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NEET PHYSICSThermal Properties of MatterMedium

Question

A black body at 227C227^\circ\text{C} radiates heat at the rate of 7 cal cm2s17 \text{ cal cm}^{-2}\text{s}^{-1}. At a temperature of 727C727^\circ\text{C}, the rate of heat radiated in the same units will be:

A

6060

B

5050

C

112112

D

8080

Step-by-Step Solution

According to the Stefan-Boltzmann law, the rate of heat radiated per unit area by a black body is directly proportional to the fourth power of its absolute temperature, i.e., ET4E \propto T^4. Given: Initial temperature, T1=227C=227+273=500 KT_1 = 227^\circ\text{C} = 227 + 273 = 500 \text{ K} Initial rate of heat radiated, E1=7 cal cm2s1E_1 = 7 \text{ cal cm}^{-2}\text{s}^{-1} Final temperature, T2=727C=727+273=1000 KT_2 = 727^\circ\text{C} = 727 + 273 = 1000 \text{ K} Therefore, the ratio of the new rate of heat radiated (E2E_2) to the initial rate (E1E_1) is: E2E1=(T2T1)4=(1000500)4=24=16\frac{E_2}{E_1} = \left(\frac{T_2}{T_1}\right)^4 = \left(\frac{1000}{500}\right)^4 = 2^4 = 16 E2=16×E1=16×7=112 cal cm2s1E_2 = 16 \times E_1 = 16 \times 7 = 112 \text{ cal cm}^{-2}\text{s}^{-1}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Thermal Properties of Matter. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

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