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A liquid in a beaker has temperature the...

A liquid in a beaker has temperature `theta(t)` at time t and `theta_0` is temperature of surroundings, then according to Newton's law of cooling the correct graph between `log_e( theta-theta_0)` and t is :

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In Newton's law of cooling (d theta)/(dt)=-k(theta-theta_(0)) the constant k is proportional to .

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