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

The problem is based on Stefan-Boltzmann Law which states that the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of its absolute temperature.

The Stefan-Boltzmann Law can be written as:

E = σT^4

where: E is the total energy radiated per unit surface area, σ is the Stefan-Boltzmann constant (approximately 5.67 x 10^-8 W m^-2 K^-4), T is the absolute temperature in Kelvin.

Given that the original temperature of the black body is 727°C, we first need to convert this to Kelvin. The conversion formula is:

T(K) = T(°C) + 273.15

So, T1 = 727°C + 273.15 = 1000 K

Let's denote E1 as the initial total radiant energy from the black body. According to the problem, the total radiant energy becomes double, so the final energy E2 = 2E1.

We need to find the final temperature T2 when E = E2. From the Stefan-Boltzmann Law, we have:

E1 = σT1^4 and E2 = σT2^4

Substituting E2 = 2E1 into the second equation gives:

2σT1^4 = σT2^4

Solving for T2 gives:

T2 = (2^(1/4)) * T1

Substituting T1 = 1000 K into the equation gives:

T2 = (2^(1/4)) * 1000 ≈ 1190 K

So, the temperature at which total radiant energy from the black body becomes double is approximately 1190 K or 917°C.