A black body at $227^{\circ} C$ radiates heat at the rate of $7\, cals / cm ^{2} s$. At a temperature of $727^{\circ} C$, the rate of heat radiated in the same units will be :

Question

Sphere of color black, red, white, and yellow are heated to the same temperature. The decreasing order of cooling is:

Answer 1

To determine the rate at which heat is radiated by the black body at a higher temperature, we can employ the Stefan-Boltzmann Law for black body radiation, which states:

$ E = \sigma T^4 $

where:

E is the energy emitted per unit area per unit time.
\sigma is the Stefan-Boltzmann constant.
T is the absolute temperature in Kelvin.

The problem gives us the rate of heat radiation at 227^{\circ} C and asks for the rate at 727^{\circ} C. First, we need to convert these temperatures to Kelvin:

227^{\circ} C = 227 + 273 = 500\, K
727^{\circ} C = 727 + 273 = 1000\, K

According to the problem, at 227^{\circ} C (or 500\, K), the radiated energy is 7\, \text{cals/cm}^2/\text{s}.

Using the ratio form of the Stefan-Boltzmann law for the same body at two temperatures:

$ \frac{E_2}{E_1} = \left(\frac{T_2}{T_1}\right)^4 $

Substitute the given values:

E_1 = 7\, \text{cals/cm}^2/\text{s}
T_1 = 500\, K
T_2 = 1000\, K

Calculate E_2:

\[ \frac{E_2}{7} = \left(\frac{1000}{500}\right)^4 = 2^4 = 16 \]

Thus,

\[ E_2 = 16 \times 7 = 112\, \text{cals/cm}^2/\text{s} \]

Therefore, the rate of heat radiated at 727^{\circ} C is 112 cals/cm²/s.

Answer 2