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Question

$\frac{-\mathrm{dT}}{\mathrm{dt}}=\mathrm{K}\left\{\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}-\mathrm{T}_{0}\right\}$

$\frac{-\mathrm{dT}}{\mathrm{dt}

Vedclass · Accepted Answer

Vedclass · Answer

$\frac{-\mathrm{dT}}{\mathrm{dt}}=\mathrm{K}\left\{\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}-\mathrm{T}_{0}\right\}$

$\frac{-\mathrm{dT}}{\mathrm{dt}}=\mathrm{K}\left\{\mathrm{T}-\mathrm{T}_{0}\right\}$

$\int \frac{-\mathrm{d} \mathrm{T}}{\mathrm{T}-\mathrm{T}_{0}}=\int \mathrm{K} \cdot \mathrm{dt}$

$\ell n$ $\left(\mathrm{T}-\mathrm{T}_{0}\right)=-\mathrm{Kt}+\mathrm{C}$

$y=-m x+c$

A liquid in a beaker has temperature $\theta \left( t \right)$ at time $t$ and ${\theta _0}$ is temperature of surroundings, then according to Newton's law of cooling the correct graph between loge ${\log _e}(\theta - {\theta _0})$ and $t$ is

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