Question

A tungsten body of diameter 2.3 cm is at 2000° C. It radiates 30% of the energy radiated by a black body

Answer 1

Answer:

r (black body)=2.1 cm

Explanation:

In ques we have given temperature of tungsten 2000°c and diameter of it is 2.3 cm

According to question 30% of energy radiated by a black body is equal to the energy radiated by tungsten so we will equate it to find the radius of black body.

Now

[30% * sigmna *(Temp of black body)^4*(4πr^2)]=Sigma *(Temp of tungsten)^4*[4π(r of tungsten)^2]

As we know TEMPERATURE OF TUNGSTEN and BLACK BODY is SAME

T(tungsten)=T(black body)=2000°c

Now,

(30/100)*r^2=(1.15)^2 ( Radis of tungsten

=2.3/2=1.15cm)

r^2=[(1.15)^2*100]/30

r^2=4.408

r=2.1cm

Answer 2

Complete question:

Tungsten body of diameter 2.3 cm is at 2000 degree Celsius it radiates 30% of heat energy radiated by a black body of the same temperature and radius find the radius of the black body which will radiate energy at the same rate at the same temperature

Answer:

The radius of the black body which will radiate energy at the same rate at the same temperature is 0.629 cm.

Explanation:

From question,

30% × σT⁴tungsten × 4π (1.15)² = σT⁴black × 4π (r)²

30% × σT⁴tungsten × (1.15)² = σT⁴black × (r)²

From question,

T⁴tungsten = T⁴black

Now,

30% × (1.15)² = (r)²

30/100 × (1.15)² = (r)²

30/100 × 1.322 = (r)²

0.3 × 1.322 = (r)²

r² = 0.396

r = √0.396

∴ r = 0.629 cm