The original temperature of a black body is 727°C. Calculate temperature at which total radiant energy from this black body becomes double

Given:
Initial temperature of the black body, T1 = 727°C = (727 + 273) K = 1000 K
Final total radiant energy = 2 × Initial total radiant energy

Using Stefan-Boltzmann law, the total radiant energy emitted per unit area of a black body is given by:

E = σT⁴

where:
E = total radiant energy
σ = Stefan-Boltzmann constant
T = absolute temperature in Kelvin

Since the total radiant energy becomes double, we write:

E₂ = 2E₁
σT₂⁴ = 2σT₁⁴

Cancelling σ from both sides:

T₂⁴ = 2T₁⁴

Taking the fourth root on both sides:

T₂ = T₁ × (2)^(1/4)

Substituting T₁ = 1000 K:

T₂ = 1000 × (2)^(1/4)
T₂ ≈ 1000 × 1.189
T₂ ≈ 1189 K

Converting back to Celsius:

T₂ = 1189 - 273
T₂ = 916°C

Final Answer: The temperature at which the total radiant energy becomes double is 916°C.