A liquid in a beaker has temperature θ(t) at time t and θ0 is temperature of surroundings, then according to Newton's law of cooling the correct graph between loge(θ

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Physics

Radiation

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A liquid in a beaker has temperature θ(t) at time t and θ₀ is temperature of surroundings, then according to Newton's law of cooling the correct graph between log_e(θ – θ₀) and t is :

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$\frac{dT}{dt} = - k (T - Ts)$

$\ln | \frac{T - T_{s}}{T_{0} - T_{s}} | = - kt$

ln(T – T_s) = ln(T₀ – T_s) – kt