Heat Transfer - Radiation: Practice Problem & Solution
A solid sphere of copper of radius ( R ) and a hollow sphere of the same material of inner radius ( r ) and outer radius ( R ) are heated to the same temperature and allowed to cool in the same environment. Assertion (A): Hollow sphere cools faster than solid sphere. Reason (R): \( \left( \frac{d\theta}{dt} \right) \propto \frac{1}{m} \).
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Heat Transfer - Radiation. Understanding the underlying formula is key to arriving at the correct answer below:
Assertion (A) is true. Hollow sphere has less mass ( m ) but same surface area ( A ) as solid sphere. Reason (R) is true, cooling rate \( \frac{dT}{dt} \propto \frac{A}{m} \). Since ( A ) is constant, \( \frac{dT}{dt} \propto \frac{1}{m} \). Lower mass ( m ) means faster cooling. Thus, both are true and (R) explains (A).
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