Question 689: Thermodynamics & KTG

Question

If a black body at 400 K surrounded by atmosphere at 300 K has rate of cooling ' $\mathrm{R}_0$ ', the same body at 900 K , surrounded by same atmosphere, will have rate of cooling nearly

Accepted Answer

Step 1: Understand Stefan's Law
The rate at which a black body cools depends on the difference between the fourth powers of its temperature and the temperature of its surroundings. This law is written as:

$$ 	ext{Rate} \propto (T^4 - T_0^4) $$

where $ T $ is the temperature of the black body, and $ T_0 $ is the temperature of the surroundings.
Step 2: Rate of Cooling for Given Temperatures
Let the temperature of the body in the first case be $ T_1 = 400\,	ext{K} $, and the temperature of the surroundings $ T_0 = 300\,	ext{K} $. For the second case, the temperature of the body is $ T_2 = 900\,	ext{K} $, but $ T_0 $ is still $ 300\,	ext{K} $.
Step 3: Formula for Each Case
The rate for the first case:

$$ R_0 = k\left(T_1^4 - T_0^4ight) $$

The rate for the second case:

$$ R_1 = k\left(T_2^4 - T_0^4ight) $$

Here, $ k $ is a constant.
Step 4: Find the Ratio of Rates
To compare, divide the two rates:

$$ \frac{R_0}{R_1} = \frac{T_1^4 - T_0^4}{T_2^4 - T_0^4} = \frac{(400)^4 - (300)^4}{(900)^4 - (300)^4} $$
Step 5: Simplify and Find the Answer
When you do the calculation,

$$ \frac{R_0}{R_1} = \frac{35}{1296} $$

So,

$$ R_1 \approx 36 R_0 $$

This means the rate of cooling at 900 K is about 36 times greater than at 400 K.

Answer

$4 \mathrm{R}_0$

Answer

$16 \mathrm{R}_0$

Answer

$36 \mathrm{R}_0$

Answer

$\frac{81 R_0}{16}$

Thermodynamics & KTG

If a black body at 400 K surrounded by atmosphere at 300 K has rate of cooling ' \mathrm{R}_0 ', the same body at 900 K , surrounded by same atmosphere, will have rate of cooling nearly

Choose the correct answer: