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# Engineering

Important formula from this concept

Answers (1)
A avinash.dongre

Rate of Loss of Heat=   R_H=\frac{dQ}{dt}=A\epsilon\sigma(\theta^{4}-\theta_{0}^{4})

Rate of Cooling=       R_{c}=\frac{\Delta \theta }{dt}=\frac{A\epsilon\sigma}{mc}(\theta^{4}-\theta_{0}^{4})

According to Newton's Law of Cooling  I.e       \frac{d\theta}{dt}\alpha(\theta-\theta_{0})      

When body Cools by Radiation from   \ \theta_1^0 C to theta \ \theta_2^0 C in time t                                                                                                                                       Then   \left[\frac{\theta_{1}-\theta_{2}}{t} \right ]=k\left[\frac{\theta_{1}+\theta_{2}}{2}-\theta_{0} \right ]                                                                                                                                                                               Where \theta_{av}=\frac{\theta_{1}+\theta_{2}}{2}

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