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Question

If the temperature of a hot body is raised by 0.5%, then the heat energy radiated would increase by :

A
0.5%
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B
1.0%
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C
1.5%
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D
2.0%
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Solution

The correct option is D 2.0%
The rate of heat energy radiated by black body is given as:
Q=σT4A where σ is the stefen-boltzmann constant, A is the area of the radiating body.
So QT4=kT4, where k is the propotionality constant that we have assumed here.
Taking log of above equation, logQ=log(kT4)=logk+4logT
Taking differential of above equation, dQQ=0+4dTT (because logk is constant).
Here dQ and dT indicate very small change in the Q and T respectively.
Multiplying by 100,
dQQ×100=4dTT×100
Percentage change in Heat rate =4× percentage change in temperature
So here, percentage change in rate of heat energy radiated =4×0.5=2%

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