Question

An object is at a temperature of 400 c At what temperature would it radaite energy twice as fast ?​

Answer 1

Answer:

from stefan-boltzman law Eb=5.67×(t/100)⁴

so applying E¹=2E²

5.67×(t/100)⁴=2×5.67×(673/100)⁴

T=800.33k

Explanation:

an object is at a temperature of 400°C . The temperature at which it can radiate energy twice as faster (neglect the temperature of surroundings)is (1/24=1.189):

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