Question

A sphere is at temperature \(600\) K. In an external environment of \(200\) K, its cooling rate is R. When the temperature of the sphere falls to \(400\) K then cooling rate R' will become

• A. \(\frac{16}{3}R\)
• B. \(\frac{16}{9}R\)
• C. \(\frac{9}{16}R\)
• D. \(\frac{3}{16}R\)

Hint:

Rate of cooling by radiation is proportional to T^4 minus T0^4.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A hot sphere in a cooler surrounding loses heat by radiation. By Stefan-Boltzmann law the net rate of heat loss is proportional to \(T^4 - T_0^4\), where \(T_0\) is the temperature of the surroundings.

Step 2: Set up the ratio:
Initial: \(R \propto 600^4 - 200^4\). Later: \(R' \propto 400^4 - 200^4\).
Take 200 as a common unit: \(600 = 3\), \(400 = 2\), \(200 = 1\).

Step 3: Compute:
\[ \frac{R'}{R} = \frac{2^4 - 1^4}{3^4 - 1^4} = \frac{16 - 1}{81 - 1} = \frac{15}{80} = \frac{3}{16} \]

Step 4: Why the other options are wrong.
\(\frac{16}{3}R\) and \(\frac{16}{9}R\) are larger than \(R\), but cooling slows as the sphere gets closer to the surrounding temperature. \(\frac{9}{16}R\) comes from taking \(\frac{3^2}{4^2}\), which is not the Stefan law.

Final Answer:
The new cooling rate is \(\frac{3}{16}R\), option (D). \[ \boxed{\frac{3}{16}R} \]