20. If the temperature of a Black body increases by 50% then amount of radiation emitted by itin
given time interval will
Answers
Answer:
Explanation:
Here,
initial thermodynamic temperature of black body, T₁ = 100
The temperature of a Black body increases by 50%. so temperature after this increase, T₂ :
T₂ = T₁ + 50 = 100 + 50 = 150.
by taking the ratio of final temperature and initial temperature, we get
T₂/ T₁ = 150/100
T₂ / T₁ = 3 / 2
T₂ = 3/2 T₁
=> According to Stefan–Boltzmann law:
U₁ = σeAT₁⁴ ...(1)
U₂ = σeAT₂⁴ ...(2)
=> By taking the ratio of eq(1) and (2), we get
U₁/U₂ = σeAT₁⁴/ σeAT₂⁴
U₁/U₂ = T₁⁴/ T₂⁴
U₁/U₂ = T₁⁴/ (3/2 T₁) ⁴
U₁ / U₂ = 16 T₁⁴/ 81 T₁⁴
U₁ / U₂ = 16/81
U₂ = 81/16 U₁
=> Increase in percentage:
increase in % = (U₂ - U₁) / U₁ * 100%
= 81/16U₁ - U₁ / U₁ * 100
= (81/16 - 1) U₁ / U₁ * 10
= 81 - 16 / 16 * 100
= 65 / 16 * 100
= 4.063 * 100
= 406.3 %
Thus, amount of radiation emitted by black body in given time interval will be 406.3 %.
Given that,
The temperature of a Black body increases by 50%.
Let the initial temperature is T.
Final temperature
The initial amount of energy of radiation emitted
Using formula of energy
The final amount of energy of radiation emitted
Using formula of energy
Put the value of T'
We need to calculate the percentage increases
Using formula for percentage increases
Put the value into the formula
Hence, The percentage increases is 406.25%.