A spherical black body with a radius of 12 cm radiates 450 watt power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be

Question

Which is an adiabatic process?

Answer 1

To solve this problem, we will use the Stefan-Boltzmann Law, which describes the power radiated by a black body in terms of its temperature. The formula is given by:

P = \sigma A T^4

Where:

P is the power radiated,
\sigma is the Stefan-Boltzmann constant (approximately 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4),
A is the surface area of the black body, and for a sphere, it is given by 4 \pi r^2,
T is the absolute temperature in Kelvin.

Initially, the power P_1 is given as 450 W with a radius r_1 = 12 \, \text{cm} = 0.12 \, \text{m} and temperature T_1 = 500 \, \text{K}.

Using the formula, the initial power radiated is:

P_1 = \sigma \cdot 4 \pi r_1^2 \cdot T_1^4

Now, the changes are:

Radius is halved, r_2 = \frac{r_1}{2} = 0.06 \, \text{m}
Temperature is doubled, T_2 = 2 \cdot T_1 = 1000 \, \text{K}

The new power P_2 will be:

P_2 = \sigma \cdot 4 \pi r_2^2 \cdot T_2^4

Substituting the new values:

P_2 = \sigma \cdot 4 \pi \left(\frac{r_1}{2}\right)^2 \cdot (2T_1)^4

Simplifying:

P_2 = \sigma \cdot 4 \pi \cdot \frac{r_1^2}{4} \cdot 16 \cdot T_1^4

P_2 = 16 \cdot \frac{1}{4} \cdot \sigma \cdot 4 \pi r_1^2 \cdot T_1^4

P_2 = 4 \cdot P_1

Since P_1 = 450 \, \text{W},

P_2 = 4 \cdot 450 \, \text{W} = 1800 \, \text{W}

Thus, the new power radiated is 1800 watts.

Answer 2