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

A 200 g sample of water at 80°C is mixed with 100 g of water at 20°C. Assuming no heat loss to the surroundings, what is the final temperature of the mixture?

Answer 1

To solve this problem, we will use the concept of black body radiation given by Stefan-Boltzmann Law. According to this law, the power radiated by a black body is directly proportional to the fourth power of its absolute temperature and directly proportional to the surface area of the body. The formula is given by:

P = \sigma A T^4, where:

P is the power radiated.
\sigma is the Stefan-Boltzmann constant.
A is the surface area of the sphere.
T is the absolute temperature in Kelvin.

First, we will calculate the surface area of the sphere with its original radius:

The surface area of a sphere is given by A = 4\pi r^2.

Original radius, r = 12 \, cm = 0.12 \, m.

Original temperature, T = 500 \, K.

Power radiated, P = 450 \, watt.

Now, we keep the Stefan-Boltzmann constant \sigma the same, we get:

450 = \sigma \times (4\pi \times (0.12)^2) \times (500)^4.

Let us calculate the power if the radius is halved and the temperature is doubled:

New radius r' = \frac{0.12}{2} = 0.06 \, m.
New temperature T' = 2 \times 500 = 1000 \, K.
New surface area A' = 4\pi (0.06)^2.

The new power radiated can be calculated as follows:

P' = \sigma \times A' \times (T')^4

P' = \sigma \times 4\pi \times (0.06)^2 \times (1000)^4

Substituting the changes:

P' = \frac{A'}{A} \times \left(\frac{T'}{T}\right)^4 \times P

P' = \frac{(0.06)^2}{(0.12)^2} \times \left(\frac{1000}{500}\right)^4 \times 450

P' = \frac{1}{4} \times 16 \times 450 = 1800 \, watts

Hence, the power radiated by the black body with the new conditions is 1800 watts.

Answer 2