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Question: A hot body is being cooled through radiation according to the law \(\frac{d\theta}{dt}\) = – K(q – q...

A hot body is being cooled through radiation according to the law dθdt\frac{d\theta}{dt} = – K(q – q0) where q and q0 have their usual meaning. Time after which body looses half its max. heat it can lose is-

A

1K\frac{1}{K}

B

ln2K\frac{\mathcal{l}n2}{K}

C

ln3K\frac{\mathcal{l}n3}{K}

D

2K\frac{2}{K}

Answer

ln2K\frac{\mathcal{l}n2}{K}

Explanation

Solution

dθdt\frac{d\theta}{dt} = – K(q – q0)

body will lose half of it maximum heat when its temperature is equal to = q0 + θ−θ02\frac{\theta - \theta_{0}}{2} = θ+θ02\frac{\theta + \theta_{0}}{2}

∫θ0(θ+θ0)/2dθθ−θ0\int_{\theta_{0}}^{(\theta + \theta_{0})/2}\frac{d\theta}{\theta - \theta_{0}} = – K ∫0tdt\int_{0}^{t}{dt}

Ž t = ln2K\frac{\mathcal{l}n2}{K}