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(B) According to Newton's law of cooling,the rate of change of temperature difference $	heta$ is proportional to the temperature difference itself:$\

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(B) According to Newton's law of cooling,the rate of change of temperature difference $	heta$ is proportional to the temperature difference itself:$\frac{d	heta}{dt} = -K	heta$Integrating this equation:$\int \frac{d	heta}{	heta} = \int -K dt$$\ln 	heta = -Kt + C$At $t = 0$,let the initial temperature difference be $	heta_i$,so $C = \ln 	heta_i$.Thus,the equation becomes:$\ln 	heta = -Kt + \ln 	heta_i$This is in the form of a linear equation $y = mx + c$,where $y = \ln 	heta$,$x = t$,slope $m = -K$ (which is negative),and intercept $c = \ln 	heta_i$. Therefore,the graph of $\ln 	heta$ versus $t$ is a straight line with a negative slope. This corresponds to the graph shown in option $(B)$.

The instantaneous temperature difference between a cooling body and its surroundings,which obeys Newton's law of cooling,is $\theta$. Which of the following graphs represents the variation of $\ln \theta$ with time $t$?

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