

A black body ...

Question

A black body at a temperature of $227^{o} C$ radiates heat energy at the rate of 5 $c a l / c m^{2} - s e c$ . At a temperature of $727^{o} C$ , the rate of heat radiated per unit area in $c a l / c m^{2}$ will be

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160

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250

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500

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Solution

The correct option is A
80

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

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A black body at a temperature of 227o C radiates heat energy at the rate of 5 cal/cm2−sec. At a temperature of 727o C, the rate of heat radiated per unit area in cal/cm2 will be

The correct option is A 80 E2E1=(T2T1)4=(273+727237+227)=(1000)4(500)4=16⇒E2=80

A black body at a temperature of $227^{o} C$ radiates heat energy at the rate of 5 $c a l / c m^{2} - s e c$ . At a temperature of $727^{o} C$ , the rate of heat radiated per unit area in $c a l / c m^{2}$ will be

The correct option is A
80

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