Question

A cube at 80°C radiates heat at a rate of 98 J/s. If the length of each side is cut in half, the rate at which it will now radiate is closest to?


Really appreciate for the help!!!

Answer 1

Answer:

The rate at which it will now radiate heat is 24.5 J/s

Explanation:

As we know that rate of heat radiation by cube is given by Stephan Boltzmann law so we have

$\frac{dQ}{dt} = \sigma A T^4$

now we know that total surface area of the cube is given as

$A = 6 a^2$

now we have

$\frac{dQ}{dt} = \sigma(6a^2)T^4$

so we have

$98 = \sigma (6a^2) T^4$

now length of each side is cut to half so we have

$\frac{dQ}{dt} = \sigma (6)(\frac{a}{2})^2 T^4$

now from above two equations

$\frac{dQ}{dt} = \frac{98}{4}$

$\frac{dQ}{dt} = 24.5 J/s$

#Learn

Topic : Radiation mode of heat transfer

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